Statistics

1. BASICS OF STATISTICS

1. STATISTICAL DATA

Statistical data are the facts which are collected for the purpose of investigation. There are two types of statistical data:

(i) Primary data: The data collected by an investigator for the first time for his own purpose are called primary data. As the primary data are collected by the user of the data, so it is more reliable and relevant.

(ii) Secondary data: The data collected by a secondary source and used by the investigator for his purpose is called secondary data. For example score of a cricket match noted from newspapers is secondary data. Thus data which are primary in the hands of one become secondary in the hands of the other. Data collected by any source also can be divided in following two types:

(i) Raw Data: Raw data are those data which are obtained from the original source but not arranged numerically. This is also called ‘ungrouped data’ for example marks of 10 students in maths are given as:  75, 96, 25, 32, 89, 62, 40, 79, 35, 55 An ‘array’ is an arrangement of raw numerical data in the ascending or descending order of magnitude. Above data can be written as  25, 32, 35, 40, 55, 62, 75, 79, 89, 96

(ii) Grouped data: An array can be placed systematically in groups or categories. For example the above data can be grouped in following manner.

GROUPS	MARKS	TOTAL NUMBER OF STUDENTS
0 to 20	–	0
21 to 40	25, 32, 35, 40	4
41 to 60	55	1
61 to 80	62, 75, 79	3
81 to 100	89, 96	2
TOTAL		10

 2. SOME BASIC DEFINITIONS

(i) Variate: Variate is a quantity that may vary from observation to observation.

(ii) Range: Range is difference between the maximum and minimum observations.

(iii) Class Interval: When data are divided in groups, each group is called a class interval.

(iv) Class Limit: Every class interval has two limits. The smallest observation of the interval is called lower limit and the largest observation of the interval is called upper limit.

(v) Class Mark: The mid value of any class is called its class mark.

$\text{ Class Mark }=\frac{\text{ Upper limit of the class + lower limit of the class }}{2}$

(vi) Class Size: Class size is defined as the difference between two successive class marks. It is also the difference between the upper and lower limits of any class interval.

(vii) Frequency: In a particular class the count of the number of observation is called its frequency. So the corresponding frequency of a class is called its class frequency.

(viii) Cumulative Frequency: The cumulative frequency of any class is obtained by adding all the frequencies successively prior to that class i.e. it is the sum of all frequencies up to that class.

(ix) True Class Limit: In the case of exclusive classes the upper and lower limits are respectively known as its true upper limits and true lower limits. In the case of inclusive classes, the true lower and upper limits are obtained by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. True upper limits and true lower limits are also known as boundaries of the class.

(x) Tally: Tally method is used to keep the chance of error at minimum in counting. A bar (|) called tally mark is put against any item when it occurs. The fifth occurrence of any item is represented by putting diagonally a cross tally (\) on the first four tallies.

3. FREQUENCY DISTRIBUTION

The tabular arrangement of data showing the frequency of each item is called a frequency distribution table. It is a method to present raw data in the form from which one can easily understand the information contained in the raw data. Frequency distribution are of two types:

i. Discrete frequency distribution:

In this type of frequency distribution, in the first column of frequency table we write all possible values of the variables from the lowest to the highest, in the second column we write tally marks and in the third column we show frequency of each item. In this method data are not divided into groups or classes.

e.g. no. of girls in 20 families is given in following data:

1, 2, 3, 1, 1, 2, 3, 3, 4, 1, 5, 1, 1, 2, 2, 3, 3, 2, 4, 1

The above data can be put in the form of a discrete frequency distribution table in the following manner:

S. No.	No. of girls	Tally Marks	Frequency
1.	1	$\bcancel{\left|\right| \left|\right|} \left|\right|$	7
2.	2	$\bcancel{\left|\right|\left|\right|}$	5
3.	3	$\bcancel{\left|\right|\left|\right|}$	5
4.	4	||	2
5.	5	|	1
TOTAL			20

ii. Continuous or Grouped Frequency Distribution

In the frequency distribution data are divided  into groups or classes. This method is used only where the values in the raw data are largely repeating and the difference between the greatest and the smallest observations is not very large.

4. CUMULATIVE FREQUENCY

Cumulative frequency table is obtained from the ordinary frequency table by successively adding the several frequencies. Thus to form a cumulative frequency table we add a column of cumulative frequency in the frequency distribution table. It is obvious that the cumulative frequency of the last class is the sum of the frequencies of all the classes.

Cumulative frequency series are of two types:

(i) Less than series        (ii) More than series

Illustration -1

The weekly saving of 30 workers working in a factory is as given below:

64, 60, 87, 75, 69, 34, 51, 78, 39, 48, 73, 54, 63, 70, 57, 88, 90, 53, 74, 44, 31, 71, 68, 72, 36, 89, 55, 67, 73, 83

(a) Taking first class interval as 30 – 40 (40 not included), form a frequency table of equal intervals.

(b) Also form a cumulative frequency table.

(c) Find the number of workers whose weekly saving is more than Rs. 60.

Solution

(i) For the given data of the weekly saving of 30 workers working in a factory, we prepare             following table:

S. No.	Class Interval (Saving in Rs.)	Tally Marks	Frequency (No. of workers)
1.	30 – 40	| | | |	4
2.	40 – 50	| |	2
3.	50 – 60	$\bcancel{\left|\right|\left|\right|}$	5
4.	60 – 70	$\bcancel{\left|\right|\left|\right|} |$	6
5.	70 – 80	$\bcancel{\left|\right|\left|\right|} \left|\right||$	8
6.	80 – 90	| | | |	4
7.	90 – 100	|	1
		TOTAL	30

(ii) For given data cumulative frequency table can be given in the following manner:

S. No.	Class Interval (Saving in Rs.)	Tally Marks	Frequency (No. of workers)	Cumulative Frequency
1.	30 – 40	| | | |	4	4
2.	40 – 50	| |	2	6
3.	50 – 60	$\bcancel{\left|\right|\left|\right|}$	5	11
4.	60 – 70	$\bcancel{\left|\right|\left|\right|} |$	6	17
5.	70 – 80	$\bcancel{\left|\right|\left|\right|} \left|\right||$	8	25
6.	80 – 90	| | | |	4	29
7.	90 – 100	|	1	30
		TOTAL	30

(iii) Number of workers whose weekly saving is Rs. 60 or more than Rs. 60  = (6 + 8 + 4 + 1) = 19

Illustration -2

Heights of seven persons in cm are 120, 125, 142, 134, 150, 155 and 128. Calculate the range.

Solution

For given data maximum height = 155 cm

and minimum height = 120 cm

Range  = maximum height – minimum height

= 155 cm – 120 cm

= 35 cm.

Illustration -3

Form a frequency table from the following table:

Marks	No. of Students
Below 10	15
Below 20	35
Below 30	60
Below 40	84
Below 50	96
Below 60	127
Below 70	198
Below 80	250

Solution

For given data we make classes 0 – 10, 10 – 20, 20 – 30, ….., 70–80.

There are 15 students who obtained below 10 marks, therefore frequency of class 0 – 10 is 15. Again number of students getting below 20 marks is 35.

This includes those students also who obtained below 10 marks.

Number of students who got marks between 10 and 20 i.e. frequency of class

10 – 20  =  35–15 = 20

Thus for given data we make following table:

S. No.	Class Interval	Comparative	Frequency
1.	0 – 10	15	15
2.	10 – 20	35	20
3.	20 – 30	60	25
4.	30 – 40	84	24
5.	40 – 50	96	12
6.	50 – 60	127	31
7.	60 – 70	198	71
8.	70 – 80	250	52
		TOTAL	250

Illustration -4

Find the unknown entries (a, b, c, d, e, f, g) from the following frequency distribution of heights of 60 students in a class:

Height (in cm)	Frequency	Cumulative frequency
160 – 165	15	a
165 – 170	b	35
170 – 175	12	c
175 – 180	e	50
180 – 185	d	55
185 – 190	5	f
Total	g

Solution

From given table, we make following cumulative frequency table:

S. No.	Height (in cm)	Frequency	Cumulative Frequency
1.	160 – 165	15	15  = a
2.	165 – 170	b	15 + b = 35
3.	170 – 175	12	15 + b + 12 = c
4.	175 – 180	d	15 + b + 12 + d = 50
5.	180 – 185	e	15 + b + 12 + b + e = 55
6.	185 – 190	5	15 + b + 12 + d + e + 5 = f
		g	60

Now, from table:

a = 15

15 + b = 35    b = 35 – 15 = 20

15 + b + 12 = c   15 + 20 + 12 = c

c = 47

15 + b + 12 + d = 50

15 + 20 + 12 + d = 50

d = 50 – 47

d = 3

15 + b + 12 + d + c = 55

15 + 20 + 12 + 3 + e = 55

e = 55 – 50

e = 5

15 + b + 12 + d + e + 5 = f

15 + 20 + 12 + 3 + 5 + 5 = f

f = 60

g = 60. Hence, a = 15, b = 20, c = 47, d = 3, e = 5, f = 60 and g = 60.

5. GRAPHICAL REPRESENTATION OF DATA

A given data can be represented in graphical way. There are various methods of graphical representation of frequency distribution.

(i) Bar Graphs

(ii) Histogram

(iii) Frequency Polygon

(iv) Pie Chart

Bar Graph

The frequency distribution of a discrete value is best represented by a bar graph. The height of the bars is proportional to the frequency of each variate-value. In a bar graph the bars must be kept distinct to show that the variate-values are distinct. The bars are of equal width and are drawn with equal spacing between them on the x-axis depicting the variable. The values of the variable are shown on the y-axis.

Illustration -5

The following table shows the number of illiterate persons in the age group (10–58 years) in a town:

Age Group (in years)	10 – 16	17 – 23	24 – 30	31 – 37	38 – 44	45 – 51	52 – 58
Number of illiterate person	175	375	100	150	250	400	525

Solution

Draw a bar graph to represent the above data.

Histogram

Histogram is a graphical representation of a grouped frequency distribution with continuous classes. It consists of a set of rectangles where heights of rectangles are proportional to their class frequencies, for equal class intervals. There is no gap between two successive rectangles. The rectangles are constructed with base as the class size and their heights representing the frequencies.

Illustration -6

Construct a histogram from the following distribution of total marks obtained by 750 students of class IX in the final examination.

Marks (mid-points)	620	660	700	740	780	820	860
Number of students	16	45	156	284	172	59	18

Solution

To draw the histogram for above data first of all we find out class limits for making class intervals from given class marks.

Here difference between second and the first class marks =  h  =  660 – 620 = 40

$\therefore \frac{h}{2}=\frac{40}{2}=20$

lower limit of first class interval =  620 – h/2

= 620 – 20

= 600

And upper limit of the first class interval = $620+\frac{h}{2}$ = 620 + 20 = 640

Similarly other class intervals would be,

(660 – 20) – (660 + 20), (700 – 20) – (700 + 20), (740 – 20) – (740 + 20), (780 – 20) – (780 + 20), (820 – 20) – (820 + 20) and (860 – 20) – (860 – 20)

i.e. 640 – 680, 680 – 720, 720 – 760, 760 – 800, 800 – 840 and 840 – 880 Using these class intervals we draw following histogram taking class intervals on X-axis and number of students on Y-axis

Frequency Polygon

A frequency polygon is a graph of frequency distribution. It is a line graph of class frequency which is plotted against class mark. A frequency polygon can be obtained by two methods:

(1) By using Histogram: A frequency polygon can be obtained by joining mid points of the top of the rectangles of a histogram. For this we obtain the mid points of the upper horizontal sides of each rectangle and then join these mid points by dotted lines to get frequency polygon. End of a frequency polygon preferably extended to the mid points of imagined class intervals adjacent to first and last class intervals.

Illustration -7

Draw the histogram and frequency polygon of the following frequency distribution:

Monthly wages (in rupees)	325-350	350-375	375-400	400-425	425-450	Total
Number of workers	30	45	75	60	55	245

Solution

Using given data we draw histogram taking monthly wages (in rupees) on X-axis and number of workers on Y-axis. Then we join the mid points of the top of the rectangles by dotted straight lines and complete it by joining the mid points of imagined class intervals adjacent to the first and last class intervals.

(2) Frequency polygon without using Histogram: Following procedure is used to make a frequency polygon without using histogram.

(i) Calculate the class marks, x₁, x₂, …., x_n of each of the given class intervals.

(ii) Mark class marks x₁, x₂, …., x_n, along X-axis and frequencies f₁, f₂, …. f_n along Y-axis.

(iii) Plot the points (x₁, f₁), (x₂, f₂), ,….., (x_n, f_n).

(iv) Obtain the mid-points of two class intervals of zero frequencies at the beginning of the first interval and at the end of the last interval.

(v) Join the points (x₁, f₁), (x₂, f₂), …, (x_n, f_n) by the line segments and complete the frequency polygon by joining the mid points of the first and last intervals to the mid points of the imagined classes adjacent to them.

Illustration -8

Represent the following distribution by a frequency polygon without using a histogram:

Scores	20-29	30-39	40-49	50-59	60-69	70-79	80-89	90-99
Frequency	2	5	3	18	20	22	12	7

Solution

Given data is in the form of inclusive classes, so first of all we shall convert them into exclusive classes. Also we have to get class marks of these class intervals for this purpose.

S. No.	Scores	True Class Limits	Class Marks	Frequency	Points to be plotted
1.	20 – 29	19.5 – 29.5	24.5	2	(24.5, 2)
2.	30 – 39	29.5 – 39.5	34.5	5	(34.5, 5)
3.	40 – 49	39.5 – 49.5	44.5	3	(44.5, 3)
4.	50 – 59	49.5 – 59.5	54.5	18	(54.5, 18)
5.	60 – 69	59.5 – 69.5	64.5	20	(64.5, 20)
6.	70 – 79	69.5 – 79.5	74.5	22	(74.5, 22)
7.	80 – 89	79.5 – 89.5	84.5	12	(84.5, 12)
8.	90 – 99	89.5 – 99.5	94.5	7	(94.5, 7)

Plotting these points and joining them by line segments we get frequency polygon. We complete the frequency polygon by extending it up to points (14.5, 0) and (104.5, 0) adjacent to first and last class marks on X-axis.

Pie Chart

In a pie-chart, various observations or components are represented by the sectors of a circle and the whole circle represents the sum of the values of all the components. Clearly, the total angle of 360o at the center of  the circle is divided according  to the values of the components. Thus we have, Central Angle for Component

= $\left(\frac{\text{ Value of the Component }}{\text{ Total Value }}\times 360\right)$

Sometimes, the value of components are expressed in percentages. In such cases, we have: Central Angle for Component = $\left(\frac{\text{ Percentage Value of the Component }}{100}\times 360\right)$

Illustration -9

A survey was conducted among a group of girls to know their preferences for the types of slippers. The given circle graph shows the results of the survey.

If the survey was conducted among a group of 1080 girls, then find out how many girls preferred to wear leather slippers?

(A) 140          (B) 280          (C) 420          (D) 560

Solution

C

In the given circle graph, it is seen that the central angle of the sector corresponding to leather slippers is 140°.

Therefore, number of girls who preferred to wear leather slippers:

$=\frac{140°}{360°}\times 1080=420$

Thus, 420 girls preferred to wear leather slippers.

6. MEASURES OF CENTRAL TENDENCY

One of the most  important objectives of statistical analysis is to get one single value that describes the characteristic of the entire data.  Such a value is called the central value or an average. The following are the important types of averages:

1. Arithmetic Mean

2. Geometric Mean

3. Harmonic mean

4. Median

5. Mode

We consider these measures in three cases (i) Individual series (i.e. each individual observation is given) (ii) discrete series (i.e., the observations along with number of times a particular observation called the frequency is given) (iii) continuous series (i.e. the class intervals along with their frequencies are given)

2. Arithmetic Mean

The average of numbers in arithmetic is known as the Arithmetic Mean of these numbers in statistics.

1. MEAN OF AN UNGROUPED DATA

The Arithmetic Mean or simply the Mean of n observations x₁, x₂, x₃, …….x_n is given by the formula:

Mean = $\frac{\left({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+\dots \dots {\mathrm{x}}_{\mathrm{n}}\right)}{\mathrm{n}}=\frac{\sum {\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}$

Where the symbol $\Sigma$, called sigma stands for the summation of the terms.

Illustration -10

Calculate the mean of the following numbers: 3, 1, 5, 6, 3, 4, 5, 3, 7, 2

Solution

Sum of the given numbers = (3 + 1 + 5 + 6 + 3 + 4 + 5 + 3 + 7 + 2) = 39.

Number of these numbers = 10.

$\therefore$ Mean of the given numbers = $\frac{39}{10}=3.9$.

Illustration -11

The weights (in kg) of 5 persons in a group are: 55, 63, 48, 59 and 61. Find their mean weight.

Solution

Sum of the weight of all persons = (55 + 63 + 48 + 59 + 61) kg = 286 kg.

Number of persons = 5. 

$\therefore$ Mean weight = $\left(\frac{286}{5}\right)$ kg = 57.2 kg.

Some Useful Result: Let the mean of x₁ x₂ x₃, ……..x_n be A. Then

(i) Mean of (x₁ + k), (x₂ + k), (x₃+ k) ………..(x_n + k) is (A + k);

(ii) Mean of (x₁ – k), (x₂ – k), (x₃ – k)………..(x_n – k) is (A – k);

(iii) Mean of kx₁, kx₂, kx₃ …….. kx_n is kA, where k $\ne$ 0.

2. MEAN OF GROUPED DATA

Direct Method

When the variates x₁, x₂, x₃, …………x_n have frequencies f₁, f₂, f₃ ……..f_n respectively, then the mean is given by the formula:

Mean = $\frac{\left(f_1{x}_1+f_2{x}_2+f_3{x}_3+\dots \dots ..+f_n{x}_n\right)}{\left(f_1+f_2+f_3+\dots \dots \dots .+f_n\right)}=\frac{\sum f_i{x}_i}{\sum f_i}$.

Illustration -12

The following table shows the weights of 15 members of an athletic team in a school.

Weight (in kg)	42	45	46	48	49
Number of athletes	4	3	5	2	1

Find the mean weight.

Solution

From the above data, we may prepare the table given below.

Weight (in kg) xi	Number of athletes (Frequency) $f_1$	$f_1{x}_i$
42 45 46 48 49	4 3 5 2 1	168 135 230 96 49
	${\mathrm{\Sigma f}}_{\mathrm{I}}$= 15	$\Sigma f_i{x}_i$ = 678

  $\therefore$Mean Weight =$\frac{\sum f_i{x}_i}{\sum f_i}=\frac{678}{15}=45.2 \mathrm{kg}$.

Illustration -13

Using short cut method, calculate the mean weekly wage from the following frequency distribution.

Weekly wages(in Rs)	950	1000	1050	1100	1250	1500	1600
Number of workers	24	18	13	15	20	11	9

Solution

Let the assumed mean be, A = 1100.

From the given data, we may prepare the table given below.

Weekly Wages (in Rs) x1	Numbers of Workers (Frequency)	di = (xi – A)     = (xi – 1100)	fidi
950 1000 1050 1100 = A 1250 1500 1600	24 18 13 15 20 11 9	–150 –100 –50 0 150 400 500	–3600 –1800 –650 0 3000 4400 4500
	$\Sigma$fI = 110		$\Sigma$fI di = 5850

$\therefore$ Mean = $\left(\mathrm{A}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)=\left(1100+\frac{5850}{110}\right)=1153.18$

Hence, Mean Wage = Rs 1153.18.

3. MEAN OF GROUPED DATA IN THE FORM OF CLASSES

I. Direct Method

Step 1: For each class, find the class mark x_i by using the relation, x_i = $\frac{1}{2}$ (lower limit + upper limit). Step 2: Use the formula, Mean = $\frac{\sum f_i{X}_i}{\sum f_i}$.

II. Short Cut Method or Deviation Method

Step 1: For each class, find the class mark x_i.

Step 2: Let A be the assumed mean.

Step 3: Find d_i = (x_i – A).

Step 4: Use the formula, Mean = $\left(\mathrm{A}\frac{\sum f_{\mathrm{i}}\mathrm{d}}{\Sigma f_{\mathrm{i}}}\right)$.

III. Step-Deviation Method

Step 1. For each class, find the class mark x₁.

Step 2: Let A be the assumed mean.

Step 3: Calculate, ${\mu }_{\mathrm{I}}=\frac{\left({\mathrm{X}}_{\mathrm{i}}–\mathrm{A}\right)}{\mathrm{C}}$, where c is the class size.

Step 4: Use the formula, Mean = $\left(\mathrm{A}+\mathrm{C}\frac{\sum f_{\mathrm{i}}{\mu }_{\mathrm{i}}}{\sum f_{\mathrm{i}}}\right)$.

Illustration -14

Using direct method, find the mean of the following frequency distribution:

Class-interval	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	6	8	12	15	10	9

Solution

From the given data, we may prepare the table given below.

Class-Interval	Class-Mark xi	Frequency fi	fixi
10-20 20-30 30-40 40-50 50-60 60-70	15 25 35 45 55 65	6 8 12 15 10 9	90 200 420 675 550 585
		$\Sigma$fI = 60	$\Sigma$fixi = 2520

$\therefore$ Mean = $\frac{\sum f_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum f_{\mathrm{i}}}=\frac{2520}{60}=42$.

Illustration -15

Using Short Cut Method find the mean for the following frequency distribution:

Class-interval	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	8	10	16	23	12	11

Solution

From the given data, we may prepare the table given below.

Class-Interval	Class-Mark xi	Frequency fi	di = (xi – A)    = (xi – 99)	fidi
84-90 90-96 96-102 102-108 108-114 114-120	87 93 99 = A 105 111 117	8 10 16 23 12 11	–12 –6 0 6 12 18	–96 –60 0 138 144 198
		$\Sigma$fI = 80		$\Sigma$fidi = 324

$\therefore$ Mean = $\frac{\Sigma f_i{d}_i}{\sum f_i}=\left(99+\frac{324}{80}\right)=(99+4.05)=103.05$

Hence, Mean = 103.05.

Illustration -16

Using Step Deviation Method, calculate the mean for the following data:

Height (in cm)	135-140	140-145	145-150	150-155	155-160	160-165	165-170	170-175
Frequency	4	9	18	28	24	10	5	2

Solution

Here, class size, c = 5. Take assumed mean, A = 152.5. Thus, from the given data, we may prepare

Class-Interval	Class-Mark   xi	Frequency  fi	$\left({\mathrm{\mu }}_{\mathrm{i}}=\frac{{\mathrm{X}}_{\mathrm{i}}–\mathrm{A}}{\mathrm{C}}\right)$	fi$\mu$i
135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175-	137.5 142.5 147.5 152.5 = A 157.5 162.5 167.5 172.5	4 9 18 28 24 10 5 2	–3 –2 –1 0 1 2 3 4	–12 –18 –18 0 24 20 15 8
		$\Sigma$fI = 100		$\Sigma$fi$\mu$i = 19

$\therefore$ Mean = $\left(\mathrm{A}+\mathrm{c}\times \frac{\sum f_{\mathrm{i}}{\mu }_{\mathrm{i}}}{\Sigma f_{\mathrm{i}}}\right)=\left(152.5+\frac{5\times 19}{100}\right)$ = (152.5 + 0.95) = 153.45 cm.

4. IMPORTANT FORMULAE FOR SOLVING ARITHMETIC MEAN

1. Arithmetic Mean of Individual Series

If x₁, x₂ , x₃,….., x_n are n values of variant x, then its Arithmetic Mean, denoted by x is

$\text{ A.M. }\left(\overline{x}\right)=\frac{x_1+x_2+x_3+\dots ..+x_n}{n}=\frac{\sum x_i}{n}$   (or)

$\text{ A.M. }\left(\overline{x}\right)=A+\frac{\sum \left(x_i–A\right)}{n}$ where A is the assumed average. (For individual series)

2. Arithmetic Mean of Discreate Series

If a variable takes values x₁, x₂ , x₃,….., x_n with corresponding frequencies f₁, f₂ , f₃,….., f_n then the arithmetic mean $\overline{x}$ is given by

$\overline{x}=\frac{f_1{x}_1+f_2{x}_2\dots .+f_n{x}_n}{f_1+f_2\dots .+f_n}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i$

where $N=\sum _{i=1}^n{f}_i$

3. Arithmetic Mean Of Continuous Series

In case of a set of data with class intervals, we cannot find the exact value of the mean because we do not know the exact values of the variables. We, therefore, try to obtain an approximate value of the mean. The method of approximate is to replace all the observed values belonging to a class by mid-value of the class.
If x₁, x₂ , x₃,….., x_n are the mid values of the class intervals having corresponding frequencies f₁, f₂ , f₃,….., f_n then we apply the same formula as in discrete series.

$\overline{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i, N=\sum _{i=1}^n{f}_i$

4. Combined Arithmetic Mean

If ${\overline{x}}_i(i=1,2,\dots .,k)$ are the means of k – series of sizes $n_i(i=1,2,3,\dots \dots \dots ..,k)$ then the combined or composite mean $\overline{x}$ can be obtained by the formula :

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2+\dots .+n_k{\overline{x}}_k}{n_1+n_2+\dots ..+n_k}=\frac{\sum n_i{\overline{x}}_i}{\sum n_i}$

5. Weighted Arithmetic Mean

If w₁, w₂ , w₃,….., w_n be the weights assigned to the values x₁, x₂ , x₃,….., x_n be respectively of a variable x, then the weighted A.M. is $\overline{x}=\frac{\sum w_i{x}_i}{\sum w_i}$.

5. PROPERTIES OF ARITHMETIC MEAN

1. Sum of all the deviations from arithmetic mean is zero i.e.,

$\sum _{i=1}^n\left(x_i–\overline{x}\right)=0$ (in case of individual series)

$\sum _{i=1}^n{f}_i\left(x_i–\overline{x}\right)=0$ (in case of discrete or continuous series)

2. If each observation is increased or decreased by a given constant K, the mean is also increased or decreased by K
The property is also known as effect of change of origin. K can be taken to be any number. However, to simplify the calculations, K should be taken as a value which is in the middle of the table.
3. Step Deviation Method or change of scale

If x₁, x₂ , x₃,….., x_n are mid values of class intervals with corresponding frequencies f₁, f₂ , f₃,….., f_n then we may change the scale by taking $d_i=\frac{x_i–A}{h}$ in this case.

$\overline{x}=A+h\times \left(\frac{1}{N}\sum f_i{d}_i\right)$ (if A is assumed mean)

A and h can be any numbers but if the lengths of class intervals are equal then h may be taken as width of the class interval.
In particular if each observation is multiplied or divided by a constant, the mean is also multiplied or divided by the same constant.
4. The sum of the squared deviation of the variate from their mean is minimum i.e., the quantity $\sum {\left(x_i–A\right)}^2\text{ or }\sum f_i{\left(x_i–A\right)}^2$ is minimum when $A=\overline{x}$.

5. E(aX + b) = aE(X) + b (where E(X) = Mean of X)

Illustration-17
The weighted mean of the first n natural numbers, the weights being the corresponding numbers, is
Solution
First n natural numbers are 1, 2, 3,…,n; whose corresponding weights are 1, 2, 3,…,n respectively.

$\therefore$ weight mean = $\frac{1\times 1+2\times 2+\dots .+n\times n}{1+2+\dots +n}$

$=\frac{1^2+2^2+\dots ..+n^2}{1+2+\dots ..+n}$

$=\frac{n(n+1)(2n+1)}{\frac{6n(n+1)}{2}}=\frac{2n+1}{3}$

Illustration-18

The weighted mean of the first n natural numbers whose weights are equal to the squares of the corresponding numbers is
Solution

weighted mean = $\frac{1.1^2+2.2^2+\dots +n·n^2}{1^2+2^2+\dots +n^2}$

$=\frac{{\mathrm{\Sigma n}}^3}{{\mathrm{\Sigma n}}^2}=\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}}=\frac{3\mathrm{n}(\mathrm{n}+1)}{2(2\mathrm{n}+1)}$

Illustration-19
The average salary of male employees in a firm is Rs. 5200 and that of females is Rs.4200. The mean salary of all the employees is Rs.5000. The percentage of male and female employees are respectively is
Solution

Let $x_1=5200, x_2=4200, \overline{x}=5000$

Also, we know that $\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$

$\Rightarrow 5000\left(n_1+n_2\right)=5200n_1+4200n_2\Rightarrow \frac{n_1}{n_2}=\frac{4}{1}$

$\therefore$ The percentage of male employees in the firm = $\frac{4}{4+1}\times 100=80\%$

and the percentage of female employees in the firm = $\frac{1}{4+1}\times 100=20\%$

Illustration-20
If the mean of 9 observations is 100 and mean of 6 observations is 80, then the mean of 15 observations is
Solution

$n_1=9, {\overline{x}}_1=100\text{ and }{n}_2=6, {\overline{x}}_2=80$

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}=\frac{9\times 100+6\times 80}{9+6}=92$

1. BASICS OF STATISTICS

1. STATISTICAL DATA

Statistical data are the facts which are collected for the purpose of investigation. There are two types of statistical data:

(i) Primary data: The data collected by an investigator for the first time for his own purpose are called primary data. As the primary data are collected by the user of the data, so it is more reliable and relevant.

(ii) Secondary data: The data collected by a secondary source and used by the investigator for his purpose is called secondary data. For example score of a cricket match noted from newspapers is secondary data. Thus data which are primary in the hands of one become secondary in the hands of the other. Data collected by any source also can be divided in following two types:

(i) Raw Data: Raw data are those data which are obtained from the original source but not arranged numerically. This is also called ‘ungrouped data’ for example marks of 10 students in maths are given as:  75, 96, 25, 32, 89, 62, 40, 79, 35, 55 An ‘array’ is an arrangement of raw numerical data in the ascending or descending order of magnitude. Above data can be written as  25, 32, 35, 40, 55, 62, 75, 79, 89, 96

(ii) Grouped data: An array can be placed systematically in groups or categories. For example the above data can be grouped in following manner.

GROUPS	MARKS	TOTAL NUMBER OF STUDENTS
0 to 20	–	0
21 to 40	25, 32, 35, 40	4
41 to 60	55	1
61 to 80	62, 75, 79	3
81 to 100	89, 96	2
TOTAL		10

 2. SOME BASIC DEFINITIONS

(i) Variate: Variate is a quantity that may vary from observation to observation.

(ii) Range: Range is difference between the maximum and minimum observations.

(iii) Class Interval: When data are divided in groups, each group is called a class interval.

(iv) Class Limit: Every class interval has two limits. The smallest observation of the interval is called lower limit and the largest observation of the interval is called upper limit.

(v) Class Mark: The mid value of any class is called its class mark.

$\text{ Class Mark }=\frac{\text{ Upper limit of the class + lower limit of the class }}{2}$

(vi) Class Size: Class size is defined as the difference between two successive class marks. It is also the difference between the upper and lower limits of any class interval.

(vii) Frequency: In a particular class the count of the number of observation is called its frequency. So the corresponding frequency of a class is called its class frequency.

(viii) Cumulative Frequency: The cumulative frequency of any class is obtained by adding all the frequencies successively prior to that class i.e. it is the sum of all frequencies up to that class.

(ix) True Class Limit: In the case of exclusive classes the upper and lower limits are respectively known as its true upper limits and true lower limits. In the case of inclusive classes, the true lower and upper limits are obtained by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. True upper limits and true lower limits are also known as boundaries of the class.

(x) Tally: Tally method is used to keep the chance of error at minimum in counting. A bar (|) called tally mark is put against any item when it occurs. The fifth occurrence of any item is represented by putting diagonally a cross tally (\) on the first four tallies.

3. FREQUENCY DISTRIBUTION

The tabular arrangement of data showing the frequency of each item is called a frequency distribution table. It is a method to present raw data in the form from which one can easily understand the information contained in the raw data. Frequency distribution are of two types:

i. Discrete frequency distribution:

In this type of frequency distribution, in the first column of frequency table we write all possible values of the variables from the lowest to the highest, in the second column we write tally marks and in the third column we show frequency of each item. In this method data are not divided into groups or classes.

e.g. no. of girls in 20 families is given in following data:

1, 2, 3, 1, 1, 2, 3, 3, 4, 1, 5, 1, 1, 2, 2, 3, 3, 2, 4, 1

The above data can be put in the form of a discrete frequency distribution table in the following manner:

S. No.	No. of girls	Tally Marks	Frequency
1.	1	$\bcancel{\left|\right| \left|\right|} \left|\right|$	7
2.	2	$\bcancel{\left|\right|\left|\right|}$	5
3.	3	$\bcancel{\left|\right|\left|\right|}$	5
4.	4	||	2
5.	5	|	1
TOTAL			20

ii. Continuous or Grouped Frequency Distribution

In the frequency distribution data are divided  into groups or classes. This method is used only where the values in the raw data are largely repeating and the difference between the greatest and the smallest observations is not very large.

4. CUMULATIVE FREQUENCY

Cumulative frequency table is obtained from the ordinary frequency table by successively adding the several frequencies. Thus to form a cumulative frequency table we add a column of cumulative frequency in the frequency distribution table. It is obvious that the cumulative frequency of the last class is the sum of the frequencies of all the classes.

Cumulative frequency series are of two types:

(i) Less than series        (ii) More than series

Illustration -1

The weekly saving of 30 workers working in a factory is as given below:

64, 60, 87, 75, 69, 34, 51, 78, 39, 48, 73, 54, 63, 70, 57, 88, 90, 53, 74, 44, 31, 71, 68, 72, 36, 89, 55, 67, 73, 83

(a) Taking first class interval as 30 – 40 (40 not included), form a frequency table of equal intervals.

(b) Also form a cumulative frequency table.

(c) Find the number of workers whose weekly saving is more than Rs. 60.

Solution

(i) For the given data of the weekly saving of 30 workers working in a factory, we prepare             following table:

S. No.	Class Interval (Saving in Rs.)	Tally Marks	Frequency (No. of workers)
1.	30 – 40	| | | |	4
2.	40 – 50	| |	2
3.	50 – 60	$\bcancel{\left|\right|\left|\right|}$	5
4.	60 – 70	$\bcancel{\left|\right|\left|\right|} |$	6
5.	70 – 80	$\bcancel{\left|\right|\left|\right|} \left|\right||$	8
6.	80 – 90	| | | |	4
7.	90 – 100	|	1
		TOTAL	30

(ii) For given data cumulative frequency table can be given in the following manner:

S. No.	Class Interval (Saving in Rs.)	Tally Marks	Frequency (No. of workers)	Cumulative Frequency
1.	30 – 40	| | | |	4	4
2.	40 – 50	| |	2	6
3.	50 – 60	$\bcancel{\left|\right|\left|\right|}$	5	11
4.	60 – 70	$\bcancel{\left|\right|\left|\right|} |$	6	17
5.	70 – 80	$\bcancel{\left|\right|\left|\right|} \left|\right||$	8	25
6.	80 – 90	| | | |	4	29
7.	90 – 100	|	1	30
		TOTAL	30

(iii) Number of workers whose weekly saving is Rs. 60 or more than Rs. 60  = (6 + 8 + 4 + 1) = 19

Illustration -2

Heights of seven persons in cm are 120, 125, 142, 134, 150, 155 and 128. Calculate the range.

Solution

For given data maximum height = 155 cm

and minimum height = 120 cm

Range  = maximum height – minimum height

= 155 cm – 120 cm

= 35 cm.

Illustration -3

Form a frequency table from the following table:

Marks	No. of Students
Below 10	15
Below 20	35
Below 30	60
Below 40	84
Below 50	96
Below 60	127
Below 70	198
Below 80	250

Solution

For given data we make classes 0 – 10, 10 – 20, 20 – 30, ….., 70–80.

There are 15 students who obtained below 10 marks, therefore frequency of class 0 – 10 is 15. Again number of students getting below 20 marks is 35.

This includes those students also who obtained below 10 marks.

Number of students who got marks between 10 and 20 i.e. frequency of class

10 – 20  =  35–15 = 20

Thus for given data we make following table:

S. No.	Class Interval	Comparative	Frequency
1.	0 – 10	15	15
2.	10 – 20	35	20
3.	20 – 30	60	25
4.	30 – 40	84	24
5.	40 – 50	96	12
6.	50 – 60	127	31
7.	60 – 70	198	71
8.	70 – 80	250	52
		TOTAL	250

Illustration -4

Find the unknown entries (a, b, c, d, e, f, g) from the following frequency distribution of heights of 60 students in a class:

Height (in cm)	Frequency	Cumulative frequency
160 – 165	15	a
165 – 170	b	35
170 – 175	12	c
175 – 180	e	50
180 – 185	d	55
185 – 190	5	f
Total	g

Solution

From given table, we make following cumulative frequency table:

S. No.	Height (in cm)	Frequency	Cumulative Frequency
1.	160 – 165	15	15  = a
2.	165 – 170	b	15 + b = 35
3.	170 – 175	12	15 + b + 12 = c
4.	175 – 180	d	15 + b + 12 + d = 50
5.	180 – 185	e	15 + b + 12 + b + e = 55
6.	185 – 190	5	15 + b + 12 + d + e + 5 = f
		g	60

Now, from table:

a = 15

15 + b = 35    b = 35 – 15 = 20

15 + b + 12 = c   15 + 20 + 12 = c

c = 47

15 + b + 12 + d = 50

15 + 20 + 12 + d = 50

d = 50 – 47

d = 3

15 + b + 12 + d + c = 55

15 + 20 + 12 + 3 + e = 55

e = 55 – 50

e = 5

15 + b + 12 + d + e + 5 = f

15 + 20 + 12 + 3 + 5 + 5 = f

f = 60

g = 60. Hence, a = 15, b = 20, c = 47, d = 3, e = 5, f = 60 and g = 60.

5. GRAPHICAL REPRESENTATION OF DATA

A given data can be represented in graphical way. There are various methods of graphical representation of frequency distribution.

(i) Bar Graphs

(ii) Histogram

(iii) Frequency Polygon

(iv) Pie Chart

Bar Graph

The frequency distribution of a discrete value is best represented by a bar graph. The height of the bars is proportional to the frequency of each variate-value. In a bar graph the bars must be kept distinct to show that the variate-values are distinct. The bars are of equal width and are drawn with equal spacing between them on the x-axis depicting the variable. The values of the variable are shown on the y-axis.

Illustration -5

The following table shows the number of illiterate persons in the age group (10–58 years) in a town:

Age Group (in years)	10 – 16	17 – 23	24 – 30	31 – 37	38 – 44	45 – 51	52 – 58
Number of illiterate person	175	375	100	150	250	400	525

Solution

Draw a bar graph to represent the above data.

Histogram

Histogram is a graphical representation of a grouped frequency distribution with continuous classes. It consists of a set of rectangles where heights of rectangles are proportional to their class frequencies, for equal class intervals. There is no gap between two successive rectangles. The rectangles are constructed with base as the class size and their heights representing the frequencies.

Illustration -6

Construct a histogram from the following distribution of total marks obtained by 750 students of class IX in the final examination.

Marks (mid-points)	620	660	700	740	780	820	860
Number of students	16	45	156	284	172	59	18

Solution

To draw the histogram for above data first of all we find out class limits for making class intervals from given class marks.

Here difference between second and the first class marks =  h  =  660 – 620 = 40

$\therefore \frac{h}{2}=\frac{40}{2}=20$

lower limit of first class interval =  620 – h/2

= 620 – 20

= 600

And upper limit of the first class interval = $620+\frac{h}{2}$ = 620 + 20 = 640

Similarly other class intervals would be,

(660 – 20) – (660 + 20), (700 – 20) – (700 + 20), (740 – 20) – (740 + 20), (780 – 20) – (780 + 20), (820 – 20) – (820 + 20) and (860 – 20) – (860 – 20)

i.e. 640 – 680, 680 – 720, 720 – 760, 760 – 800, 800 – 840 and 840 – 880 Using these class intervals we draw following histogram taking class intervals on X-axis and number of students on Y-axis

Frequency Polygon

A frequency polygon is a graph of frequency distribution. It is a line graph of class frequency which is plotted against class mark. A frequency polygon can be obtained by two methods:

(1) By using Histogram: A frequency polygon can be obtained by joining mid points of the top of the rectangles of a histogram. For this we obtain the mid points of the upper horizontal sides of each rectangle and then join these mid points by dotted lines to get frequency polygon. End of a frequency polygon preferably extended to the mid points of imagined class intervals adjacent to first and last class intervals.

Illustration -7

Draw the histogram and frequency polygon of the following frequency distribution:

Monthly wages (in rupees)	325-350	350-375	375-400	400-425	425-450	Total
Number of workers	30	45	75	60	55	245

Solution

Using given data we draw histogram taking monthly wages (in rupees) on X-axis and number of workers on Y-axis. Then we join the mid points of the top of the rectangles by dotted straight lines and complete it by joining the mid points of imagined class intervals adjacent to the first and last class intervals.

(2) Frequency polygon without using Histogram: Following procedure is used to make a frequency polygon without using histogram.

(i) Calculate the class marks, x₁, x₂, …., x_n of each of the given class intervals.

(ii) Mark class marks x₁, x₂, …., x_n, along X-axis and frequencies f₁, f₂, …. f_n along Y-axis.

(iii) Plot the points (x₁, f₁), (x₂, f₂), ,….., (x_n, f_n).

(iv) Obtain the mid-points of two class intervals of zero frequencies at the beginning of the first interval and at the end of the last interval.

(v) Join the points (x₁, f₁), (x₂, f₂), …, (x_n, f_n) by the line segments and complete the frequency polygon by joining the mid points of the first and last intervals to the mid points of the imagined classes adjacent to them.

Illustration -8

Represent the following distribution by a frequency polygon without using a histogram:

Scores	20-29	30-39	40-49	50-59	60-69	70-79	80-89	90-99
Frequency	2	5	3	18	20	22	12	7

Solution

Given data is in the form of inclusive classes, so first of all we shall convert them into exclusive classes. Also we have to get class marks of these class intervals for this purpose.

S. No.	Scores	True Class Limits	Class Marks	Frequency	Points to be plotted
1.	20 – 29	19.5 – 29.5	24.5	2	(24.5, 2)
2.	30 – 39	29.5 – 39.5	34.5	5	(34.5, 5)
3.	40 – 49	39.5 – 49.5	44.5	3	(44.5, 3)
4.	50 – 59	49.5 – 59.5	54.5	18	(54.5, 18)
5.	60 – 69	59.5 – 69.5	64.5	20	(64.5, 20)
6.	70 – 79	69.5 – 79.5	74.5	22	(74.5, 22)
7.	80 – 89	79.5 – 89.5	84.5	12	(84.5, 12)
8.	90 – 99	89.5 – 99.5	94.5	7	(94.5, 7)

Plotting these points and joining them by line segments we get frequency polygon. We complete the frequency polygon by extending it up to points (14.5, 0) and (104.5, 0) adjacent to first and last class marks on X-axis.

Pie Chart

In a pie-chart, various observations or components are represented by the sectors of a circle and the whole circle represents the sum of the values of all the components. Clearly, the total angle of 360o at the center of  the circle is divided according  to the values of the components. Thus we have, Central Angle for Component

= $\left(\frac{\text{ Value of the Component }}{\text{ Total Value }}\times 360\right)$

Sometimes, the value of components are expressed in percentages. In such cases, we have: Central Angle for Component = $\left(\frac{\text{ Percentage Value of the Component }}{100}\times 360\right)$

Illustration -9

A survey was conducted among a group of girls to know their preferences for the types of slippers. The given circle graph shows the results of the survey.

If the survey was conducted among a group of 1080 girls, then find out how many girls preferred to wear leather slippers?

(A) 140          (B) 280          (C) 420          (D) 560

Solution

C

In the given circle graph, it is seen that the central angle of the sector corresponding to leather slippers is 140°.

Therefore, number of girls who preferred to wear leather slippers:

$=\frac{140°}{360°}\times 1080=420$

Thus, 420 girls preferred to wear leather slippers.

6. MEASURES OF CENTRAL TENDENCY

One of the most  important objectives of statistical analysis is to get one single value that describes the characteristic of the entire data.  Such a value is called the central value or an average. The following are the important types of averages:

1. Arithmetic Mean

2. Geometric Mean

3. Harmonic mean

4. Median

5. Mode

We consider these measures in three cases (i) Individual series (i.e. each individual observation is given) (ii) discrete series (i.e., the observations along with number of times a particular observation called the frequency is given) (iii) continuous series (i.e. the class intervals along with their frequencies are given)

2. Arithmetic Mean

The average of numbers in arithmetic is known as the Arithmetic Mean of these numbers in statistics.

1. MEAN OF AN UNGROUPED DATA

The Arithmetic Mean or simply the Mean of n observations x₁, x₂, x₃, …….x_n is given by the formula:

Mean = $\frac{\left({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+\dots \dots {\mathrm{x}}_{\mathrm{n}}\right)}{\mathrm{n}}=\frac{\sum {\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}$

Where the symbol $\Sigma$, called sigma stands for the summation of the terms.

Illustration -10

Calculate the mean of the following numbers: 3, 1, 5, 6, 3, 4, 5, 3, 7, 2

Solution

Sum of the given numbers = (3 + 1 + 5 + 6 + 3 + 4 + 5 + 3 + 7 + 2) = 39.

Number of these numbers = 10.

$\therefore$ Mean of the given numbers = $\frac{39}{10}=3.9$.

Illustration -11

The weights (in kg) of 5 persons in a group are: 55, 63, 48, 59 and 61. Find their mean weight.

Solution

Sum of the weight of all persons = (55 + 63 + 48 + 59 + 61) kg = 286 kg.

Number of persons = 5. 

$\therefore$ Mean weight = $\left(\frac{286}{5}\right)$ kg = 57.2 kg.

Some Useful Result: Let the mean of x₁ x₂ x₃, ……..x_n be A. Then

(i) Mean of (x₁ + k), (x₂ + k), (x₃+ k) ………..(x_n + k) is (A + k);

(ii) Mean of (x₁ – k), (x₂ – k), (x₃ – k)………..(x_n – k) is (A – k);

(iii) Mean of kx₁, kx₂, kx₃ …….. kx_n is kA, where k $\ne$ 0.

2. MEAN OF GROUPED DATA

Direct Method

When the variates x₁, x₂, x₃, …………x_n have frequencies f₁, f₂, f₃ ……..f_n respectively, then the mean is given by the formula:

Mean = $\frac{\left(f_1{x}_1+f_2{x}_2+f_3{x}_3+\dots \dots ..+f_n{x}_n\right)}{\left(f_1+f_2+f_3+\dots \dots \dots .+f_n\right)}=\frac{\sum f_i{x}_i}{\sum f_i}$.

Illustration -12

The following table shows the weights of 15 members of an athletic team in a school.

Weight (in kg)	42	45	46	48	49
Number of athletes	4	3	5	2	1

Find the mean weight.

Solution

From the above data, we may prepare the table given below.

Weight (in kg) xi	Number of athletes (Frequency) $f_1$	$f_1{x}_i$
42 45 46 48 49	4 3 5 2 1	168 135 230 96 49
	${\mathrm{\Sigma f}}_{\mathrm{I}}$= 15	$\Sigma f_i{x}_i$ = 678

  $\therefore$Mean Weight =$\frac{\sum f_i{x}_i}{\sum f_i}=\frac{678}{15}=45.2 \mathrm{kg}$.

Illustration -13

Using short cut method, calculate the mean weekly wage from the following frequency distribution.

Weekly wages(in Rs)	950	1000	1050	1100	1250	1500	1600
Number of workers	24	18	13	15	20	11	9

Solution

Let the assumed mean be, A = 1100.

From the given data, we may prepare the table given below.

Weekly Wages (in Rs) x1	Numbers of Workers (Frequency)	di = (xi – A)     = (xi – 1100)	fidi
950 1000 1050 1100 = A 1250 1500 1600	24 18 13 15 20 11 9	–150 –100 –50 0 150 400 500	–3600 –1800 –650 0 3000 4400 4500
	$\Sigma$fI = 110		$\Sigma$fI di = 5850

$\therefore$ Mean = $\left(\mathrm{A}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)=\left(1100+\frac{5850}{110}\right)=1153.18$

Hence, Mean Wage = Rs 1153.18.

3. MEAN OF GROUPED DATA IN THE FORM OF CLASSES

I. Direct Method

Step 1: For each class, find the class mark x_i by using the relation, x_i = $\frac{1}{2}$ (lower limit + upper limit). Step 2: Use the formula, Mean = $\frac{\sum f_i{X}_i}{\sum f_i}$.

II. Short Cut Method or Deviation Method

Step 1: For each class, find the class mark x_i.

Step 2: Let A be the assumed mean.

Step 3: Find d_i = (x_i – A).

Step 4: Use the formula, Mean = $\left(\mathrm{A}\frac{\sum f_{\mathrm{i}}\mathrm{d}}{\Sigma f_{\mathrm{i}}}\right)$.

III. Step-Deviation Method

Step 1. For each class, find the class mark x₁.

Step 2: Let A be the assumed mean.

Step 3: Calculate, ${\mu }_{\mathrm{I}}=\frac{\left({\mathrm{X}}_{\mathrm{i}}–\mathrm{A}\right)}{\mathrm{C}}$, where c is the class size.

Step 4: Use the formula, Mean = $\left(\mathrm{A}+\mathrm{C}\frac{\sum f_{\mathrm{i}}{\mu }_{\mathrm{i}}}{\sum f_{\mathrm{i}}}\right)$.

Illustration -14

Using direct method, find the mean of the following frequency distribution:

Class-interval	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	6	8	12	15	10	9

Solution

From the given data, we may prepare the table given below.

Class-Interval	Class-Mark xi	Frequency fi	fixi
10-20 20-30 30-40 40-50 50-60 60-70	15 25 35 45 55 65	6 8 12 15 10 9	90 200 420 675 550 585
		$\Sigma$fI = 60	$\Sigma$fixi = 2520

$\therefore$ Mean = $\frac{\sum f_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum f_{\mathrm{i}}}=\frac{2520}{60}=42$.

Illustration -15

Using Short Cut Method find the mean for the following frequency distribution:

Class-interval	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	8	10	16	23	12	11

Solution

From the given data, we may prepare the table given below.

Class-Interval	Class-Mark xi	Frequency fi	di = (xi – A)    = (xi – 99)	fidi
84-90 90-96 96-102 102-108 108-114 114-120	87 93 99 = A 105 111 117	8 10 16 23 12 11	–12 –6 0 6 12 18	–96 –60 0 138 144 198
		$\Sigma$fI = 80		$\Sigma$fidi = 324

$\therefore$ Mean = $\frac{\Sigma f_i{d}_i}{\sum f_i}=\left(99+\frac{324}{80}\right)=(99+4.05)=103.05$

Hence, Mean = 103.05.

Illustration -16

Using Step Deviation Method, calculate the mean for the following data:

Height (in cm)	135-140	140-145	145-150	150-155	155-160	160-165	165-170	170-175
Frequency	4	9	18	28	24	10	5	2

Solution

Here, class size, c = 5. Take assumed mean, A = 152.5. Thus, from the given data, we may prepare

Class-Interval	Class-Mark   xi	Frequency  fi	$\left({\mathrm{\mu }}_{\mathrm{i}}=\frac{{\mathrm{X}}_{\mathrm{i}}–\mathrm{A}}{\mathrm{C}}\right)$	fi$\mu$i
135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175-	137.5 142.5 147.5 152.5 = A 157.5 162.5 167.5 172.5	4 9 18 28 24 10 5 2	–3 –2 –1 0 1 2 3 4	–12 –18 –18 0 24 20 15 8
		$\Sigma$fI = 100		$\Sigma$fi$\mu$i = 19

$\therefore$ Mean = $\left(\mathrm{A}+\mathrm{c}\times \frac{\sum f_{\mathrm{i}}{\mu }_{\mathrm{i}}}{\Sigma f_{\mathrm{i}}}\right)=\left(152.5+\frac{5\times 19}{100}\right)$ = (152.5 + 0.95) = 153.45 cm.

4. IMPORTANT FORMULAE FOR SOLVING ARITHMETIC MEAN

1. Arithmetic Mean of Individual Series

If x₁, x₂ , x₃,….., x_n are n values of variant x, then its Arithmetic Mean, denoted by x is

$\text{ A.M. }\left(\overline{x}\right)=\frac{x_1+x_2+x_3+\dots ..+x_n}{n}=\frac{\sum x_i}{n}$   (or)

$\text{ A.M. }\left(\overline{x}\right)=A+\frac{\sum \left(x_i–A\right)}{n}$ where A is the assumed average. (For individual series)

2. Arithmetic Mean of Discreate Series

If a variable takes values x₁, x₂ , x₃,….., x_n with corresponding frequencies f₁, f₂ , f₃,….., f_n then the arithmetic mean $\overline{x}$ is given by

$\overline{x}=\frac{f_1{x}_1+f_2{x}_2\dots .+f_n{x}_n}{f_1+f_2\dots .+f_n}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i$

where $N=\sum _{i=1}^n{f}_i$

3. Arithmetic Mean Of Continuous Series

In case of a set of data with class intervals, we cannot find the exact value of the mean because we do not know the exact values of the variables. We, therefore, try to obtain an approximate value of the mean. The method of approximate is to replace all the observed values belonging to a class by mid-value of the class.
If x₁, x₂ , x₃,….., x_n are the mid values of the class intervals having corresponding frequencies f₁, f₂ , f₃,….., f_n then we apply the same formula as in discrete series.

$\overline{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i, N=\sum _{i=1}^n{f}_i$

4. Combined Arithmetic Mean

If ${\overline{x}}_i(i=1,2,\dots .,k)$ are the means of k – series of sizes $n_i(i=1,2,3,\dots \dots \dots ..,k)$ then the combined or composite mean $\overline{x}$ can be obtained by the formula :

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2+\dots .+n_k{\overline{x}}_k}{n_1+n_2+\dots ..+n_k}=\frac{\sum n_i{\overline{x}}_i}{\sum n_i}$

5. Weighted Arithmetic Mean

If w₁, w₂ , w₃,….., w_n be the weights assigned to the values x₁, x₂ , x₃,….., x_n be respectively of a variable x, then the weighted A.M. is $\overline{x}=\frac{\sum w_i{x}_i}{\sum w_i}$.

5. PROPERTIES OF ARITHMETIC MEAN

1. Sum of all the deviations from arithmetic mean is zero i.e.,

$\sum _{i=1}^n\left(x_i–\overline{x}\right)=0$ (in case of individual series)

$\sum _{i=1}^n{f}_i\left(x_i–\overline{x}\right)=0$ (in case of discrete or continuous series)

2. If each observation is increased or decreased by a given constant K, the mean is also increased or decreased by K
The property is also known as effect of change of origin. K can be taken to be any number. However, to simplify the calculations, K should be taken as a value which is in the middle of the table.
3. Step Deviation Method or change of scale

If x₁, x₂ , x₃,….., x_n are mid values of class intervals with corresponding frequencies f₁, f₂ , f₃,….., f_n then we may change the scale by taking $d_i=\frac{x_i–A}{h}$ in this case.

$\overline{x}=A+h\times \left(\frac{1}{N}\sum f_i{d}_i\right)$ (if A is assumed mean)

A and h can be any numbers but if the lengths of class intervals are equal then h may be taken as width of the class interval.
In particular if each observation is multiplied or divided by a constant, the mean is also multiplied or divided by the same constant.
4. The sum of the squared deviation of the variate from their mean is minimum i.e., the quantity $\sum {\left(x_i–A\right)}^2\text{ or }\sum f_i{\left(x_i–A\right)}^2$ is minimum when $A=\overline{x}$.

5. E(aX + b) = aE(X) + b (where E(X) = Mean of X)

Illustration-17
The weighted mean of the first n natural numbers, the weights being the corresponding numbers, is
Solution
First n natural numbers are 1, 2, 3,…,n; whose corresponding weights are 1, 2, 3,…,n respectively.

$\therefore$ weight mean = $\frac{1\times 1+2\times 2+\dots .+n\times n}{1+2+\dots +n}$

$=\frac{1^2+2^2+\dots ..+n^2}{1+2+\dots ..+n}$

$=\frac{n(n+1)(2n+1)}{\frac{6n(n+1)}{2}}=\frac{2n+1}{3}$

Illustration-18

The weighted mean of the first n natural numbers whose weights are equal to the squares of the corresponding numbers is
Solution

weighted mean = $\frac{1.1^2+2.2^2+\dots +n·n^2}{1^2+2^2+\dots +n^2}$

$=\frac{{\mathrm{\Sigma n}}^3}{{\mathrm{\Sigma n}}^2}=\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}}=\frac{3\mathrm{n}(\mathrm{n}+1)}{2(2\mathrm{n}+1)}$

Illustration-19
The average salary of male employees in a firm is Rs. 5200 and that of females is Rs.4200. The mean salary of all the employees is Rs.5000. The percentage of male and female employees are respectively is
Solution

Let $x_1=5200, x_2=4200, \overline{x}=5000$

Also, we know that $\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$

$\Rightarrow 5000\left(n_1+n_2\right)=5200n_1+4200n_2\Rightarrow \frac{n_1}{n_2}=\frac{4}{1}$

$\therefore$ The percentage of male employees in the firm = $\frac{4}{4+1}\times 100=80\%$

and the percentage of female employees in the firm = $\frac{1}{4+1}\times 100=20\%$

Illustration-20
If the mean of 9 observations is 100 and mean of 6 observations is 80, then the mean of 15 observations is
Solution

$n_1=9, {\overline{x}}_1=100\text{ and }{n}_2=6, {\overline{x}}_2=80$

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}=\frac{9\times 100+6\times 80}{9+6}=92$

Illustration-21

If a variate X is expressed as a linear function of two variates U and V in the form X = aU + bV then the mean $\overline{X}$ of X is

Solution

We have $\Sigma X=a\Sigma U+b\Sigma V$

$\overline{X}=\frac{1}{n}\Sigma X=a.\frac{1}{n}\Sigma U+b.\frac{1}{n}\Sigma V$

$\Rightarrow \overline{X}=a\overline{U}+b\overline{V}$

Illustration-22

If the arithmetic mean of the numbers x₁, x₂ , x₃,….., x_n is $\overline{x}$, then the arithmetic mean of the numbers $a{x}_1+b,a{x}_2+b,a{x}_3+b,\dots .a{x}_n+b$, where a, b are two constants, would be

Solution

Required mean $=\frac{\left(a{x}_1+b\right)+\left(a{x}_2+b\right)+\dots .+\left(a{x}_n+b\right)}{n}$

$=\frac{a\left(x_1+x_2+\dots .+x_n\right)}{n}+b=a\overline{x}+b \left(\because \frac{x_1+x_2+\dots +x_n}{n}=\overline{x}\right)$

Illustration-23

If the mean of the numbers 27+x, 31+x, 89+x , 107+x, 156+x is 82, then the mean of 130+x, 126+x, 68+x, 50+x, 1+x is

Solution

Given

$82=\frac{(27+x)+(31+x)+(89+x)+(107+x)+(156+x)}{5}$

$\Rightarrow 82\times 5=410+5x\Rightarrow 410–410=5x\Rightarrow x=0$

Therefore, the required mean is

$\overline{x}=\frac{130+x+126+x+68+x+50+x+1+x}{5}$

$=\frac{375+5x}{7}=75$

Illustration-24

A student obtain 75%, 80% and 85% in three subjects. If the marks of another subject is added, then his average cannot be less than

Solution

Marks obtained from three subjects out of 300 is 75+80+85 = 240

If the marks of another subject is added, then the marks will be $\ge$240 out of 400

$\therefore$ minimum average marks = $\frac{240}{4}=60\%$

[when marks in the fourth subject = 0]

Illustration-25

The mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is

Solution

Sum of 100 items = 49 × 100 = 4900

sum of items added = 60+ 70+80=210

new sum = 4900+210–110=5000

$\therefore$ correct mean = $\frac{5000}{100}=50$

Illustration-26

The mean weight per student in a group of seven students is 55kg. If the individual weights of six students are 52, 58, 55, 53, 56 and 54, then the weight of the seventh student is

Solution

The total weight of seven students is 55×7 = 385kg

The sum of the weights of six students is 52+58+55+53+56+54 = 328kg

Hence, the weight of the seventh student is = 385 – 328 = 57kg

3. GEOMETRIC MEAN

1. GEOMETRIC MEAN IN INDIVIDUAL SERIES

1. If x₁, x₂ , x₃,….., x_n are n values of a variate x, none of them being zero, then geometric mean (G.M.) is given by G.M = ${\left(x_1–x_2–x_3\dots \dots x_n\right)}^{1/n}$

2. log (G.M.) = $\frac{1}{n}\left(\log x_1+\log x_2+\dots ..+\log x_n\right)$

2. GEOMETRIC MEAN IN DISCRETE SERIES

In case of frequency distribution, G.M. of n values x₁, x₂ , x₃,….., x_n of a variate x occurring with frequency f₁, f₂ , f₃,….., f_n is given by G.M = ${\left(x_1^{f_1}.x_2^{f_2}\dots ..x_n^{f_n}\right)}^{1/N}$, where $N=f_1+f_2+\dots ..+f_n$.

Illustration-27

The geometric mean of the numbers $3,3^2,3^3,\dots ,3^n$ is

Solution

$\therefore G.M={\left(3.3^2\dots \dots 3^n\right)}^{1/n}$

$=3^{\frac{1+2+\dots +n}{n}}=3^{\frac{n(n+1)}{2n}}=3^{\frac{n+1}{2}}$

Illustration-28

Find the geometric mean of the following values:

Solution

Given data, 15, 12, 13, 19, 10

X	Log x
15	1.1761
12	1.0792
13	1.1139
19	1.2788
10	1.0000
Total	5.648

$\Sigma$ log x =5.648 and n = 5

log(G.M.) = $\frac{1}{n}\left(\log x_1+\log x_2+\dots ..+\log x_n\right)$.

= Antilog $\frac{5.648}{5}$

= Antilog 1.1296

= 13.48

Answer: Geometric Mean = 13.48

4. HARMONIC MEAN

The harmonic mean is based on the reciprocals of the value of the variable

$\mathrm{H}.\mathrm{M}=\frac{1}{\frac{1}{n}\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots +\frac{1}{x_n}\right)}\text{ or }\frac{1}{H}=\frac{1}{n}\sum _{i=1}^n\frac{1}{x_i}$ (Incase of Individual series) and $\frac{1}{H}=\frac{1}{N}\sum _{i=1}^n{f}_i\frac{1}{x_i}$ (in case of discrete series or continuous series)

Relation among Arithmetic Mean, Geometric Mean and Harmonic Mean

The Arithmetic Mean (A. M.), Geometric Mean (G.M.) and Harmonic Mean (H.M.) for a given set of observations x₁, x₂ , x₃,….., x_n > 0 of a series are related as under:

$\mathrm{A}.\mathrm{M}\ge \mathrm{G}.\mathrm{M}\ge \mathrm{H}.\mathrm{M}$

Illustration-29

The harmonic mean of 3, 7, 8, 10, 14 is

(A) $\frac{3+7+8+10+14}{5}$     (B) $\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{14}$     (C) $\frac{\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{14}}{4}$     (D) $\frac{5}{\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{14}}$

Solution

D

H.M. = $\frac{1}{\frac{1}{n}\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots .+\frac{1}{x_n}\right)}=\frac{5}{\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{14}}$

Illustration-30

Find the harmonic mean of $\frac{1}{2},\frac{2}{3},\frac{3}{4},\dots .,\frac{n}{n+1}$, occurring with frequencies 1, 2, 3,…..n, respectively.

Solution

We know that, Harmonic mean = $\frac{\sum f}{\sum \left(\frac{f}{x}\right)}$

$\sum f=1+2+3+\dots .+n=\frac{n(n+1)}{2}$

and $\sum \frac{f}{x}=\frac{1}{1/2}+\frac{2}{2/3}+\frac{3}{3/4}+\dots .+\frac{n}{n/(n+1)}$

$=2+\frac{3\times 2}{2}+\frac{4\times 3}{3}+\dots +\frac{n(n+1)}{n}$

$=2+3+4+\dots +n+(n+1)$

Which is an arithmetic progression with a = 2 and d = 1.

By the formula of sum of n term of an A.P, $\sum \left(\frac{f}{x}\right)=\left[\frac{n}{2}\{2a+(n–1\left)d\right\}\right]$

we have = $\frac{n}{2}\{2\times 2+n–1\}=\frac{n}{2}(3+n)$

$\therefore$Harmonic mean

$=\frac{2}{n(3+n)}=\frac{n(n+1)\times 2}{n(3+n)\times 2}=\frac{n+1}{3+n}$

5. MEDIAN AND MODE

1. MEDIAN
The median is the central value of the set of observations provided all the observations are arranged in the ascending or descending orders. It is generally used, when effect of extreme items is to be kept out.
Calculation of median
(i) Individual series: If the data is raw, arrange in ascending or descending order. Let n be the number of observations.

If n is odd, Median = value of ${\left(\frac{n+1}{2}\right)}^{th}$ item.

If n is even, Median = $\frac{1}{2}\left[\text{ value of }{\left(\frac{n}{2}\right)}^{\text{th }}\text{ item + value of }{\left(\frac{n}{2}+1\right)}^{\text{th }}\text{ item }\right]$

(ii) Discrete series : In this case, we first find the cumulative frequencies of the variables arranged in ascending or descending order and the median is given by Median = ${\left(\frac{n+1}{2}\right)}^{\text{th }}$ observation, where n is the cumulative frequency.

(iii) For grouped or continuous distributions : In this case, following formula can be used.

(a) For series in ascending order, Median = $I+\frac{\left(\frac{N}{2}–C\right)}{f}\times i$

Where

l = Lower limit of the median class
f = Frequency of the median class
N = The sum of all frequencies
i = The width of the median class
C = The cumulative frequency of the class preceding to median class.

(b) For series in descending order

Median = $u–\left(\frac{\frac{N}{2}–C}{f}\right)\times i$, where u = upper limit of the median class, $N=\sum _{i=1}^n{f}_i$

As median divides a distribution into two equal parts, similarly the quartiles, quantiles, deciles and percentiles divide the distribution respectively into 4, 5, 10 and 100 equal parts. The j^th quartile is given by $Q_j=I+\left(\frac{j\frac{N}{4}–C}{f}\right)i; j=1,2,3$. Q₁ is the lower quartile, Q₂ is the median and Q₃ is called the upper quartile.

(2) Lower quartile

(i) Discrete series : $Q_1=\text{ size of }{\left(\frac{n+1}{4}\right)}^{\text{th }}\text{ item }$

(ii) Continuous series : $Q_1=I+\frac{\left(\frac{N}{4}–C\right)}{f}\times i$

(3) Upper quartile

(i) Discrete series : $Q_3=\text{ size of }{\left[\frac{3(n+1)}{4}\right]}^{\text{th }}\text{ item }$

(ii) Continuous series : $Q_3=I+\frac{\left(\frac{3N}{4}–C\right)}{f}\times i$

(4) Decile : Decile divide total frequencies N into ten equal parts.

$D_j=I+\frac{\frac{N\times j}{10}–C}{f}\times i$,     [where j = 1, 2, 3, 4, 5, 6, 7, 8, 9]

If j = 5, then $D_5=I+\frac{\frac{N}{2}–C}{f}\times i$. Hence D₅ is also known as median.

(5) Percentile : Percentile divide total frequencies N into hundred equal parts and $P_k=I+\frac{\frac{N\times k}{100}–C}{f}\times i$,  where k = 1, 2, 3, 4, 5,…….,99.

Illustration-31

The median of 10, 14, 11, 9, 8, 12, 6 is

(A) 10     (B) 12     (C) 14     (D) 11

Solution

(A)

Arrange the items in ascending order i.e., 6, 8, 9, 10, 11, 12, 14.

If n is odd then, Median = value of ${\left(\frac{n+1}{2}\right)}^{th}$ term

$\therefore$ Median = ${\left(\frac{7+1}{2}\right)}^{th}$ term = 4^th term = 10.

Illustration-32
The median of a set of nine distinct observations is 20.5. If each of the last four observations of the set is increased by 2, then the median of the new set is
Solution

Since n = 9, median term = ${\left(\frac{9+1}{2}\right)}^{th}$ = 5^th term.

Now, the last four observations are increased by 2. Since the median is the 5th observation, which remains unchanged, there will be no change in median.

Illustration-33
If a variable takes the discrete value $\alpha –4, \alpha –\frac{7}{2}, \alpha –\frac{5}{2}, \alpha –3, \alpha –2, \alpha +\frac{1}{2}, \alpha –\frac{1}{2}, \alpha +5(a>0),$ then the median is

Solution

Arrange the data as follows:

$\alpha –\frac{7}{2}, \alpha –3, \alpha –\frac{5}{2}, \alpha –2, \alpha –\frac{1}{2}, \alpha +\frac{1}{2}, \alpha –4, \alpha +5$

median = $\frac{1}{2}$[value of 4th item+value of 5th item]

$\therefore$ median = $\frac{\alpha –2+\alpha –\frac{1}{2}}{2}=\alpha –\frac{5}{4}$

Illustration-34

The median of distribution 83, 54, 78, 64, 90, 59, 67, 72, 70, 73 is

Solution

On arranging in ascending order, we get 54, 59, 64, 67, 70, 72, 73, 78, 83, 90

n = 10

$\therefore \text{ median }=\frac{\text{ value of }{\frac{10}{2}}^{\mathrm{th}}\text{ term }+\text{ valueof }{\left(\frac{10}{2}+1\right)}^{\text{th }}\text{ term }}{2}$

= $\frac{\text{ value of }{5}^{\text{th}}\text{ term }+\text{ value of }{6}^{\text{th}}\text{ term }}{2}$

$=\frac{70+72}{2}=71$

2. MODE

The mode or model value of a distribution is that value of the variable for which the frequency is maximum. For continuous series, mode is calculated as,

Mode = $I_1+\left[\frac{f_1–f_0}{2f_1–f_0–f_2}\right]\times i$

Where, l₁ = The lower limit of the model class

f₁ = The frequency of the model class

f₀ = The frequency of the class preceding the model class

f₂ = The frequency of the class succeeding the model class

i = The size of the model class.

Symmetric distribution
A distribution is a symmetric distribution if the values of mean, mode and median coincide. In a
symmetric distribution frequencies are symmetrically distributed on both sides of the centre point of the frequency curve.

A distribution which is not symmetric is called a skewed-distribution. In a moderately asymmetric
distribution, the interval between the mean and the median is approximately one-third of the interval between the mean and the mode i.e., we have the following empirical relation between them,

Mean – Mode = 3(Mean – Median) $\Rightarrow$ Mode = 3 Median – 2 Mean. It is known as Empirical relation.

Illustration-35

The mode of the following distribution is

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	8	7	12	28	20	10	10

Solution

Here, maximum frequency is 28. Thus, the class 40-50 is the modal class.

Mode = $l+\frac{f_1–f_0}{2f_1–f_2–f_0}\times h$

$=40+\frac{10(28–12)}{(2\times 28–12–20)}$

= 40+6.666 = 46.67(approx.)

Illustration-36
If in a frequency distribution, the mean and median are 20 and 21 respectively, then its mode is approximately
Solution
mode = 3 median – 2 mean = 3 (21) – 2(20) = 23.

Illustration-37
If in a moderately asymmetrical distribution the mode and the mean of the data are 6$\lambda$ and 9$\lambda$ , respectively, then the median is
Solution

For a moderately skewed distribution,
mode = 3median – 2 mean
$\Rightarrow 6\lambda =3\text{ median }–18\lambda \Rightarrow \text{ median }=8\lambda$

6. MEASURES OF DISPERSION

The degree to which numerical data tend to spread about an average value is called the dispersion of the data. The four measure of dispersion are

(1) Range     (2) Mean deviation     (3) Standard deviation     (4) Square deviation

1. RANGE
It is the difference between the values of extreme items in a series. Range = $X_{\max }–X_{\min }$

The coefficient of range (scatter) = $\frac{x_{\max }–x_{\min }}{x_{\max }+x_{\min }}$.

Range is not the measure of central tendency. Range is widely used in statistical series relating to
quality control in production.

Range is commonly used measures of dispersion in case of changes in interest rates, exchange rate, share prices and like statistical information. It helps us to determine changes in the qualities of the goods produced in factories.

i. Inter-quartile range
We know that quartiles are the magnitudes of the items which divide the distribution into four equal parts. The inter-quartile range is found by taking the difference between third and first quartiles and is given by the following formula,

Inter-quartile range = Q₃ – Q₁,

where Q₁ = First quartile or lower quartile and Q₃ = Third quartile or upper quartile.

ii. Percentile range

This is measured by the following formula,

Percentile range = P₉₀ – P₁₀,

where P₉₀ – 90^th percentile and P₁₀ = 10^th percentile.

Percentile range is considered better than range as well as inter-quartile range.

iii. Quartile deviation or semi inter-quartile range

It is one-half of the difference between the third quartile and first quartile i.e., Q.D. = $\frac{Q_3–Q_1}{2}$ and coefficient of quartile deviation = $\frac{Q_3–Q_1}{Q_3+Q_1}$, where Q₃ is the third or upper quartile and Q₁ is the lowest or first quartile.

2. MEAN DEVIATION
The arithmetic average of the deviations (all taking positive) from the mean, median or mode is
known as mean deviation.

Mean deviation is used for calculating dispersion of the series relating to economic and social inequalities. Dispersion in the distribution of income and wealth is measured in term of mean deviation.

i. Mean deviation from ungrouped data (or individual series)

Mean deviation = $\frac{\sum |x–M|}{n}$.

where |x – M| means the modulus of the deviation of the variate from the mean (mean, median or
mode) and n is the number of terms.

ii. Mean deviation from continuous series

Here first of all we find the mean from which deviation is to be taken. Then we find the deviation
dM =|x – M| of each variate from the mean M so obtained.
Next we multiply these deviations by the corresponding frequency and find the product f.dM and then the sum $\Sigma$f dM of these products.

Lastly we use the formula, mean deviation = $\frac{\sum f|x–M|}{n}=\frac{\sum fdM}{n},\text{ where }n=\mathrm{Sf}\text{.}$

3. STANDARD DEVIATION

Standard deviation (or S.D.) is the square root of the arithmetic mean of the square of deviations of various values from their arithmetic mean and is generally denoted by s read as sigma. It is used in statistical analysis.
i. Coefficient of standard deviation
To compare the dispersion of two frequency distributions the relative measure of standard deviation is computed which is known as coefficient of standard deviation and is given by

Coefficient of A.D. = $\frac{\sigma }{\overline{x}}$, where $\overline{x}$ is the A.M.

ii. Standard deviation from individual series

$\sigma =\sqrt{\frac{\sum (x–\overline{X}{)}^2}{N}}$

where, $\overline{x}$ = The arithmetic mean of series

N = The total frequency.

iii. Standard deviation from continuous series

$\sigma =\sqrt{\frac{\sum f_i{\left(x_i–\overline{x}\right)}^2}{N}}$

where, $\overline{x}$ = Arithmetic mean of series
x_i = Mid value of the class
f_i = Frequency of the corresponding
N = Sf = The total frequency

Short cut method

(i) $\sigma =\sqrt{\frac{\sum f{d}^2}{N}–{\left(\frac{\sum fd}{N}\right)}^2}$          (ii) $\sigma =\sqrt{\frac{\sum d^2}{N}–{\left(\frac{\sum d}{N}\right)}^2}$

where, d = x – A = Deviation from the assumed mean A
f = Frequency of the item
N = f = Sum of frequencies

4. SQUARE DEVIATION
i. Root mean square deviation

$S=\sqrt{\frac{1}{N}\sum _{i=1}^n{f}_i{\left(x_i–A\right)}^2}$

where A is any arbitrary number and S is called mean square deviation.

ii. Relation between S.D. and root mean square deviation

If s be the standard deviation and S be the root mean square deviation.

Then, $S^2={\sigma }^2+d^2$,

Obviously, S² will be least when d = 0 i.e., $\overline{x}=A$

Hence, mean square deviation and consequently root mean square deviation is least, if the deviations are taken from the mean.

VARIANCE

The square of standard deviation is called the variance.

Coefficient of standard deviation and variance

The coefficient of standard deviation is the ratio of the S.D. to A.M. i.e., $\frac{\sigma }{x}$.

Coefficient of variance = coefficient of S.D. $\times 100=\frac{\sigma }{\overline{x}}\times 100$.

Variance of the combined series

If $n_1, n_2$ are the sizes, $\overline{x_1}, \overline{x_2}$ the means and ${\sigma }_1, {\sigma }_2$ the standard deviation of two series, then

${\sigma }^2=\frac{1}{n_1+n_2}\left[n_1\left({\sigma }_1^2+d_1^2\right)+n_2\left({\sigma }_2^2+d_2^2\right)\right]$

where $d_1={\overline{x}}_1–\overline{x}, d_2={\overline{x}}_2–\overline{x}$ and $\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$.

Note :1

if V (X) is variance of X then

V(X + a) = V(X)

V(aX) = a²V(X)

V(aX + b) = a²V(X)

V (aX + bY) = a²v (X) + b²v (Y)

Note :2

For a, a + d, a + 2d, ….., a + (n – 1)d,

$\overline{x}=a+\frac{(n–1)d}{2}; {\sigma }^2=\frac{n^2–1}{12}{d}^2$

Note :3

1) Q.D < M.D < S.D

2) $\frac{\mathrm{Q}.\mathrm{D}}{10}=\frac{\mathrm{M}.\mathrm{D}}{12}=\frac{\mathrm{S}.\mathrm{D}}{15}$

SKEWNESS

“Skewness” measures the lack of symmetry. It is measured by ${\gamma }_1=\frac{\sum {\left(x_i–\mu \right)}^3}{{\left\{\Sigma \left(x_i–{\mu }^2\right)\right\}}^{3/2}}$ and is denoted by ${\gamma }_1$.

The distribution is skewed if,

(i) Mean $\ne$ Median $\ne$ Mode

(ii) Quartiles are not equidistant from the median

(iii) The frequency curve is stretched more to one side than to the other.

1. Distribution

There are three types of distributions.

(i) Normal distribution :

When ${\gamma }_1=0$, the distribution is said to be normal. In this case, Mean = Median = Mode

(ii) Positively skewed distribution : When ${\gamma }_1>0$, the distribution is said to be positively skewed. In this case, Mean > Median > Mode

(iii) Negative skewed distribution : When ${\gamma }_1<0$, the distribution is said to be negatively skewed. In this case, Mean < Median < Mode

Measures of skewness

(i) Absolute measures of skewness : Various measures of skewness are

(a) $S_K=M–M_d$          (b) $S_K=M–M_0$          (c) $S_K=Q_3+Q_1–2M_d$

where, M_d = median, M₀ = mode, M = mean.

Absolute measures of skewness are not useful to compare two series, therefore relative measure of
dispersion are used, as they are pure numbers.

3. Relative measures of skewness

(i) Karl Pearson’s coefficient of skewness

$S_k=\frac{M–M_o}{\sigma }=3\frac{\left(M–M_d\right)}{\sigma },–3\le S_k\le 3$

where s is standard deviation.

(ii) Bowley’s coefficient of skewness :

$S_k=\frac{Q_3+Q_1–2M_d}{Q_3–Q_1}$

Bowley’s coefficient of skewness lies between –1 and 1.

(iii) Kelly’s coefficient of skewness

$S_K=\frac{P_{10}+P_{90}–2M_d}{P_{90}–P_{10}}=\frac{D_1+D_9–2M_d}{D_9–D_1}$

Illustration-38

The quartile deviation of daily wages of in (Rs.) of 11 persons given below 140, 145, 130, 165, 160, 125, 150, 170, 175, 120, 180

Solution

The given data in ascending order of magnitude is 120, 125, 130, 140, 145, 150, 160, 165, 170, 175, 180

Here, $Q_1={\left(\frac{n+1}{4}\right)}^{th}\text{ term }=\frac{11+1}{4}\text{ term }=3^{rd}\text{ term }$ = 130

$Q_3=\frac{3(n+1{)}^{th}}{4}=\frac{3\times (11+1)}{4}=9^{th}=170$

$QD=\frac{Q_3–Q_1}{2}=\frac{170–130}{2}=\frac{40}{2}=20$

Illustration-39

The variance of the first ‘n’ natural numbers is

Solution

Variance

$=(SD{)}^2=\frac{1}{n}\Sigma x^2–{\left(\frac{\Sigma x}{n}\right)}^2, \left(\because \overline{x}=\frac{\Sigma x}{n}\right)$

$=\frac{n(n+1)(2n+1)}{6n}–{\left(\frac{n(n+1)}{2n}\right)}^2=\frac{n^2–1}{12}$

Illustration-40

If the M.D is 12, the value of S.D will be

Solution

We know that $\mathrm{Q}.\mathrm{D}=\frac{5}{6}\times M.D=\frac{5}{6}\times 12=10$

$\therefore S.D=\frac{3}{2}\times Q.D=\frac{3}{2}\times 10=15$

Illustration-41

The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, then the other two observations are

Solution

Let the two unknown items be x and y, then

Mean = 4 $\Rightarrow \frac{1+2+6+x+y}{5}=4$

$\Rightarrow$ x + y = 11…..(1) and variance = 5.2

$\Rightarrow \frac{1^2+2^2+6^2+x^2+y^2}{5}–(\text{ mean }{)}^2=5.2$

$41+x^2+y^2=5\left[5.2+(4{)}^2\right]$

$41+x^2+y^2=106$

$x^2+y^2=65\dots \dots \left(2\right)$

Solving (1) and (2) for x and y, we get

x = 4, y = 7 or x = 7, y = 4.

1. BASICS OF STATISTICS

1. STATISTICAL DATA

Statistical data are the facts which are collected for the purpose of investigation. There are two types of statistical data:

(i) Primary data: The data collected by an investigator for the first time for his own purpose are called primary data. As the primary data are collected by the user of the data, so it is more reliable and relevant.

(ii) Secondary data: The data collected by a secondary source and used by the investigator for his purpose is called secondary data. For example score of a cricket match noted from newspapers is secondary data. Thus data which are primary in the hands of one become secondary in the hands of the other. Data collected by any source also can be divided in following two types:

(i) Raw Data: Raw data are those data which are obtained from the original source but not arranged numerically. This is also called ‘ungrouped data’ for example marks of 10 students in maths are given as:  75, 96, 25, 32, 89, 62, 40, 79, 35, 55 An ‘array’ is an arrangement of raw numerical data in the ascending or descending order of magnitude. Above data can be written as  25, 32, 35, 40, 55, 62, 75, 79, 89, 96

(ii) Grouped data: An array can be placed systematically in groups or categories. For example the above data can be grouped in following manner.

GROUPS	MARKS	TOTAL NUMBER OF STUDENTS
0 to 20	–	0
21 to 40	25, 32, 35, 40	4
41 to 60	55	1
61 to 80	62, 75, 79	3
81 to 100	89, 96	2
TOTAL		10

 2. SOME BASIC DEFINITIONS

(i) Variate: Variate is a quantity that may vary from observation to observation.

(ii) Range: Range is difference between the maximum and minimum observations.

(iii) Class Interval: When data are divided in groups, each group is called a class interval.

(iv) Class Limit: Every class interval has two limits. The smallest observation of the interval is called lower limit and the largest observation of the interval is called upper limit.

(v) Class Mark: The mid value of any class is called its class mark.

$\text{ Class Mark }=\frac{\text{ Upper limit of the class + lower limit of the class }}{2}$

(vi) Class Size: Class size is defined as the difference between two successive class marks. It is also the difference between the upper and lower limits of any class interval.

(vii) Frequency: In a particular class the count of the number of observation is called its frequency. So the corresponding frequency of a class is called its class frequency.

(viii) Cumulative Frequency: The cumulative frequency of any class is obtained by adding all the frequencies successively prior to that class i.e. it is the sum of all frequencies up to that class.

(ix) True Class Limit: In the case of exclusive classes the upper and lower limits are respectively known as its true upper limits and true lower limits. In the case of inclusive classes, the true lower and upper limits are obtained by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. True upper limits and true lower limits are also known as boundaries of the class.

(x) Tally: Tally method is used to keep the chance of error at minimum in counting. A bar (|) called tally mark is put against any item when it occurs. The fifth occurrence of any item is represented by putting diagonally a cross tally (\) on the first four tallies.

3. FREQUENCY DISTRIBUTION

The tabular arrangement of data showing the frequency of each item is called a frequency distribution table. It is a method to present raw data in the form from which one can easily understand the information contained in the raw data. Frequency distribution are of two types:

i. Discrete frequency distribution:

In this type of frequency distribution, in the first column of frequency table we write all possible values of the variables from the lowest to the highest, in the second column we write tally marks and in the third column we show frequency of each item. In this method data are not divided into groups or classes.

e.g. no. of girls in 20 families is given in following data:

1, 2, 3, 1, 1, 2, 3, 3, 4, 1, 5, 1, 1, 2, 2, 3, 3, 2, 4, 1

The above data can be put in the form of a discrete frequency distribution table in the following manner:

S. No.	No. of girls	Tally Marks	Frequency
1.	1	$\bcancel{\left|\right| \left|\right|} \left|\right|$	7
2.	2	$\bcancel{\left|\right|\left|\right|}$	5
3.	3	$\bcancel{\left|\right|\left|\right|}$	5
4.	4	||	2
5.	5	|	1
TOTAL			20

ii. Continuous or Grouped Frequency Distribution

In the frequency distribution data are divided  into groups or classes. This method is used only where the values in the raw data are largely repeating and the difference between the greatest and the smallest observations is not very large.

4. CUMULATIVE FREQUENCY

Cumulative frequency table is obtained from the ordinary frequency table by successively adding the several frequencies. Thus to form a cumulative frequency table we add a column of cumulative frequency in the frequency distribution table. It is obvious that the cumulative frequency of the last class is the sum of the frequencies of all the classes.

Cumulative frequency series are of two types:

(i) Less than series        (ii) More than series

Illustration -1

The weekly saving of 30 workers working in a factory is as given below:

64, 60, 87, 75, 69, 34, 51, 78, 39, 48, 73, 54, 63, 70, 57, 88, 90, 53, 74, 44, 31, 71, 68, 72, 36, 89, 55, 67, 73, 83

(a) Taking first class interval as 30 – 40 (40 not included), form a frequency table of equal intervals.

(b) Also form a cumulative frequency table.

(c) Find the number of workers whose weekly saving is more than Rs. 60.

Solution

(i) For the given data of the weekly saving of 30 workers working in a factory, we prepare             following table:

S. No.	Class Interval (Saving in Rs.)	Tally Marks	Frequency (No. of workers)
1.	30 – 40	| | | |	4
2.	40 – 50	| |	2
3.	50 – 60	$\bcancel{\left|\right|\left|\right|}$	5
4.	60 – 70	$\bcancel{\left|\right|\left|\right|} |$	6
5.	70 – 80	$\bcancel{\left|\right|\left|\right|} \left|\right||$	8
6.	80 – 90	| | | |	4
7.	90 – 100	|	1
		TOTAL	30

(ii) For given data cumulative frequency table can be given in the following manner:

S. No.	Class Interval (Saving in Rs.)	Tally Marks	Frequency (No. of workers)	Cumulative Frequency
1.	30 – 40	| | | |	4	4
2.	40 – 50	| |	2	6
3.	50 – 60	$\bcancel{\left|\right|\left|\right|}$	5	11
4.	60 – 70	$\bcancel{\left|\right|\left|\right|} |$	6	17
5.	70 – 80	$\bcancel{\left|\right|\left|\right|} \left|\right||$	8	25
6.	80 – 90	| | | |	4	29
7.	90 – 100	|	1	30
		TOTAL	30

(iii) Number of workers whose weekly saving is Rs. 60 or more than Rs. 60  = (6 + 8 + 4 + 1) = 19

Illustration -2

Heights of seven persons in cm are 120, 125, 142, 134, 150, 155 and 128. Calculate the range.

Solution

For given data maximum height = 155 cm

and minimum height = 120 cm

Range  = maximum height – minimum height

= 155 cm – 120 cm

= 35 cm.

Illustration -3

Form a frequency table from the following table:

Marks	No. of Students
Below 10	15
Below 20	35
Below 30	60
Below 40	84
Below 50	96
Below 60	127
Below 70	198
Below 80	250

Solution

For given data we make classes 0 – 10, 10 – 20, 20 – 30, ….., 70–80.

There are 15 students who obtained below 10 marks, therefore frequency of class 0 – 10 is 15. Again number of students getting below 20 marks is 35.

This includes those students also who obtained below 10 marks.

Number of students who got marks between 10 and 20 i.e. frequency of class

10 – 20  =  35–15 = 20

Thus for given data we make following table:

S. No.	Class Interval	Comparative	Frequency
1.	0 – 10	15	15
2.	10 – 20	35	20
3.	20 – 30	60	25
4.	30 – 40	84	24
5.	40 – 50	96	12
6.	50 – 60	127	31
7.	60 – 70	198	71
8.	70 – 80	250	52
		TOTAL	250

Illustration -4

Find the unknown entries (a, b, c, d, e, f, g) from the following frequency distribution of heights of 60 students in a class:

Height (in cm)	Frequency	Cumulative frequency
160 – 165	15	a
165 – 170	b	35
170 – 175	12	c
175 – 180	e	50
180 – 185	d	55
185 – 190	5	f
Total	g

Solution

From given table, we make following cumulative frequency table:

S. No.	Height (in cm)	Frequency	Cumulative Frequency
1.	160 – 165	15	15  = a
2.	165 – 170	b	15 + b = 35
3.	170 – 175	12	15 + b + 12 = c
4.	175 – 180	d	15 + b + 12 + d = 50
5.	180 – 185	e	15 + b + 12 + b + e = 55
6.	185 – 190	5	15 + b + 12 + d + e + 5 = f
		g	60

Now, from table:

a = 15

15 + b = 35    b = 35 – 15 = 20

15 + b + 12 = c   15 + 20 + 12 = c

c = 47

15 + b + 12 + d = 50

15 + 20 + 12 + d = 50

d = 50 – 47

d = 3

15 + b + 12 + d + c = 55

15 + 20 + 12 + 3 + e = 55

e = 55 – 50

e = 5

15 + b + 12 + d + e + 5 = f

15 + 20 + 12 + 3 + 5 + 5 = f

f = 60

g = 60. Hence, a = 15, b = 20, c = 47, d = 3, e = 5, f = 60 and g = 60.

5. GRAPHICAL REPRESENTATION OF DATA

A given data can be represented in graphical way. There are various methods of graphical representation of frequency distribution.

(i) Bar Graphs

(ii) Histogram

(iii) Frequency Polygon

(iv) Pie Chart

Bar Graph

The frequency distribution of a discrete value is best represented by a bar graph. The height of the bars is proportional to the frequency of each variate-value. In a bar graph the bars must be kept distinct to show that the variate-values are distinct. The bars are of equal width and are drawn with equal spacing between them on the x-axis depicting the variable. The values of the variable are shown on the y-axis.

Illustration -5

The following table shows the number of illiterate persons in the age group (10–58 years) in a town:

Age Group (in years)	10 – 16	17 – 23	24 – 30	31 – 37	38 – 44	45 – 51	52 – 58
Number of illiterate person	175	375	100	150	250	400	525

Solution

Draw a bar graph to represent the above data.

Histogram

Histogram is a graphical representation of a grouped frequency distribution with continuous classes. It consists of a set of rectangles where heights of rectangles are proportional to their class frequencies, for equal class intervals. There is no gap between two successive rectangles. The rectangles are constructed with base as the class size and their heights representing the frequencies.

Illustration -6

Construct a histogram from the following distribution of total marks obtained by 750 students of class IX in the final examination.

Marks (mid-points)	620	660	700	740	780	820	860
Number of students	16	45	156	284	172	59	18

Solution

To draw the histogram for above data first of all we find out class limits for making class intervals from given class marks.

Here difference between second and the first class marks =  h  =  660 – 620 = 40

$\therefore \frac{h}{2}=\frac{40}{2}=20$

lower limit of first class interval =  620 – h/2

= 620 – 20

= 600

And upper limit of the first class interval = $620+\frac{h}{2}$ = 620 + 20 = 640

Similarly other class intervals would be,

(660 – 20) – (660 + 20), (700 – 20) – (700 + 20), (740 – 20) – (740 + 20), (780 – 20) – (780 + 20), (820 – 20) – (820 + 20) and (860 – 20) – (860 – 20)

i.e. 640 – 680, 680 – 720, 720 – 760, 760 – 800, 800 – 840 and 840 – 880 Using these class intervals we draw following histogram taking class intervals on X-axis and number of students on Y-axis

Frequency Polygon

A frequency polygon is a graph of frequency distribution. It is a line graph of class frequency which is plotted against class mark. A frequency polygon can be obtained by two methods:

(1) By using Histogram: A frequency polygon can be obtained by joining mid points of the top of the rectangles of a histogram. For this we obtain the mid points of the upper horizontal sides of each rectangle and then join these mid points by dotted lines to get frequency polygon. End of a frequency polygon preferably extended to the mid points of imagined class intervals adjacent to first and last class intervals.

Illustration -7

Draw the histogram and frequency polygon of the following frequency distribution:

Monthly wages (in rupees)	325-350	350-375	375-400	400-425	425-450	Total
Number of workers	30	45	75	60	55	245

Solution

Using given data we draw histogram taking monthly wages (in rupees) on X-axis and number of workers on Y-axis. Then we join the mid points of the top of the rectangles by dotted straight lines and complete it by joining the mid points of imagined class intervals adjacent to the first and last class intervals.

(2) Frequency polygon without using Histogram: Following procedure is used to make a frequency polygon without using histogram.

(i) Calculate the class marks, x₁, x₂, …., x_n of each of the given class intervals.

(ii) Mark class marks x₁, x₂, …., x_n, along X-axis and frequencies f₁, f₂, …. f_n along Y-axis.

(iii) Plot the points (x₁, f₁), (x₂, f₂), ,….., (x_n, f_n).

(iv) Obtain the mid-points of two class intervals of zero frequencies at the beginning of the first interval and at the end of the last interval.

(v) Join the points (x₁, f₁), (x₂, f₂), …, (x_n, f_n) by the line segments and complete the frequency polygon by joining the mid points of the first and last intervals to the mid points of the imagined classes adjacent to them.

Illustration -8

Represent the following distribution by a frequency polygon without using a histogram:

Scores	20-29	30-39	40-49	50-59	60-69	70-79	80-89	90-99
Frequency	2	5	3	18	20	22	12	7

Solution

Given data is in the form of inclusive classes, so first of all we shall convert them into exclusive classes. Also we have to get class marks of these class intervals for this purpose.

S. No.	Scores	True Class Limits	Class Marks	Frequency	Points to be plotted
1.	20 – 29	19.5 – 29.5	24.5	2	(24.5, 2)
2.	30 – 39	29.5 – 39.5	34.5	5	(34.5, 5)
3.	40 – 49	39.5 – 49.5	44.5	3	(44.5, 3)
4.	50 – 59	49.5 – 59.5	54.5	18	(54.5, 18)
5.	60 – 69	59.5 – 69.5	64.5	20	(64.5, 20)
6.	70 – 79	69.5 – 79.5	74.5	22	(74.5, 22)
7.	80 – 89	79.5 – 89.5	84.5	12	(84.5, 12)
8.	90 – 99	89.5 – 99.5	94.5	7	(94.5, 7)

Plotting these points and joining them by line segments we get frequency polygon. We complete the frequency polygon by extending it up to points (14.5, 0) and (104.5, 0) adjacent to first and last class marks on X-axis.

Pie Chart

In a pie-chart, various observations or components are represented by the sectors of a circle and the whole circle represents the sum of the values of all the components. Clearly, the total angle of 360o at the center of  the circle is divided according  to the values of the components. Thus we have, Central Angle for Component

= $\left(\frac{\text{ Value of the Component }}{\text{ Total Value }}\times 360\right)$

Sometimes, the value of components are expressed in percentages. In such cases, we have: Central Angle for Component = $\left(\frac{\text{ Percentage Value of the Component }}{100}\times 360\right)$

Illustration -9

A survey was conducted among a group of girls to know their preferences for the types of slippers. The given circle graph shows the results of the survey.

If the survey was conducted among a group of 1080 girls, then find out how many girls preferred to wear leather slippers?

(A) 140          (B) 280          (C) 420          (D) 560

Solution

C

In the given circle graph, it is seen that the central angle of the sector corresponding to leather slippers is 140°.

Therefore, number of girls who preferred to wear leather slippers:

$=\frac{140°}{360°}\times 1080=420$

Thus, 420 girls preferred to wear leather slippers.

6. MEASURES OF CENTRAL TENDENCY

One of the most  important objectives of statistical analysis is to get one single value that describes the characteristic of the entire data.  Such a value is called the central value or an average. The following are the important types of averages:

1. Arithmetic Mean

2. Geometric Mean

3. Harmonic mean

4. Median

5. Mode

We consider these measures in three cases (i) Individual series (i.e. each individual observation is given) (ii) discrete series (i.e., the observations along with number of times a particular observation called the frequency is given) (iii) continuous series (i.e. the class intervals along with their frequencies are given)

2. Arithmetic Mean

The average of numbers in arithmetic is known as the Arithmetic Mean of these numbers in statistics.

1. MEAN OF AN UNGROUPED DATA

The Arithmetic Mean or simply the Mean of n observations x₁, x₂, x₃, …….x_n is given by the formula:

Mean = $\frac{\left({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+\dots \dots {\mathrm{x}}_{\mathrm{n}}\right)}{\mathrm{n}}=\frac{\sum {\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}$

Where the symbol $\Sigma$, called sigma stands for the summation of the terms.

Illustration -10

Calculate the mean of the following numbers: 3, 1, 5, 6, 3, 4, 5, 3, 7, 2

Solution

Sum of the given numbers = (3 + 1 + 5 + 6 + 3 + 4 + 5 + 3 + 7 + 2) = 39.

Number of these numbers = 10.

$\therefore$ Mean of the given numbers = $\frac{39}{10}=3.9$.

Illustration -11

The weights (in kg) of 5 persons in a group are: 55, 63, 48, 59 and 61. Find their mean weight.

Solution

Sum of the weight of all persons = (55 + 63 + 48 + 59 + 61) kg = 286 kg.

Number of persons = 5. 

$\therefore$ Mean weight = $\left(\frac{286}{5}\right)$ kg = 57.2 kg.

Some Useful Result: Let the mean of x₁ x₂ x₃, ……..x_n be A. Then

(i) Mean of (x₁ + k), (x₂ + k), (x₃+ k) ………..(x_n + k) is (A + k);

(ii) Mean of (x₁ – k), (x₂ – k), (x₃ – k)………..(x_n – k) is (A – k);

(iii) Mean of kx₁, kx₂, kx₃ …….. kx_n is kA, where k $\ne$ 0.

2. MEAN OF GROUPED DATA

Direct Method

When the variates x₁, x₂, x₃, …………x_n have frequencies f₁, f₂, f₃ ……..f_n respectively, then the mean is given by the formula:

Mean = $\frac{\left(f_1{x}_1+f_2{x}_2+f_3{x}_3+\dots \dots ..+f_n{x}_n\right)}{\left(f_1+f_2+f_3+\dots \dots \dots .+f_n\right)}=\frac{\sum f_i{x}_i}{\sum f_i}$.

Illustration -12

The following table shows the weights of 15 members of an athletic team in a school.

Weight (in kg)	42	45	46	48	49
Number of athletes	4	3	5	2	1

Find the mean weight.

Solution

From the above data, we may prepare the table given below.

Weight (in kg) xi	Number of athletes (Frequency) $f_1$	$f_1{x}_i$
42 45 46 48 49	4 3 5 2 1	168 135 230 96 49
	${\mathrm{\Sigma f}}_{\mathrm{I}}$= 15	$\Sigma f_i{x}_i$ = 678

  $\therefore$Mean Weight =$\frac{\sum f_i{x}_i}{\sum f_i}=\frac{678}{15}=45.2 \mathrm{kg}$.

Illustration -13

Using short cut method, calculate the mean weekly wage from the following frequency distribution.

Weekly wages(in Rs)	950	1000	1050	1100	1250	1500	1600
Number of workers	24	18	13	15	20	11	9

Solution

Let the assumed mean be, A = 1100.

From the given data, we may prepare the table given below.

Weekly Wages (in Rs) x1	Numbers of Workers (Frequency)	di = (xi – A)     = (xi – 1100)	fidi
950 1000 1050 1100 = A 1250 1500 1600	24 18 13 15 20 11 9	–150 –100 –50 0 150 400 500	–3600 –1800 –650 0 3000 4400 4500
	$\Sigma$fI = 110		$\Sigma$fI di = 5850

$\therefore$ Mean = $\left(\mathrm{A}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)=\left(1100+\frac{5850}{110}\right)=1153.18$

Hence, Mean Wage = Rs 1153.18.

3. MEAN OF GROUPED DATA IN THE FORM OF CLASSES

I. Direct Method

Step 1: For each class, find the class mark x_i by using the relation, x_i = $\frac{1}{2}$ (lower limit + upper limit). Step 2: Use the formula, Mean = $\frac{\sum f_i{X}_i}{\sum f_i}$.

II. Short Cut Method or Deviation Method

Step 1: For each class, find the class mark x_i.

Step 2: Let A be the assumed mean.

Step 3: Find d_i = (x_i – A).

Step 4: Use the formula, Mean = $\left(\mathrm{A}\frac{\sum f_{\mathrm{i}}\mathrm{d}}{\Sigma f_{\mathrm{i}}}\right)$.

III. Step-Deviation Method

Step 1. For each class, find the class mark x₁.

Step 2: Let A be the assumed mean.

Step 3: Calculate, ${\mu }_{\mathrm{I}}=\frac{\left({\mathrm{X}}_{\mathrm{i}}–\mathrm{A}\right)}{\mathrm{C}}$, where c is the class size.

Step 4: Use the formula, Mean = $\left(\mathrm{A}+\mathrm{C}\frac{\sum f_{\mathrm{i}}{\mu }_{\mathrm{i}}}{\sum f_{\mathrm{i}}}\right)$.

Illustration -14

Using direct method, find the mean of the following frequency distribution:

Class-interval	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	6	8	12	15	10	9

Solution

From the given data, we may prepare the table given below.

Class-Interval	Class-Mark xi	Frequency fi	fixi
10-20 20-30 30-40 40-50 50-60 60-70	15 25 35 45 55 65	6 8 12 15 10 9	90 200 420 675 550 585
		$\Sigma$fI = 60	$\Sigma$fixi = 2520

$\therefore$ Mean = $\frac{\sum f_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum f_{\mathrm{i}}}=\frac{2520}{60}=42$.

Illustration -15

Using Short Cut Method find the mean for the following frequency distribution:

Class-interval	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	8	10	16	23	12	11

Solution

From the given data, we may prepare the table given below.

Class-Interval	Class-Mark xi	Frequency fi	di = (xi – A)    = (xi – 99)	fidi
84-90 90-96 96-102 102-108 108-114 114-120	87 93 99 = A 105 111 117	8 10 16 23 12 11	–12 –6 0 6 12 18	–96 –60 0 138 144 198
		$\Sigma$fI = 80		$\Sigma$fidi = 324

$\therefore$ Mean = $\frac{\Sigma f_i{d}_i}{\sum f_i}=\left(99+\frac{324}{80}\right)=(99+4.05)=103.05$

Hence, Mean = 103.05.

Illustration -16

Using Step Deviation Method, calculate the mean for the following data:

Height (in cm)	135-140	140-145	145-150	150-155	155-160	160-165	165-170	170-175
Frequency	4	9	18	28	24	10	5	2

Solution

Here, class size, c = 5. Take assumed mean, A = 152.5. Thus, from the given data, we may prepare

Class-Interval	Class-Mark   xi	Frequency  fi	$\left({\mathrm{\mu }}_{\mathrm{i}}=\frac{{\mathrm{X}}_{\mathrm{i}}–\mathrm{A}}{\mathrm{C}}\right)$	fi$\mu$i
135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175-	137.5 142.5 147.5 152.5 = A 157.5 162.5 167.5 172.5	4 9 18 28 24 10 5 2	–3 –2 –1 0 1 2 3 4	–12 –18 –18 0 24 20 15 8
		$\Sigma$fI = 100		$\Sigma$fi$\mu$i = 19

$\therefore$ Mean = $\left(\mathrm{A}+\mathrm{c}\times \frac{\sum f_{\mathrm{i}}{\mu }_{\mathrm{i}}}{\Sigma f_{\mathrm{i}}}\right)=\left(152.5+\frac{5\times 19}{100}\right)$ = (152.5 + 0.95) = 153.45 cm.

4. IMPORTANT FORMULAE FOR SOLVING ARITHMETIC MEAN

1. Arithmetic Mean of Individual Series

If x₁, x₂ , x₃,….., x_n are n values of variant x, then its Arithmetic Mean, denoted by x is

$\text{ A.M. }\left(\overline{x}\right)=\frac{x_1+x_2+x_3+\dots ..+x_n}{n}=\frac{\sum x_i}{n}$   (or)

$\text{ A.M. }\left(\overline{x}\right)=A+\frac{\sum \left(x_i–A\right)}{n}$ where A is the assumed average. (For individual series)

2. Arithmetic Mean of Discreate Series

If a variable takes values x₁, x₂ , x₃,….., x_n with corresponding frequencies f₁, f₂ , f₃,….., f_n then the arithmetic mean $\overline{x}$ is given by

$\overline{x}=\frac{f_1{x}_1+f_2{x}_2\dots .+f_n{x}_n}{f_1+f_2\dots .+f_n}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i$

where $N=\sum _{i=1}^n{f}_i$

3. Arithmetic Mean Of Continuous Series

In case of a set of data with class intervals, we cannot find the exact value of the mean because we do not know the exact values of the variables. We, therefore, try to obtain an approximate value of the mean. The method of approximate is to replace all the observed values belonging to a class by mid-value of the class.
If x₁, x₂ , x₃,….., x_n are the mid values of the class intervals having corresponding frequencies f₁, f₂ , f₃,….., f_n then we apply the same formula as in discrete series.

$\overline{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i, N=\sum _{i=1}^n{f}_i$

4. Combined Arithmetic Mean

If ${\overline{x}}_i(i=1,2,\dots .,k)$ are the means of k – series of sizes $n_i(i=1,2,3,\dots \dots \dots ..,k)$ then the combined or composite mean $\overline{x}$ can be obtained by the formula :

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2+\dots .+n_k{\overline{x}}_k}{n_1+n_2+\dots ..+n_k}=\frac{\sum n_i{\overline{x}}_i}{\sum n_i}$

5. Weighted Arithmetic Mean

If w₁, w₂ , w₃,….., w_n be the weights assigned to the values x₁, x₂ , x₃,….., x_n be respectively of a variable x, then the weighted A.M. is $\overline{x}=\frac{\sum w_i{x}_i}{\sum w_i}$.

5. PROPERTIES OF ARITHMETIC MEAN

1. Sum of all the deviations from arithmetic mean is zero i.e.,

$\sum _{i=1}^n\left(x_i–\overline{x}\right)=0$ (in case of individual series)

$\sum _{i=1}^n{f}_i\left(x_i–\overline{x}\right)=0$ (in case of discrete or continuous series)

2. If each observation is increased or decreased by a given constant K, the mean is also increased or decreased by K
The property is also known as effect of change of origin. K can be taken to be any number. However, to simplify the calculations, K should be taken as a value which is in the middle of the table.
3. Step Deviation Method or change of scale

If x₁, x₂ , x₃,….., x_n are mid values of class intervals with corresponding frequencies f₁, f₂ , f₃,….., f_n then we may change the scale by taking $d_i=\frac{x_i–A}{h}$ in this case.

$\overline{x}=A+h\times \left(\frac{1}{N}\sum f_i{d}_i\right)$ (if A is assumed mean)

A and h can be any numbers but if the lengths of class intervals are equal then h may be taken as width of the class interval.
In particular if each observation is multiplied or divided by a constant, the mean is also multiplied or divided by the same constant.
4. The sum of the squared deviation of the variate from their mean is minimum i.e., the quantity $\sum {\left(x_i–A\right)}^2\text{ or }\sum f_i{\left(x_i–A\right)}^2$ is minimum when $A=\overline{x}$.

5. E(aX + b) = aE(X) + b (where E(X) = Mean of X)

Illustration-17
The weighted mean of the first n natural numbers, the weights being the corresponding numbers, is
Solution
First n natural numbers are 1, 2, 3,…,n; whose corresponding weights are 1, 2, 3,…,n respectively.

$\therefore$ weight mean = $\frac{1\times 1+2\times 2+\dots .+n\times n}{1+2+\dots +n}$

$=\frac{1^2+2^2+\dots ..+n^2}{1+2+\dots ..+n}$

$=\frac{n(n+1)(2n+1)}{\frac{6n(n+1)}{2}}=\frac{2n+1}{3}$

Illustration-18

The weighted mean of the first n natural numbers whose weights are equal to the squares of the corresponding numbers is
Solution

weighted mean = $\frac{1.1^2+2.2^2+\dots +n·n^2}{1^2+2^2+\dots +n^2}$

$=\frac{{\mathrm{\Sigma n}}^3}{{\mathrm{\Sigma n}}^2}=\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}}=\frac{3\mathrm{n}(\mathrm{n}+1)}{2(2\mathrm{n}+1)}$

Illustration-19
The average salary of male employees in a firm is Rs. 5200 and that of females is Rs.4200. The mean salary of all the employees is Rs.5000. The percentage of male and female employees are respectively is
Solution

Let $x_1=5200, x_2=4200, \overline{x}=5000$

Also, we know that $\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$

$\Rightarrow 5000\left(n_1+n_2\right)=5200n_1+4200n_2\Rightarrow \frac{n_1}{n_2}=\frac{4}{1}$

$\therefore$ The percentage of male employees in the firm = $\frac{4}{4+1}\times 100=80\%$

and the percentage of female employees in the firm = $\frac{1}{4+1}\times 100=20\%$

Illustration-20
If the mean of 9 observations is 100 and mean of 6 observations is 80, then the mean of 15 observations is
Solution

$n_1=9, {\overline{x}}_1=100\text{ and }{n}_2=6, {\overline{x}}_2=80$

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}=\frac{9\times 100+6\times 80}{9+6}=92$

1. BASICS OF STATISTICS

1. STATISTICAL DATA

Statistical data are the facts which are collected for the purpose of investigation. There are two types of statistical data:

(i) Primary data: The data collected by an investigator for the first time for his own purpose are called primary data. As the primary data are collected by the user of the data, so it is more reliable and relevant.

(ii) Secondary data: The data collected by a secondary source and used by the investigator for his purpose is called secondary data. For example score of a cricket match noted from newspapers is secondary data. Thus data which are primary in the hands of one become secondary in the hands of the other. Data collected by any source also can be divided in following two types:

(i) Raw Data: Raw data are those data which are obtained from the original source but not arranged numerically. This is also called ‘ungrouped data’ for example marks of 10 students in maths are given as:  75, 96, 25, 32, 89, 62, 40, 79, 35, 55 An ‘array’ is an arrangement of raw numerical data in the ascending or descending order of magnitude. Above data can be written as  25, 32, 35, 40, 55, 62, 75, 79, 89, 96

(ii) Grouped data: An array can be placed systematically in groups or categories. For example the above data can be grouped in following manner.

GROUPS	MARKS	TOTAL NUMBER OF STUDENTS
0 to 20	–	0
21 to 40	25, 32, 35, 40	4
41 to 60	55	1
61 to 80	62, 75, 79	3
81 to 100	89, 96	2
TOTAL		10

 2. SOME BASIC DEFINITIONS

(i) Variate: Variate is a quantity that may vary from observation to observation.

(ii) Range: Range is difference between the maximum and minimum observations.

(iii) Class Interval: When data are divided in groups, each group is called a class interval.

(iv) Class Limit: Every class interval has two limits. The smallest observation of the interval is called lower limit and the largest observation of the interval is called upper limit.

(v) Class Mark: The mid value of any class is called its class mark.

$\text{ Class Mark }=\frac{\text{ Upper limit of the class + lower limit of the class }}{2}$

(vi) Class Size: Class size is defined as the difference between two successive class marks. It is also the difference between the upper and lower limits of any class interval.

(vii) Frequency: In a particular class the count of the number of observation is called its frequency. So the corresponding frequency of a class is called its class frequency.

(viii) Cumulative Frequency: The cumulative frequency of any class is obtained by adding all the frequencies successively prior to that class i.e. it is the sum of all frequencies up to that class.

(ix) True Class Limit: In the case of exclusive classes the upper and lower limits are respectively known as its true upper limits and true lower limits. In the case of inclusive classes, the true lower and upper limits are obtained by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. True upper limits and true lower limits are also known as boundaries of the class.

(x) Tally: Tally method is used to keep the chance of error at minimum in counting. A bar (|) called tally mark is put against any item when it occurs. The fifth occurrence of any item is represented by putting diagonally a cross tally (\) on the first four tallies.

3. FREQUENCY DISTRIBUTION

The tabular arrangement of data showing the frequency of each item is called a frequency distribution table. It is a method to present raw data in the form from which one can easily understand the information contained in the raw data. Frequency distribution are of two types:

i. Discrete frequency distribution:

In this type of frequency distribution, in the first column of frequency table we write all possible values of the variables from the lowest to the highest, in the second column we write tally marks and in the third column we show frequency of each item. In this method data are not divided into groups or classes.

e.g. no. of girls in 20 families is given in following data:

1, 2, 3, 1, 1, 2, 3, 3, 4, 1, 5, 1, 1, 2, 2, 3, 3, 2, 4, 1

The above data can be put in the form of a discrete frequency distribution table in the following manner:

S. No.	No. of girls	Tally Marks	Frequency
1.	1	$\bcancel{\left|\right| \left|\right|} \left|\right|$	7
2.	2	$\bcancel{\left|\right|\left|\right|}$	5
3.	3	$\bcancel{\left|\right|\left|\right|}$	5
4.	4	||	2
5.	5	|	1
TOTAL			20

ii. Continuous or Grouped Frequency Distribution

In the frequency distribution data are divided  into groups or classes. This method is used only where the values in the raw data are largely repeating and the difference between the greatest and the smallest observations is not very large.

4. CUMULATIVE FREQUENCY

Cumulative frequency table is obtained from the ordinary frequency table by successively adding the several frequencies. Thus to form a cumulative frequency table we add a column of cumulative frequency in the frequency distribution table. It is obvious that the cumulative frequency of the last class is the sum of the frequencies of all the classes.

Cumulative frequency series are of two types:

(i) Less than series        (ii) More than series

Illustration -1

The weekly saving of 30 workers working in a factory is as given below:

64, 60, 87, 75, 69, 34, 51, 78, 39, 48, 73, 54, 63, 70, 57, 88, 90, 53, 74, 44, 31, 71, 68, 72, 36, 89, 55, 67, 73, 83

(a) Taking first class interval as 30 – 40 (40 not included), form a frequency table of equal intervals.

(b) Also form a cumulative frequency table.

(c) Find the number of workers whose weekly saving is more than Rs. 60.

Solution

(i) For the given data of the weekly saving of 30 workers working in a factory, we prepare             following table:

S. No.	Class Interval (Saving in Rs.)	Tally Marks	Frequency (No. of workers)
1.	30 – 40	| | | |	4
2.	40 – 50	| |	2
3.	50 – 60	$\bcancel{\left|\right|\left|\right|}$	5
4.	60 – 70	$\bcancel{\left|\right|\left|\right|} |$	6
5.	70 – 80	$\bcancel{\left|\right|\left|\right|} \left|\right||$	8
6.	80 – 90	| | | |	4
7.	90 – 100	|	1
		TOTAL	30

(ii) For given data cumulative frequency table can be given in the following manner:

S. No.	Class Interval (Saving in Rs.)	Tally Marks	Frequency (No. of workers)	Cumulative Frequency
1.	30 – 40	| | | |	4	4
2.	40 – 50	| |	2	6
3.	50 – 60	$\bcancel{\left|\right|\left|\right|}$	5	11
4.	60 – 70	$\bcancel{\left|\right|\left|\right|} |$	6	17
5.	70 – 80	$\bcancel{\left|\right|\left|\right|} \left|\right||$	8	25
6.	80 – 90	| | | |	4	29
7.	90 – 100	|	1	30
		TOTAL	30

(iii) Number of workers whose weekly saving is Rs. 60 or more than Rs. 60  = (6 + 8 + 4 + 1) = 19

Illustration -2

Heights of seven persons in cm are 120, 125, 142, 134, 150, 155 and 128. Calculate the range.

Solution

For given data maximum height = 155 cm

and minimum height = 120 cm

Range  = maximum height – minimum height

= 155 cm – 120 cm

= 35 cm.

Illustration -3

Form a frequency table from the following table:

Marks	No. of Students
Below 10	15
Below 20	35
Below 30	60
Below 40	84
Below 50	96
Below 60	127
Below 70	198
Below 80	250

Solution

For given data we make classes 0 – 10, 10 – 20, 20 – 30, ….., 70–80.

There are 15 students who obtained below 10 marks, therefore frequency of class 0 – 10 is 15. Again number of students getting below 20 marks is 35.

This includes those students also who obtained below 10 marks.

Number of students who got marks between 10 and 20 i.e. frequency of class

10 – 20  =  35–15 = 20

Thus for given data we make following table:

S. No.	Class Interval	Comparative	Frequency
1.	0 – 10	15	15
2.	10 – 20	35	20
3.	20 – 30	60	25
4.	30 – 40	84	24
5.	40 – 50	96	12
6.	50 – 60	127	31
7.	60 – 70	198	71
8.	70 – 80	250	52
		TOTAL	250

Illustration -4

Find the unknown entries (a, b, c, d, e, f, g) from the following frequency distribution of heights of 60 students in a class:

Height (in cm)	Frequency	Cumulative frequency
160 – 165	15	a
165 – 170	b	35
170 – 175	12	c
175 – 180	e	50
180 – 185	d	55
185 – 190	5	f
Total	g

Solution

From given table, we make following cumulative frequency table:

S. No.	Height (in cm)	Frequency	Cumulative Frequency
1.	160 – 165	15	15  = a
2.	165 – 170	b	15 + b = 35
3.	170 – 175	12	15 + b + 12 = c
4.	175 – 180	d	15 + b + 12 + d = 50
5.	180 – 185	e	15 + b + 12 + b + e = 55
6.	185 – 190	5	15 + b + 12 + d + e + 5 = f
		g	60

Now, from table:

a = 15

15 + b = 35    b = 35 – 15 = 20

15 + b + 12 = c   15 + 20 + 12 = c

c = 47

15 + b + 12 + d = 50

15 + 20 + 12 + d = 50

d = 50 – 47

d = 3

15 + b + 12 + d + c = 55

15 + 20 + 12 + 3 + e = 55

e = 55 – 50

e = 5

15 + b + 12 + d + e + 5 = f

15 + 20 + 12 + 3 + 5 + 5 = f

f = 60

g = 60. Hence, a = 15, b = 20, c = 47, d = 3, e = 5, f = 60 and g = 60.

5. GRAPHICAL REPRESENTATION OF DATA

A given data can be represented in graphical way. There are various methods of graphical representation of frequency distribution.

(i) Bar Graphs

(ii) Histogram

(iii) Frequency Polygon

(iv) Pie Chart

Bar Graph

The frequency distribution of a discrete value is best represented by a bar graph. The height of the bars is proportional to the frequency of each variate-value. In a bar graph the bars must be kept distinct to show that the variate-values are distinct. The bars are of equal width and are drawn with equal spacing between them on the x-axis depicting the variable. The values of the variable are shown on the y-axis.

Illustration -5

The following table shows the number of illiterate persons in the age group (10–58 years) in a town:

Age Group (in years)	10 – 16	17 – 23	24 – 30	31 – 37	38 – 44	45 – 51	52 – 58
Number of illiterate person	175	375	100	150	250	400	525

Solution

Draw a bar graph to represent the above data.

Histogram

Histogram is a graphical representation of a grouped frequency distribution with continuous classes. It consists of a set of rectangles where heights of rectangles are proportional to their class frequencies, for equal class intervals. There is no gap between two successive rectangles. The rectangles are constructed with base as the class size and their heights representing the frequencies.

Illustration -6

Construct a histogram from the following distribution of total marks obtained by 750 students of class IX in the final examination.

Marks (mid-points)	620	660	700	740	780	820	860
Number of students	16	45	156	284	172	59	18

Solution

To draw the histogram for above data first of all we find out class limits for making class intervals from given class marks.

Here difference between second and the first class marks =  h  =  660 – 620 = 40

$\therefore \frac{h}{2}=\frac{40}{2}=20$

lower limit of first class interval =  620 – h/2

= 620 – 20

= 600

And upper limit of the first class interval = $620+\frac{h}{2}$ = 620 + 20 = 640

Similarly other class intervals would be,

(660 – 20) – (660 + 20), (700 – 20) – (700 + 20), (740 – 20) – (740 + 20), (780 – 20) – (780 + 20), (820 – 20) – (820 + 20) and (860 – 20) – (860 – 20)

i.e. 640 – 680, 680 – 720, 720 – 760, 760 – 800, 800 – 840 and 840 – 880 Using these class intervals we draw following histogram taking class intervals on X-axis and number of students on Y-axis

Frequency Polygon

A frequency polygon is a graph of frequency distribution. It is a line graph of class frequency which is plotted against class mark. A frequency polygon can be obtained by two methods:

(1) By using Histogram: A frequency polygon can be obtained by joining mid points of the top of the rectangles of a histogram. For this we obtain the mid points of the upper horizontal sides of each rectangle and then join these mid points by dotted lines to get frequency polygon. End of a frequency polygon preferably extended to the mid points of imagined class intervals adjacent to first and last class intervals.

Illustration -7

Draw the histogram and frequency polygon of the following frequency distribution:

Monthly wages (in rupees)	325-350	350-375	375-400	400-425	425-450	Total
Number of workers	30	45	75	60	55	245

Solution

Using given data we draw histogram taking monthly wages (in rupees) on X-axis and number of workers on Y-axis. Then we join the mid points of the top of the rectangles by dotted straight lines and complete it by joining the mid points of imagined class intervals adjacent to the first and last class intervals.

(2) Frequency polygon without using Histogram: Following procedure is used to make a frequency polygon without using histogram.

(i) Calculate the class marks, x₁, x₂, …., x_n of each of the given class intervals.

(ii) Mark class marks x₁, x₂, …., x_n, along X-axis and frequencies f₁, f₂, …. f_n along Y-axis.

(iii) Plot the points (x₁, f₁), (x₂, f₂), ,….., (x_n, f_n).

(iv) Obtain the mid-points of two class intervals of zero frequencies at the beginning of the first interval and at the end of the last interval.

(v) Join the points (x₁, f₁), (x₂, f₂), …, (x_n, f_n) by the line segments and complete the frequency polygon by joining the mid points of the first and last intervals to the mid points of the imagined classes adjacent to them.

Illustration -8

Represent the following distribution by a frequency polygon without using a histogram:

Scores	20-29	30-39	40-49	50-59	60-69	70-79	80-89	90-99
Frequency	2	5	3	18	20	22	12	7

Solution

Given data is in the form of inclusive classes, so first of all we shall convert them into exclusive classes. Also we have to get class marks of these class intervals for this purpose.

S. No.	Scores	True Class Limits	Class Marks	Frequency	Points to be plotted
1.	20 – 29	19.5 – 29.5	24.5	2	(24.5, 2)
2.	30 – 39	29.5 – 39.5	34.5	5	(34.5, 5)
3.	40 – 49	39.5 – 49.5	44.5	3	(44.5, 3)
4.	50 – 59	49.5 – 59.5	54.5	18	(54.5, 18)
5.	60 – 69	59.5 – 69.5	64.5	20	(64.5, 20)
6.	70 – 79	69.5 – 79.5	74.5	22	(74.5, 22)
7.	80 – 89	79.5 – 89.5	84.5	12	(84.5, 12)
8.	90 – 99	89.5 – 99.5	94.5	7	(94.5, 7)

Plotting these points and joining them by line segments we get frequency polygon. We complete the frequency polygon by extending it up to points (14.5, 0) and (104.5, 0) adjacent to first and last class marks on X-axis.

Pie Chart

In a pie-chart, various observations or components are represented by the sectors of a circle and the whole circle represents the sum of the values of all the components. Clearly, the total angle of 360o at the center of  the circle is divided according  to the values of the components. Thus we have, Central Angle for Component

= $\left(\frac{\text{ Value of the Component }}{\text{ Total Value }}\times 360\right)$

Sometimes, the value of components are expressed in percentages. In such cases, we have: Central Angle for Component = $\left(\frac{\text{ Percentage Value of the Component }}{100}\times 360\right)$

Illustration -9

A survey was conducted among a group of girls to know their preferences for the types of slippers. The given circle graph shows the results of the survey.

If the survey was conducted among a group of 1080 girls, then find out how many girls preferred to wear leather slippers?

(A) 140          (B) 280          (C) 420          (D) 560

Solution

C

In the given circle graph, it is seen that the central angle of the sector corresponding to leather slippers is 140°.

Therefore, number of girls who preferred to wear leather slippers:

$=\frac{140°}{360°}\times 1080=420$

Thus, 420 girls preferred to wear leather slippers.

6. MEASURES OF CENTRAL TENDENCY

One of the most  important objectives of statistical analysis is to get one single value that describes the characteristic of the entire data.  Such a value is called the central value or an average. The following are the important types of averages:

1. Arithmetic Mean

2. Geometric Mean

3. Harmonic mean

4. Median

5. Mode

We consider these measures in three cases (i) Individual series (i.e. each individual observation is given) (ii) discrete series (i.e., the observations along with number of times a particular observation called the frequency is given) (iii) continuous series (i.e. the class intervals along with their frequencies are given)

2. Arithmetic Mean

The average of numbers in arithmetic is known as the Arithmetic Mean of these numbers in statistics.

1. MEAN OF AN UNGROUPED DATA

The Arithmetic Mean or simply the Mean of n observations x₁, x₂, x₃, …….x_n is given by the formula:

Mean = $\frac{\left({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+\dots \dots {\mathrm{x}}_{\mathrm{n}}\right)}{\mathrm{n}}=\frac{\sum {\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}$

Where the symbol $\Sigma$, called sigma stands for the summation of the terms.

Illustration -10

Calculate the mean of the following numbers: 3, 1, 5, 6, 3, 4, 5, 3, 7, 2

Solution

Sum of the given numbers = (3 + 1 + 5 + 6 + 3 + 4 + 5 + 3 + 7 + 2) = 39.

Number of these numbers = 10.

$\therefore$ Mean of the given numbers = $\frac{39}{10}=3.9$.

Illustration -11

The weights (in kg) of 5 persons in a group are: 55, 63, 48, 59 and 61. Find their mean weight.

Solution

Sum of the weight of all persons = (55 + 63 + 48 + 59 + 61) kg = 286 kg.

Number of persons = 5. 

$\therefore$ Mean weight = $\left(\frac{286}{5}\right)$ kg = 57.2 kg.

Some Useful Result: Let the mean of x₁ x₂ x₃, ……..x_n be A. Then

(i) Mean of (x₁ + k), (x₂ + k), (x₃+ k) ………..(x_n + k) is (A + k);

(ii) Mean of (x₁ – k), (x₂ – k), (x₃ – k)………..(x_n – k) is (A – k);

(iii) Mean of kx₁, kx₂, kx₃ …….. kx_n is kA, where k $\ne$ 0.

2. MEAN OF GROUPED DATA

Direct Method

When the variates x₁, x₂, x₃, …………x_n have frequencies f₁, f₂, f₃ ……..f_n respectively, then the mean is given by the formula:

Mean = $\frac{\left(f_1{x}_1+f_2{x}_2+f_3{x}_3+\dots \dots ..+f_n{x}_n\right)}{\left(f_1+f_2+f_3+\dots \dots \dots .+f_n\right)}=\frac{\sum f_i{x}_i}{\sum f_i}$.

Illustration -12

The following table shows the weights of 15 members of an athletic team in a school.

Weight (in kg)	42	45	46	48	49
Number of athletes	4	3	5	2	1

Find the mean weight.

Solution

From the above data, we may prepare the table given below.

Weight (in kg) xi	Number of athletes (Frequency) $f_1$	$f_1{x}_i$
42 45 46 48 49	4 3 5 2 1	168 135 230 96 49
	${\mathrm{\Sigma f}}_{\mathrm{I}}$= 15	$\Sigma f_i{x}_i$ = 678

  $\therefore$Mean Weight =$\frac{\sum f_i{x}_i}{\sum f_i}=\frac{678}{15}=45.2 \mathrm{kg}$.

Illustration -13

Using short cut method, calculate the mean weekly wage from the following frequency distribution.

Weekly wages(in Rs)	950	1000	1050	1100	1250	1500	1600
Number of workers	24	18	13	15	20	11	9

Solution

Let the assumed mean be, A = 1100.

From the given data, we may prepare the table given below.

Weekly Wages (in Rs) x1	Numbers of Workers (Frequency)	di = (xi – A)     = (xi – 1100)	fidi
950 1000 1050 1100 = A 1250 1500 1600	24 18 13 15 20 11 9	–150 –100 –50 0 150 400 500	–3600 –1800 –650 0 3000 4400 4500
	$\Sigma$fI = 110		$\Sigma$fI di = 5850

$\therefore$ Mean = $\left(\mathrm{A}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)=\left(1100+\frac{5850}{110}\right)=1153.18$

Hence, Mean Wage = Rs 1153.18.

3. MEAN OF GROUPED DATA IN THE FORM OF CLASSES

I. Direct Method

Step 1: For each class, find the class mark x_i by using the relation, x_i = $\frac{1}{2}$ (lower limit + upper limit). Step 2: Use the formula, Mean = $\frac{\sum f_i{X}_i}{\sum f_i}$.

II. Short Cut Method or Deviation Method

Step 1: For each class, find the class mark x_i.

Step 2: Let A be the assumed mean.

Step 3: Find d_i = (x_i – A).

Step 4: Use the formula, Mean = $\left(\mathrm{A}\frac{\sum f_{\mathrm{i}}\mathrm{d}}{\Sigma f_{\mathrm{i}}}\right)$.

III. Step-Deviation Method

Step 1. For each class, find the class mark x₁.

Step 2: Let A be the assumed mean.

Step 3: Calculate, ${\mu }_{\mathrm{I}}=\frac{\left({\mathrm{X}}_{\mathrm{i}}–\mathrm{A}\right)}{\mathrm{C}}$, where c is the class size.

Step 4: Use the formula, Mean = $\left(\mathrm{A}+\mathrm{C}\frac{\sum f_{\mathrm{i}}{\mu }_{\mathrm{i}}}{\sum f_{\mathrm{i}}}\right)$.

Illustration -14

Using direct method, find the mean of the following frequency distribution:

Class-interval	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	6	8	12	15	10	9

Solution

From the given data, we may prepare the table given below.

Class-Interval	Class-Mark xi	Frequency fi	fixi
10-20 20-30 30-40 40-50 50-60 60-70	15 25 35 45 55 65	6 8 12 15 10 9	90 200 420 675 550 585
		$\Sigma$fI = 60	$\Sigma$fixi = 2520

$\therefore$ Mean = $\frac{\sum f_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum f_{\mathrm{i}}}=\frac{2520}{60}=42$.

Illustration -15

Using Short Cut Method find the mean for the following frequency distribution:

Class-interval	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	8	10	16	23	12	11

Solution

From the given data, we may prepare the table given below.

Class-Interval	Class-Mark xi	Frequency fi	di = (xi – A)    = (xi – 99)	fidi
84-90 90-96 96-102 102-108 108-114 114-120	87 93 99 = A 105 111 117	8 10 16 23 12 11	–12 –6 0 6 12 18	–96 –60 0 138 144 198
		$\Sigma$fI = 80		$\Sigma$fidi = 324

$\therefore$ Mean = $\frac{\Sigma f_i{d}_i}{\sum f_i}=\left(99+\frac{324}{80}\right)=(99+4.05)=103.05$

Hence, Mean = 103.05.

Illustration -16

Using Step Deviation Method, calculate the mean for the following data:

Height (in cm)	135-140	140-145	145-150	150-155	155-160	160-165	165-170	170-175
Frequency	4	9	18	28	24	10	5	2

Solution

Here, class size, c = 5. Take assumed mean, A = 152.5. Thus, from the given data, we may prepare

Class-Interval	Class-Mark   xi	Frequency  fi	$\left({\mathrm{\mu }}_{\mathrm{i}}=\frac{{\mathrm{X}}_{\mathrm{i}}–\mathrm{A}}{\mathrm{C}}\right)$	fi$\mu$i
135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175-	137.5 142.5 147.5 152.5 = A 157.5 162.5 167.5 172.5	4 9 18 28 24 10 5 2	–3 –2 –1 0 1 2 3 4	–12 –18 –18 0 24 20 15 8
		$\Sigma$fI = 100		$\Sigma$fi$\mu$i = 19

$\therefore$ Mean = $\left(\mathrm{A}+\mathrm{c}\times \frac{\sum f_{\mathrm{i}}{\mu }_{\mathrm{i}}}{\Sigma f_{\mathrm{i}}}\right)=\left(152.5+\frac{5\times 19}{100}\right)$ = (152.5 + 0.95) = 153.45 cm.

4. IMPORTANT FORMULAE FOR SOLVING ARITHMETIC MEAN

1. Arithmetic Mean of Individual Series

If x₁, x₂ , x₃,….., x_n are n values of variant x, then its Arithmetic Mean, denoted by x is

$\text{ A.M. }\left(\overline{x}\right)=\frac{x_1+x_2+x_3+\dots ..+x_n}{n}=\frac{\sum x_i}{n}$   (or)

$\text{ A.M. }\left(\overline{x}\right)=A+\frac{\sum \left(x_i–A\right)}{n}$ where A is the assumed average. (For individual series)

2. Arithmetic Mean of Discreate Series

If a variable takes values x₁, x₂ , x₃,….., x_n with corresponding frequencies f₁, f₂ , f₃,….., f_n then the arithmetic mean $\overline{x}$ is given by

$\overline{x}=\frac{f_1{x}_1+f_2{x}_2\dots .+f_n{x}_n}{f_1+f_2\dots .+f_n}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i$

where $N=\sum _{i=1}^n{f}_i$

3. Arithmetic Mean Of Continuous Series

In case of a set of data with class intervals, we cannot find the exact value of the mean because we do not know the exact values of the variables. We, therefore, try to obtain an approximate value of the mean. The method of approximate is to replace all the observed values belonging to a class by mid-value of the class.
If x₁, x₂ , x₃,….., x_n are the mid values of the class intervals having corresponding frequencies f₁, f₂ , f₃,….., f_n then we apply the same formula as in discrete series.

$\overline{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i, N=\sum _{i=1}^n{f}_i$

4. Combined Arithmetic Mean

If ${\overline{x}}_i(i=1,2,\dots .,k)$ are the means of k – series of sizes $n_i(i=1,2,3,\dots \dots \dots ..,k)$ then the combined or composite mean $\overline{x}$ can be obtained by the formula :

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2+\dots .+n_k{\overline{x}}_k}{n_1+n_2+\dots ..+n_k}=\frac{\sum n_i{\overline{x}}_i}{\sum n_i}$

5. Weighted Arithmetic Mean

If w₁, w₂ , w₃,….., w_n be the weights assigned to the values x₁, x₂ , x₃,….., x_n be respectively of a variable x, then the weighted A.M. is $\overline{x}=\frac{\sum w_i{x}_i}{\sum w_i}$.

5. PROPERTIES OF ARITHMETIC MEAN

1. Sum of all the deviations from arithmetic mean is zero i.e.,

$\sum _{i=1}^n\left(x_i–\overline{x}\right)=0$ (in case of individual series)

$\sum _{i=1}^n{f}_i\left(x_i–\overline{x}\right)=0$ (in case of discrete or continuous series)

2. If each observation is increased or decreased by a given constant K, the mean is also increased or decreased by K
The property is also known as effect of change of origin. K can be taken to be any number. However, to simplify the calculations, K should be taken as a value which is in the middle of the table.
3. Step Deviation Method or change of scale

If x₁, x₂ , x₃,….., x_n are mid values of class intervals with corresponding frequencies f₁, f₂ , f₃,….., f_n then we may change the scale by taking $d_i=\frac{x_i–A}{h}$ in this case.

$\overline{x}=A+h\times \left(\frac{1}{N}\sum f_i{d}_i\right)$ (if A is assumed mean)

A and h can be any numbers but if the lengths of class intervals are equal then h may be taken as width of the class interval.
In particular if each observation is multiplied or divided by a constant, the mean is also multiplied or divided by the same constant.
4. The sum of the squared deviation of the variate from their mean is minimum i.e., the quantity $\sum {\left(x_i–A\right)}^2\text{ or }\sum f_i{\left(x_i–A\right)}^2$ is minimum when $A=\overline{x}$.

5. E(aX + b) = aE(X) + b (where E(X) = Mean of X)

Illustration-17
The weighted mean of the first n natural numbers, the weights being the corresponding numbers, is
Solution
First n natural numbers are 1, 2, 3,…,n; whose corresponding weights are 1, 2, 3,…,n respectively.

$\therefore$ weight mean = $\frac{1\times 1+2\times 2+\dots .+n\times n}{1+2+\dots +n}$

$=\frac{1^2+2^2+\dots ..+n^2}{1+2+\dots ..+n}$

$=\frac{n(n+1)(2n+1)}{\frac{6n(n+1)}{2}}=\frac{2n+1}{3}$

Illustration-18

The weighted mean of the first n natural numbers whose weights are equal to the squares of the corresponding numbers is
Solution

weighted mean = $\frac{1.1^2+2.2^2+\dots +n·n^2}{1^2+2^2+\dots +n^2}$

$=\frac{{\mathrm{\Sigma n}}^3}{{\mathrm{\Sigma n}}^2}=\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}}=\frac{3\mathrm{n}(\mathrm{n}+1)}{2(2\mathrm{n}+1)}$

Illustration-19
The average salary of male employees in a firm is Rs. 5200 and that of females is Rs.4200. The mean salary of all the employees is Rs.5000. The percentage of male and female employees are respectively is
Solution

Let $x_1=5200, x_2=4200, \overline{x}=5000$

Also, we know that $\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$

$\Rightarrow 5000\left(n_1+n_2\right)=5200n_1+4200n_2\Rightarrow \frac{n_1}{n_2}=\frac{4}{1}$

$\therefore$ The percentage of male employees in the firm = $\frac{4}{4+1}\times 100=80\%$

and the percentage of female employees in the firm = $\frac{1}{4+1}\times 100=20\%$

Illustration-20
If the mean of 9 observations is 100 and mean of 6 observations is 80, then the mean of 15 observations is
Solution

$n_1=9, {\overline{x}}_1=100\text{ and }{n}_2=6, {\overline{x}}_2=80$

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}=\frac{9\times 100+6\times 80}{9+6}=92$

Illustration-21

If a variate X is expressed as a linear function of two variates U and V in the form X = aU + bV then the mean $\overline{X}$ of X is

Solution

We have $\Sigma X=a\Sigma U+b\Sigma V$

$\overline{X}=\frac{1}{n}\Sigma X=a.\frac{1}{n}\Sigma U+b.\frac{1}{n}\Sigma V$

$\Rightarrow \overline{X}=a\overline{U}+b\overline{V}$

Illustration-22

If the arithmetic mean of the numbers x₁, x₂ , x₃,….., x_n is $\overline{x}$, then the arithmetic mean of the numbers $a{x}_1+b,a{x}_2+b,a{x}_3+b,\dots .a{x}_n+b$, where a, b are two constants, would be

Solution

Required mean $=\frac{\left(a{x}_1+b\right)+\left(a{x}_2+b\right)+\dots .+\left(a{x}_n+b\right)}{n}$

$=\frac{a\left(x_1+x_2+\dots .+x_n\right)}{n}+b=a\overline{x}+b \left(\because \frac{x_1+x_2+\dots +x_n}{n}=\overline{x}\right)$

Illustration-23

If the mean of the numbers 27+x, 31+x, 89+x , 107+x, 156+x is 82, then the mean of 130+x, 126+x, 68+x, 50+x, 1+x is

Solution

Given

$82=\frac{(27+x)+(31+x)+(89+x)+(107+x)+(156+x)}{5}$

$\Rightarrow 82\times 5=410+5x\Rightarrow 410–410=5x\Rightarrow x=0$

Therefore, the required mean is

$\overline{x}=\frac{130+x+126+x+68+x+50+x+1+x}{5}$

$=\frac{375+5x}{7}=75$

Illustration-24

A student obtain 75%, 80% and 85% in three subjects. If the marks of another subject is added, then his average cannot be less than

Solution

Marks obtained from three subjects out of 300 is 75+80+85 = 240

If the marks of another subject is added, then the marks will be $\ge$240 out of 400

$\therefore$ minimum average marks = $\frac{240}{4}=60\%$

[when marks in the fourth subject = 0]

Illustration-25

The mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is

Solution

Sum of 100 items = 49 × 100 = 4900

sum of items added = 60+ 70+80=210

new sum = 4900+210–110=5000

$\therefore$ correct mean = $\frac{5000}{100}=50$

Illustration-26

The mean weight per student in a group of seven students is 55kg. If the individual weights of six students are 52, 58, 55, 53, 56 and 54, then the weight of the seventh student is

Solution

The total weight of seven students is 55×7 = 385kg

The sum of the weights of six students is 52+58+55+53+56+54 = 328kg

Hence, the weight of the seventh student is = 385 – 328 = 57kg

3. GEOMETRIC MEAN

1. GEOMETRIC MEAN IN INDIVIDUAL SERIES

1. If x₁, x₂ , x₃,….., x_n are n values of a variate x, none of them being zero, then geometric mean (G.M.) is given by G.M = ${\left(x_1–x_2–x_3\dots \dots x_n\right)}^{1/n}$

2. log (G.M.) = $\frac{1}{n}\left(\log x_1+\log x_2+\dots ..+\log x_n\right)$

2. GEOMETRIC MEAN IN DISCRETE SERIES

In case of frequency distribution, G.M. of n values x₁, x₂ , x₃,….., x_n of a variate x occurring with frequency f₁, f₂ , f₃,….., f_n is given by G.M = ${\left(x_1^{f_1}.x_2^{f_2}\dots ..x_n^{f_n}\right)}^{1/N}$, where $N=f_1+f_2+\dots ..+f_n$.

Illustration-27

The geometric mean of the numbers $3,3^2,3^3,\dots ,3^n$ is

Solution

$\therefore G.M={\left(3.3^2\dots \dots 3^n\right)}^{1/n}$

$=3^{\frac{1+2+\dots +n}{n}}=3^{\frac{n(n+1)}{2n}}=3^{\frac{n+1}{2}}$

Illustration-28

Find the geometric mean of the following values:

Solution

Given data, 15, 12, 13, 19, 10

X	Log x
15	1.1761
12	1.0792
13	1.1139
19	1.2788
10	1.0000
Total	5.648

$\Sigma$ log x =5.648 and n = 5

log(G.M.) = $\frac{1}{n}\left(\log x_1+\log x_2+\dots ..+\log x_n\right)$.

= Antilog $\frac{5.648}{5}$

= Antilog 1.1296

= 13.48

Answer: Geometric Mean = 13.48

4. HARMONIC MEAN

The harmonic mean is based on the reciprocals of the value of the variable

$\mathrm{H}.\mathrm{M}=\frac{1}{\frac{1}{n}\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots +\frac{1}{x_n}\right)}\text{ or }\frac{1}{H}=\frac{1}{n}\sum _{i=1}^n\frac{1}{x_i}$ (Incase of Individual series) and $\frac{1}{H}=\frac{1}{N}\sum _{i=1}^n{f}_i\frac{1}{x_i}$ (in case of discrete series or continuous series)

Relation among Arithmetic Mean, Geometric Mean and Harmonic Mean

The Arithmetic Mean (A. M.), Geometric Mean (G.M.) and Harmonic Mean (H.M.) for a given set of observations x₁, x₂ , x₃,….., x_n > 0 of a series are related as under:

$\mathrm{A}.\mathrm{M}\ge \mathrm{G}.\mathrm{M}\ge \mathrm{H}.\mathrm{M}$

Illustration-29

The harmonic mean of 3, 7, 8, 10, 14 is

(A) $\frac{3+7+8+10+14}{5}$     (B) $\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{14}$     (C) $\frac{\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{14}}{4}$     (D) $\frac{5}{\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{14}}$

Solution

D

H.M. = $\frac{1}{\frac{1}{n}\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots .+\frac{1}{x_n}\right)}=\frac{5}{\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{14}}$

Illustration-30

Find the harmonic mean of $\frac{1}{2},\frac{2}{3},\frac{3}{4},\dots .,\frac{n}{n+1}$, occurring with frequencies 1, 2, 3,…..n, respectively.

Solution

We know that, Harmonic mean = $\frac{\sum f}{\sum \left(\frac{f}{x}\right)}$

$\sum f=1+2+3+\dots .+n=\frac{n(n+1)}{2}$

and $\sum \frac{f}{x}=\frac{1}{1/2}+\frac{2}{2/3}+\frac{3}{3/4}+\dots .+\frac{n}{n/(n+1)}$

$=2+\frac{3\times 2}{2}+\frac{4\times 3}{3}+\dots +\frac{n(n+1)}{n}$

$=2+3+4+\dots +n+(n+1)$

Which is an arithmetic progression with a = 2 and d = 1.

By the formula of sum of n term of an A.P, $\sum \left(\frac{f}{x}\right)=\left[\frac{n}{2}\{2a+(n–1\left)d\right\}\right]$

we have = $\frac{n}{2}\{2\times 2+n–1\}=\frac{n}{2}(3+n)$

$\therefore$Harmonic mean

$=\frac{2}{n(3+n)}=\frac{n(n+1)\times 2}{n(3+n)\times 2}=\frac{n+1}{3+n}$

5. MEDIAN AND MODE

1. MEDIAN
The median is the central value of the set of observations provided all the observations are arranged in the ascending or descending orders. It is generally used, when effect of extreme items is to be kept out.
Calculation of median
(i) Individual series: If the data is raw, arrange in ascending or descending order. Let n be the number of observations.

If n is odd, Median = value of ${\left(\frac{n+1}{2}\right)}^{th}$ item.

If n is even, Median = $\frac{1}{2}\left[\text{ value of }{\left(\frac{n}{2}\right)}^{\text{th }}\text{ item + value of }{\left(\frac{n}{2}+1\right)}^{\text{th }}\text{ item }\right]$

(ii) Discrete series : In this case, we first find the cumulative frequencies of the variables arranged in ascending or descending order and the median is given by Median = ${\left(\frac{n+1}{2}\right)}^{\text{th }}$ observation, where n is the cumulative frequency.

(iii) For grouped or continuous distributions : In this case, following formula can be used.

(a) For series in ascending order, Median = $I+\frac{\left(\frac{N}{2}–C\right)}{f}\times i$

Where

l = Lower limit of the median class
f = Frequency of the median class
N = The sum of all frequencies
i = The width of the median class
C = The cumulative frequency of the class preceding to median class.

(b) For series in descending order

Median = $u–\left(\frac{\frac{N}{2}–C}{f}\right)\times i$, where u = upper limit of the median class, $N=\sum _{i=1}^n{f}_i$

As median divides a distribution into two equal parts, similarly the quartiles, quantiles, deciles and percentiles divide the distribution respectively into 4, 5, 10 and 100 equal parts. The j^th quartile is given by $Q_j=I+\left(\frac{j\frac{N}{4}–C}{f}\right)i; j=1,2,3$. Q₁ is the lower quartile, Q₂ is the median and Q₃ is called the upper quartile.

(2) Lower quartile

(i) Discrete series : $Q_1=\text{ size of }{\left(\frac{n+1}{4}\right)}^{\text{th }}\text{ item }$

(ii) Continuous series : $Q_1=I+\frac{\left(\frac{N}{4}–C\right)}{f}\times i$

(3) Upper quartile

(i) Discrete series : $Q_3=\text{ size of }{\left[\frac{3(n+1)}{4}\right]}^{\text{th }}\text{ item }$

(ii) Continuous series : $Q_3=I+\frac{\left(\frac{3N}{4}–C\right)}{f}\times i$

(4) Decile : Decile divide total frequencies N into ten equal parts.

$D_j=I+\frac{\frac{N\times j}{10}–C}{f}\times i$,     [where j = 1, 2, 3, 4, 5, 6, 7, 8, 9]

If j = 5, then $D_5=I+\frac{\frac{N}{2}–C}{f}\times i$. Hence D₅ is also known as median.

(5) Percentile : Percentile divide total frequencies N into hundred equal parts and $P_k=I+\frac{\frac{N\times k}{100}–C}{f}\times i$,  where k = 1, 2, 3, 4, 5,…….,99.

Illustration-31

The median of 10, 14, 11, 9, 8, 12, 6 is

(A) 10     (B) 12     (C) 14     (D) 11

Solution

(A)

Arrange the items in ascending order i.e., 6, 8, 9, 10, 11, 12, 14.

If n is odd then, Median = value of ${\left(\frac{n+1}{2}\right)}^{th}$ term

$\therefore$ Median = ${\left(\frac{7+1}{2}\right)}^{th}$ term = 4^th term = 10.

Illustration-32
The median of a set of nine distinct observations is 20.5. If each of the last four observations of the set is increased by 2, then the median of the new set is
Solution

Since n = 9, median term = ${\left(\frac{9+1}{2}\right)}^{th}$ = 5^th term.

Now, the last four observations are increased by 2. Since the median is the 5th observation, which remains unchanged, there will be no change in median.

Illustration-33
If a variable takes the discrete value $\alpha –4, \alpha –\frac{7}{2}, \alpha –\frac{5}{2}, \alpha –3, \alpha –2, \alpha +\frac{1}{2}, \alpha –\frac{1}{2}, \alpha +5(a>0),$ then the median is

Solution

Arrange the data as follows:

$\alpha –\frac{7}{2}, \alpha –3, \alpha –\frac{5}{2}, \alpha –2, \alpha –\frac{1}{2}, \alpha +\frac{1}{2}, \alpha –4, \alpha +5$

median = $\frac{1}{2}$[value of 4th item+value of 5th item]

$\therefore$ median = $\frac{\alpha –2+\alpha –\frac{1}{2}}{2}=\alpha –\frac{5}{4}$

Illustration-34

The median of distribution 83, 54, 78, 64, 90, 59, 67, 72, 70, 73 is

Solution

On arranging in ascending order, we get 54, 59, 64, 67, 70, 72, 73, 78, 83, 90

n = 10

$\therefore \text{ median }=\frac{\text{ value of }{\frac{10}{2}}^{\mathrm{th}}\text{ term }+\text{ valueof }{\left(\frac{10}{2}+1\right)}^{\text{th }}\text{ term }}{2}$

= $\frac{\text{ value of }{5}^{\text{th}}\text{ term }+\text{ value of }{6}^{\text{th}}\text{ term }}{2}$

$=\frac{70+72}{2}=71$

2. MODE

The mode or model value of a distribution is that value of the variable for which the frequency is maximum. For continuous series, mode is calculated as,

Mode = $I_1+\left[\frac{f_1–f_0}{2f_1–f_0–f_2}\right]\times i$

Where, l₁ = The lower limit of the model class

f₁ = The frequency of the model class

f₀ = The frequency of the class preceding the model class

f₂ = The frequency of the class succeeding the model class

i = The size of the model class.

Symmetric distribution
A distribution is a symmetric distribution if the values of mean, mode and median coincide. In a
symmetric distribution frequencies are symmetrically distributed on both sides of the centre point of the frequency curve.

A distribution which is not symmetric is called a skewed-distribution. In a moderately asymmetric
distribution, the interval between the mean and the median is approximately one-third of the interval between the mean and the mode i.e., we have the following empirical relation between them,

Mean – Mode = 3(Mean – Median) $\Rightarrow$ Mode = 3 Median – 2 Mean. It is known as Empirical relation.

Illustration-35

The mode of the following distribution is

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	8	7	12	28	20	10	10

Solution

Here, maximum frequency is 28. Thus, the class 40-50 is the modal class.

Mode = $l+\frac{f_1–f_0}{2f_1–f_2–f_0}\times h$

$=40+\frac{10(28–12)}{(2\times 28–12–20)}$

= 40+6.666 = 46.67(approx.)

Illustration-36
If in a frequency distribution, the mean and median are 20 and 21 respectively, then its mode is approximately
Solution
mode = 3 median – 2 mean = 3 (21) – 2(20) = 23.

Illustration-37
If in a moderately asymmetrical distribution the mode and the mean of the data are 6$\lambda$ and 9$\lambda$ , respectively, then the median is
Solution

For a moderately skewed distribution,
mode = 3median – 2 mean
$\Rightarrow 6\lambda =3\text{ median }–18\lambda \Rightarrow \text{ median }=8\lambda$

6. MEASURES OF DISPERSION

The degree to which numerical data tend to spread about an average value is called the dispersion of the data. The four measure of dispersion are

(1) Range     (2) Mean deviation     (3) Standard deviation     (4) Square deviation

1. RANGE
It is the difference between the values of extreme items in a series. Range = $X_{\max }–X_{\min }$

The coefficient of range (scatter) = $\frac{x_{\max }–x_{\min }}{x_{\max }+x_{\min }}$.

Range is not the measure of central tendency. Range is widely used in statistical series relating to
quality control in production.

Range is commonly used measures of dispersion in case of changes in interest rates, exchange rate, share prices and like statistical information. It helps us to determine changes in the qualities of the goods produced in factories.

i. Inter-quartile range
We know that quartiles are the magnitudes of the items which divide the distribution into four equal parts. The inter-quartile range is found by taking the difference between third and first quartiles and is given by the following formula,

Inter-quartile range = Q₃ – Q₁,

where Q₁ = First quartile or lower quartile and Q₃ = Third quartile or upper quartile.

ii. Percentile range

This is measured by the following formula,

Percentile range = P₉₀ – P₁₀,

where P₉₀ – 90^th percentile and P₁₀ = 10^th percentile.

Percentile range is considered better than range as well as inter-quartile range.

iii. Quartile deviation or semi inter-quartile range

It is one-half of the difference between the third quartile and first quartile i.e., Q.D. = $\frac{Q_3–Q_1}{2}$ and coefficient of quartile deviation = $\frac{Q_3–Q_1}{Q_3+Q_1}$, where Q₃ is the third or upper quartile and Q₁ is the lowest or first quartile.

2. MEAN DEVIATION
The arithmetic average of the deviations (all taking positive) from the mean, median or mode is
known as mean deviation.

Mean deviation is used for calculating dispersion of the series relating to economic and social inequalities. Dispersion in the distribution of income and wealth is measured in term of mean deviation.

i. Mean deviation from ungrouped data (or individual series)

Mean deviation = $\frac{\sum |x–M|}{n}$.

where |x – M| means the modulus of the deviation of the variate from the mean (mean, median or
mode) and n is the number of terms.

ii. Mean deviation from continuous series

Here first of all we find the mean from which deviation is to be taken. Then we find the deviation
dM =|x – M| of each variate from the mean M so obtained.
Next we multiply these deviations by the corresponding frequency and find the product f.dM and then the sum $\Sigma$f dM of these products.

Lastly we use the formula, mean deviation = $\frac{\sum f|x–M|}{n}=\frac{\sum fdM}{n},\text{ where }n=\mathrm{Sf}\text{.}$

3. STANDARD DEVIATION

Standard deviation (or S.D.) is the square root of the arithmetic mean of the square of deviations of various values from their arithmetic mean and is generally denoted by s read as sigma. It is used in statistical analysis.
i. Coefficient of standard deviation
To compare the dispersion of two frequency distributions the relative measure of standard deviation is computed which is known as coefficient of standard deviation and is given by

Coefficient of A.D. = $\frac{\sigma }{\overline{x}}$, where $\overline{x}$ is the A.M.

ii. Standard deviation from individual series

$\sigma =\sqrt{\frac{\sum (x–\overline{X}{)}^2}{N}}$

where, $\overline{x}$ = The arithmetic mean of series

N = The total frequency.

iii. Standard deviation from continuous series

$\sigma =\sqrt{\frac{\sum f_i{\left(x_i–\overline{x}\right)}^2}{N}}$

where, $\overline{x}$ = Arithmetic mean of series
x_i = Mid value of the class
f_i = Frequency of the corresponding
N = Sf = The total frequency

Short cut method

(i) $\sigma =\sqrt{\frac{\sum f{d}^2}{N}–{\left(\frac{\sum fd}{N}\right)}^2}$          (ii) $\sigma =\sqrt{\frac{\sum d^2}{N}–{\left(\frac{\sum d}{N}\right)}^2}$

where, d = x – A = Deviation from the assumed mean A
f = Frequency of the item
N = f = Sum of frequencies

4. SQUARE DEVIATION
i. Root mean square deviation

$S=\sqrt{\frac{1}{N}\sum _{i=1}^n{f}_i{\left(x_i–A\right)}^2}$

where A is any arbitrary number and S is called mean square deviation.

ii. Relation between S.D. and root mean square deviation

If s be the standard deviation and S be the root mean square deviation.

Then, $S^2={\sigma }^2+d^2$,

Obviously, S² will be least when d = 0 i.e., $\overline{x}=A$

Hence, mean square deviation and consequently root mean square deviation is least, if the deviations are taken from the mean.

VARIANCE

The square of standard deviation is called the variance.

Coefficient of standard deviation and variance

The coefficient of standard deviation is the ratio of the S.D. to A.M. i.e., $\frac{\sigma }{x}$.

Coefficient of variance = coefficient of S.D. $\times 100=\frac{\sigma }{\overline{x}}\times 100$.

Variance of the combined series

If $n_1, n_2$ are the sizes, $\overline{x_1}, \overline{x_2}$ the means and ${\sigma }_1, {\sigma }_2$ the standard deviation of two series, then

${\sigma }^2=\frac{1}{n_1+n_2}\left[n_1\left({\sigma }_1^2+d_1^2\right)+n_2\left({\sigma }_2^2+d_2^2\right)\right]$

where $d_1={\overline{x}}_1–\overline{x}, d_2={\overline{x}}_2–\overline{x}$ and $\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$.

Note :1

if V (X) is variance of X then

V(X + a) = V(X)

V(aX) = a²V(X)

V(aX + b) = a²V(X)

V (aX + bY) = a²v (X) + b²v (Y)

Note :2

For a, a + d, a + 2d, ….., a + (n – 1)d,

$\overline{x}=a+\frac{(n–1)d}{2}; {\sigma }^2=\frac{n^2–1}{12}{d}^2$

Note :3

1) Q.D < M.D < S.D

2) $\frac{\mathrm{Q}.\mathrm{D}}{10}=\frac{\mathrm{M}.\mathrm{D}}{12}=\frac{\mathrm{S}.\mathrm{D}}{15}$

SKEWNESS

“Skewness” measures the lack of symmetry. It is measured by ${\gamma }_1=\frac{\sum {\left(x_i–\mu \right)}^3}{{\left\{\Sigma \left(x_i–{\mu }^2\right)\right\}}^{3/2}}$ and is denoted by ${\gamma }_1$.

The distribution is skewed if,

(i) Mean $\ne$ Median $\ne$ Mode

(ii) Quartiles are not equidistant from the median

(iii) The frequency curve is stretched more to one side than to the other.

1. Distribution

There are three types of distributions.

(i) Normal distribution :

When ${\gamma }_1=0$, the distribution is said to be normal. In this case, Mean = Median = Mode

(ii) Positively skewed distribution : When ${\gamma }_1>0$, the distribution is said to be positively skewed. In this case, Mean > Median > Mode

(iii) Negative skewed distribution : When ${\gamma }_1<0$, the distribution is said to be negatively skewed. In this case, Mean < Median < Mode

Measures of skewness

(i) Absolute measures of skewness : Various measures of skewness are

(a) $S_K=M–M_d$          (b) $S_K=M–M_0$          (c) $S_K=Q_3+Q_1–2M_d$

where, M_d = median, M₀ = mode, M = mean.

Absolute measures of skewness are not useful to compare two series, therefore relative measure of
dispersion are used, as they are pure numbers.

3. Relative measures of skewness

(i) Karl Pearson’s coefficient of skewness

$S_k=\frac{M–M_o}{\sigma }=3\frac{\left(M–M_d\right)}{\sigma },–3\le S_k\le 3$

where s is standard deviation.

(ii) Bowley’s coefficient of skewness :

$S_k=\frac{Q_3+Q_1–2M_d}{Q_3–Q_1}$

Bowley’s coefficient of skewness lies between –1 and 1.

(iii) Kelly’s coefficient of skewness

$S_K=\frac{P_{10}+P_{90}–2M_d}{P_{90}–P_{10}}=\frac{D_1+D_9–2M_d}{D_9–D_1}$

Illustration-38

The quartile deviation of daily wages of in (Rs.) of 11 persons given below 140, 145, 130, 165, 160, 125, 150, 170, 175, 120, 180

Solution

The given data in ascending order of magnitude is 120, 125, 130, 140, 145, 150, 160, 165, 170, 175, 180

Here, $Q_1={\left(\frac{n+1}{4}\right)}^{th}\text{ term }=\frac{11+1}{4}\text{ term }=3^{rd}\text{ term }$ = 130

$Q_3=\frac{3(n+1{)}^{th}}{4}=\frac{3\times (11+1)}{4}=9^{th}=170$

$QD=\frac{Q_3–Q_1}{2}=\frac{170–130}{2}=\frac{40}{2}=20$

Illustration-39

The variance of the first ‘n’ natural numbers is

Solution

Variance

$=(SD{)}^2=\frac{1}{n}\Sigma x^2–{\left(\frac{\Sigma x}{n}\right)}^2, \left(\because \overline{x}=\frac{\Sigma x}{n}\right)$

$=\frac{n(n+1)(2n+1)}{6n}–{\left(\frac{n(n+1)}{2n}\right)}^2=\frac{n^2–1}{12}$

Illustration-40

If the M.D is 12, the value of S.D will be

Solution

We know that $\mathrm{Q}.\mathrm{D}=\frac{5}{6}\times M.D=\frac{5}{6}\times 12=10$

$\therefore S.D=\frac{3}{2}\times Q.D=\frac{3}{2}\times 10=15$

Illustration-41

The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, then the other two observations are

Solution

Let the two unknown items be x and y, then

Mean = 4 $\Rightarrow \frac{1+2+6+x+y}{5}=4$

$\Rightarrow$ x + y = 11…..(1) and variance = 5.2

$\Rightarrow \frac{1^2+2^2+6^2+x^2+y^2}{5}–(\text{ mean }{)}^2=5.2$

$41+x^2+y^2=5\left[5.2+(4{)}^2\right]$

$41+x^2+y^2=106$

$x^2+y^2=65\dots \dots \left(2\right)$

Solving (1) and (2) for x and y, we get

x = 4, y = 7 or x = 7, y = 4.