MEASURES OF CENTRAL TENDENCY OR AVERAGES · MEASURES OF CENTRAL TENDENCY OR AVERAGES MEASURES OF CENTRAL TENDENCY OR LOCATION A single expression, representing the whole group, is

MEASURES OF CENTRALTENDENCY OR AVERAGES

MEASURES OF CENTRAL TENDENCY OR LOCATION

A single expression, representing the whole group, is selected whichmay convey a fairly adequate idea about the whole group. This singleexpression in statistics is known as the average. Averages are,generally, the central part of the distribution and, therefore, they arealso called the measures of central tendency. There are five types ofmeasures of central tendency or averages which are commonly used.These are:

1. Arithmetic Mean (A.M.) 2. Median3. Mode

4. Geometric Mean (G.M.) 5. Harmonic Mean (H.M.)

FEATURES OF A GOOD AVERAGE

According to Professor G.U. Yule, a good average must have thefollowing characteristics:

1. It should be rigidly defined so that different persons may notinterpret it differently.

2. It should be easy to understand and easy to calculate.

3. It should be based on all the observations of the data.

4. It should be easily subjected to further mathematical calculations.2MEASURES OF CENTRAL TENDENCY OR AVERAGES 87

5. It should be least affected by the fluctuations of the sampling.

6. It should not be unduly affected by the extreme values.

7. It should be easy to interpret.

8. It should have sampling stability. It means that Vthe average is computed for similar groups,the result should also be similar.

ARITHMETIC MEAN

Arithmetic Mean of Raw Data

We know that in an ungrouped or raw data, we are given individual items. We also know thatthe average ol n numbers is obtained by finding their sum (by adding) and then dividing it by n.Let x1, x2 ...... xn , be n numbers, then their average or arithmetic mean is given by

n

x

nx.....xxxx

n

lii

n321

Example 1. Find the arithmetic mean of the marks obtained by 10 students of class X in Mathematicsin a certain examination. The marks obtained are:

25, 30, 21, 55, 47, 10, 15, 17, 45, 35.

Solution. Let x be the average marks.

Sumofalltheobservations= 25 + 30 + 21 + 55 + 47 + 10 + 15 + 17 + 45 + 35 = 300.

Number of students = 10.

Arithmetic mean = 3010300

nxi

PROPERTIES OF ARITHMETIC MEAN

Property 1. If the mean of n observations x1, x2,........ xn is x , then

0)xX(......)xx()xx( n2i i.e

n

li0)xx(

88 STATISTICS (CPT)

Proof. i 2 n 1 2 n(x x) (x x) ...... (X x) (x x ..... x ) nx xnxn 0

Thus, the algebraic sum of deviations from mean is zero.

Property II. If the mean of n observations, x1, x2 ....... xn is x then the mean of the observations(x1 + a), (x2 + a), (x3 + a) ....... (xn + a) is x + a.

i.e., if each observation is increased by ‘a’, then the mean is also increased by a.

Property III. If the mean of n observations x1, x2, x3, ..... xn is x , then the mean of theobservations (x1 - a), (x2 - a), (x3 - a), ........ (xn - a) is x - a.

i.e., if each observation is decreased by ‘a’, then the new mean is also decreased by a.

Property IV. The mean of n observations x1, x2, x3, ..... xn is x . If each observation is multipliedby p, p 0, then the mean of new observations is p x .

Property V. The mean of n observationsx1, x2, x3, ..... xn is x . If each observation is divided

by p, then the mean of new observations is PX

. (P # 0)

Example 2. The mean of 68 numbers is 18. If each number is divided by 6 & find the new mean.Find the relation between the new mean and old mean.

Solution. Mean of 68 numbers = 18. Sum of 68 numbers = 68 x 18.

If each number is divided by 6, the sum is also divided by 6.

New sum = 68 18

6

New Mean = 68 186 68

= 3. [New Mean = 6anOriginalme

.

Example 3. The mean weight of 150 students in a class is 60 kg. The mean weight of the boys is 70kg while that of the girls is 55 kg. Find the number of boys and that of the girls in the class.

Solution. Let the number of boys be x. Then, the number of girls = (150 - x).

Total weight of 150 students = (total weight of x boys) + [total weight of (150 - x) girls]

= 150 x 60 = 70x + 55 x (150 - x). This gives, 15x = 750 or x = 50.

The number of boys 50 and the number of girls = (150 - 50) = 100.

MEASURES OF CENTRAL TENDENCY OR AVERAGES 89

Example 4. In a school 85 boys and 35 girls appeared in a public examination. The mean marks ofboys was found to be 40 %. whereas the mean marks of girls was 60 %. Determine the averagemarks percentage of the school.

Solution. Total score of 85 boys = 85 x 40 = 3400 [Total = Number x Mean]

Total score of 35 girls = 35 x 60 = 2100.

Total score of 120 students (85 boys and 35 girls) = 3400 + 2100 = 5400.

Average marks percentage of the school = 1205400

= 45%.

METHODS TO CALCULATE ARITHMETIC MEAN

Arithmetic Mean of Grouped Data

Direct Method. If n observations in the raw data consist of n distinct values denoted by x1, x2, x3,........., xn of the observed variable x occurring with frequencies f1, f2, f3, ........., fn respectively,then the arithmetic mean of the variate x is given by x , where

x = n321

nn332211f....fff

xf....xfxfxf

=

i

n

1iii

f

xf =

N

xfn

1iii

where N =

n

1iif = f1 + f2 + ..... + fn Sum of frequencies

Example 5. Find the Arithmetic Mean from the following frequency table:

Marks 52 58 60 65 68 70 75

No.ofStudents 7 5 4 6 3 3 2

Solution. Let x be the marks andfbe the frequency, so that we have the following table:

Computation Table

x : 52 58 60 65 68 70 75 Total

f : 7 5 4 6 3 3 2 30

f x x : 360 290 240 390 204 210 150 1848

Here N = f = 30 and fx = 1848.

90 STATISTICS (CPT)

Mean : = x = 10616

301848

Nfx

= 61.6

Concealed Frequency

Example 6. The following table gives the distribution of 31 accidents in New Delhi during sevendays of a week of a given month. During that month there were S Mondays, 5 Tuesdays and 5Wednesdays and only four each of the other days. Calculate the number of accidents per day.

Day : Sunday Monday Tuesday Wednesday Thursday Friday Saturday

No.of 26 16 12 10 8 10 18

Accidents :

Solution. Let us prepare the following table.

Table : Computation of Mean

Day No. of Accidents (x) Number (f) f x x

Sunday 26 4 104

Monday 16 5 80

Tuesday 12 5 60

Wednesday 10 5 50

Thursday 8 4 32

Friday 10 4 40

Saturday 18 4 72

f=31 fx = 438

Now Mean : x = ffx

= 31438

= 14.13.

Arithmetic Mean of a Grouped Data

Direct Method. In this case the raw data is presented in the form of a Frequency Distributionwith Class Intervals. The arithmetic mean is calculated by the Direct Method. The arithmeticmean is defined as:

Mean : n

xfx

n

1iii

where n =

k

1iif


and xi = class mark (or mid-value or mid-point) of the ith class interval and is given by :

xi = lower limit of i th class interval + upper limit of i the class interval

2

Example 7. The data on number of patients attending a hospital in a month are given below. Findthe average number of patients attending the hospital in a day.

Number of patients 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60

Number days 2 6 9 7 4 2

attending the hospital

Solution. Table : Calculation of Mean

No. of patients Frequency Mid-value Fi xi

(Class Interval) xi

0 -10 2 5 2 x 5 =10

10 - 20 6 15 6 x 15 = 90

20 - 30 9 25 9 x 25 = 225

30 - 40 7 35 7 x 35 = 245

40 - 50 4 45 4 x 45 =180

50 - 60 2 55 2 x 54 =110

Total fi = 30 fi x xi = 860

Average = x = i

iifxf

= 30860

= 28.67 29 patients per day.

Arithmetic Mean by Short-Cut Method

CASE I. Short-cut Method for Ungrouped Data

For the mean of ungrouped data, the short-cut method is applied when the frequencies and thevalues of the variables are quite large and it becomes very difficult to compute the arithmeticmean ps in the case of above example. In a frequency table of such a type, the provisional mean ‘a’is taken as that value of x (mid-value of the class interval) which corresponds to the highestfrequency or which comes near the middle value of the frequency distribution. This number iscalled the provisional Mean or Assumed Mean. Also find the deviations of the variates from thisprovisional mean. Then the arithmetic mean is given by the formula.

(i) In the case of ungrouped data

92 STATISTICS (CPT)

x = a + nd

, where a = assumed mean, n = number of items.

d = x - a = deviations of the variate ‘x’ from a.

CASE II. Short-cut Method for Grouped Data

In the case of grouped data, the mean is given by

x = a + Nfd

, where

fd = product of the frequency and the corresponding deviation,

N = f = the sum of all the frequencies.

Example 8. Ten coins were tossed together and the number of the resulting from them wereobserved. The operation was performed 1050 times and the frequencies thus obtained for dVferentnumber of tails (x) are shown in the following table. Calculate the arithmetic mean by the shortcut method.

x : 0 1 2 3 4 5 6 7 8 9 10

f : 2 8 43 133 207 260 213 120 54 9 1

Solution. Let 5 be the assumed mean, i.e., a = 5. Let us prepare the following table in order tocalculate the arithmetic mean:

x f d = x - a = x - 5 fd

0 2 - 5 - 10

1 8 - 4 - 32

2 43 - 3 - 129

3 133 -2 - 266

4 207 -1 - 207

5 260 0 0

6 213 1 +213

7 120 2 +240

8 54 3 +162

9 9 4 +36

10 1 5 +5

f= 1050 fd=12

Arithmetic Mean : x = a + ffd

= 5 + 105012

= 5 + 0.0114 = 5.0114.


Example 9. For the following frequency table, find the mean.

Class: 100 - 120 120 -140 140 - 160 160 - 180 180 - 200 200 - 220 220 - 240

Frequency: 10 8 4 4 3 1 2.

Solution. Let a = 170, then we have the following table :

Table: Computation of Mean

Class Frequency Mid-value d = x - a

(f) (x) = x - 170 fd

100—120 10 110 -60 - 600

120—140 8 130 -40 - 320

140—160 4 150 -20 - 80

160—180 4 170 0 0

180—200 3 190 20 60

200—220 1 210 40 40

220—240 2 230 60 120

f = 32 fd = -780

Now780170 170 24 375 145 62532

fdMean a . .

f

Arithmetic Mean by Step-Deviation Method

Step Deviation Method:

When the class intervals in a grouped data are equal, then the calculation can be simplified furtherby taking out the common factor from the deviations. This common factor is equal to the width ofthe class-interval. In such cases, the deviation of variate x from the assumed mean ‘a’ (i.e., d = x —a) are divided by the common factor. The arithmetic mean is then obtained by the followingformula:

fdx a i,

f

where a = assumed mean or provisional mean,

d = x a

i

= the deviation of the variate x from a,

i = the width of the class-interval;N = the number of observations.

94 STATISTICS (CPT)

Example 10. Calculate, by step deviation method, the arithmetic mean of the following marksobtained by students in English.

Marks 5 10 15 20 25 30 35 40 45 50

No. of Students20 43 75 67 72 45 39 9 8 6

Solution. Let a = 30 and i = 5.

Table: Computation of Mean

x f dx = x - 30 d’x=dx/i fd’x

5 20 - 25 - 5 - 100

10 43 - 20 - 4 - 172

15 75 - 15 - 3 - 225

20 67 -10 - 2 - 134

25 72 -5 - 1 - 72

30 45 0 0 0

35 39 5 1 39

40 9 10 2 18

45 8 15 3 24

50 6 20 4 24

f = 384 598fd' x

Nowfd' x

x a if

.

= 30 - 598384 x 5 = 30 - 1.56 x 5 = 30 - 7.80 = 22.2

Example 11. In a study on patients, the following data were obtained. Find the arithmetic mean.

Age (in years) : 10—19 20—29 30—39 40—49 50—59 60—69 70—79 80—89

jjVo. of cases: 0 1 1 10 17 38 9 3

Solution. The data is presented in the form of an inclusive series. We have to transform, theinclusive series into exclusive series. It can be transformed as follows:


We find the distance between the lower limit of the second class interval and the upper

limit of the first class-interval. This is equal to 20 - 19 = 1. We subtract 12 of this distance (i.e., 0.5)

from the lower limit and add it to the upper limit. The new classes will be formed as follows:

The new data is 10-0.5 = 9.5; 19 + 0.5 = 19.5 and so on.

Let a = 44.5, and d = (x - a) / i = (x - 44.5) / 10, where i = 10.


Age No. of cases Mid-valuex - 44.5

10 fd

(in years) (f) (x)

9.-19.5 1 14.5 -3 -3

19.5-29.5 0 24.5 - 2 0

29.5-39.5 1 34.5 - 1 - 1

39.5-49.5 10 44.5 0 0

49.5-59.5 17 54.5 1 17

59.5-69.5 38 64.5 2 76

69.5-79.5 9 74.5 3 27

79.5-89.5 3 84.5 4 12

N = 79 f =128

Now fd

x a iN

Here a = 44.5 ; i = 10 ; N = 79.

128x = 44.5 + ×10 = 44.5 +16.2 = 60.779

Example 12. Following is the distribution of marks obtained by 60 students in Economics test.

Marks No. of students Marks No. of students

More than 0 60 More than 30 20



Calculate the arithmetic mean.

96 STATISTICS (CPT)

Solution. We are given the cumulative frequency series. All the 60 students have secured morethan 0 marks; but 56 students have secured more than 10 marks and so on. It means 4 studentshave secured marks between 0-10. We can similarly rearrange the distribution as follows:


Marks No. of Students Mid-valuex - 35d10

fd

0-10 4 5 -3 -12

10-20 16 15 -2 -32

20-30 20 25 -1 -20

30-40 10 35 0 0

40-50 7 45 1 7

50-60 3 55 2 6

N=60 fd = - 51Here a = 35, N = 60, fd = - 51 , i = 10.

Nowfd

x a iN

Hence, (-51)x = 35 + ×1060 = 35 - 8.5 = 26.5.

Missing Frequencies

Example 13. Find the value of p for the following distribution whose mean is 16.6.

f : 12 16 20 24 16 8 4.

x : 8 12 15 p 20 25 30

Solution. Here i if x = 8 x 12 + 12 x 16 + 15 x 20 + p x 24 + 20 x 16 + 25 x 8 + 30 x 4

= 96 + 192 + 300 + 24p + 320 + 200 + 120 = 24p + 1228

Also if = 12 + 16 + 20 + 24 + 16 + 8 + 4 =100.

Mean = i i

i

f xf

=

24p +1228100 = 16.6 [Given]

24 p = 1660 - 1228 = 432 p = 18.


WEIGHTED ARITHMETIC MEAN

If w1, w2, w3, ..............., wn are the weights assigned to the values x1, x2, x3,.............., xn respectively,

then the weighted average is defined as:

Weighted Arithmetic Mean = 1 1 2 2 n n

1 2 n

wxw x + w x +......+ w x = .w + w +.........+ w w

It is, generally, denoted by wx or wX .

Example 14. A candidate obtains the following percentages in an examination. English 46%,Mathematics 67%, Sanskrit 72%, Economics 58%, Political Science 53%. It is agreed to give doubleweights to marks in English and Mathematics as compared to other subjects. What is the weightedmean?

Solution. Since it is agreed to give double weights to marks in English and Mathematics, we havethe following data :

Subjects Marks Weights wx

(x) (w)

English 46 2 92

Mathematics 67 2 134

Sanskrit 72 1 72

Economics 58 1 58

Political Science 53 1 53

7w 409wx

Weighted mean = wxw

=

4097 = 58.43.

CORRECTED MEAN

Incorrect Ex Incorrect item + Correct item

FORMULA: Correct Mean = Incorrect x Incorrect item Correct item

n

Example 15. The mean marks scored by 100 students was found to be 40. Later on, it was discoveredthat a score of 53 was misread as 83. Find the correct mean.

98 STATISTICS (CPT)

Solution. Here X = 40, n = 100. Also X =x

n 40 =

10x x = 4000

or Incorrect value of x = 4000.

Correct value of x = Incorrect value of x - Incorrect item + correct item

= 4000 — 83 + 53 = 3970.

Correct mean = Correct value of x

n =

3970100 = 397.

Example 16. Mean of 25 observations was found to be 78.4. But later on it was found that 96 wasmisread as 69. Find the correct mean.

Solution. We know that the mean is given by : X = x

n or x = n X .

Here X =78.4, n = 25 x = 25 x 78.4 = 1960.

But this is incorrect as 96 was misread as 69.

Correct x = 1960 - 69 + 96 = 1987.

Correct Mean = Correct value of x

n =

198725 = 79.48

COMBINED MEAN

If we are given the mean of two series and their sizes, then the combined mean for the resultantseries can be obtained by the formula:

1 1 2 2

1 2

n x + n xX =n + n

where 12X or X = Combined mean of the two series

1X = Mean of the first series

2X = Mean of the second series

n1 = Size of the first series

n2 = Size of the second series

Example 17. A firm of readymade garments make both men ‘s and women ‘s shirts. Its profitaverage is 6% of sales. Its profits in men & shirts average 8% of sales; and women shirts comprise60% of output. What is the average profit per sales rupee in women shirts.


Solution. Here X = 6, 1X = 8, n1 = 40, n2 = 60. Assuming that the total output is 100, we arerequired to find out 2X . We know that

X = 1 1 2 2

1 2

n x + n xn + n =

240 8 6040 60

x

6 = 2320 + 60x

100

2x = 600 - 320

60 = 28060 =

143 = 4.670

Thus, the average profits in women’s shirt is 4.66 per cent of sales, or Re. 0.0467 per salerupee.

Example 18. The average score of girls in class X examination in a school is 67 and that of boys is63. The average score for the whole class is 64.5, find the percentage of girls and boys in the class.

Solution. Let n1 and n2 respectively be the number of girls and boys.

1X = Average score of girls = 67; 2X = Average score of boys = 63;

X = Average score of the whole class = 64.5;

X = 1 21 2

1 2

n X + n Xn +n 64.5 =

1 2

1 2

67n + 63nn +n

64.5 n1 + 64.5 n2 = 67n1 + 63n2 2.5n1 = 1.5n2 25n1 = 15n2 5n1 = 3n2.

Total number of students in the class = n1 + n2.

Percentage of girls = 1

1 2

nn +n ×100 = 15 3

1

1

n 100n / n

[ 5n1 = 3n2]

= 1

1 1

3n ×1003n +5n =

38 100 = 37.5%

and Percentage of boys = 2

1 2

n ×100n +n = 25 3

2

2

n 100/ n n

=

2

2 2

5n ×1003n + 5n = 62.5%

Hence, there are 37.5% girls and 62.5% boys in the class.

Example 19. There are 50 students in a class of which 40 are boys and rest girls. The average

100 STATISTICS (CPT)

weight of the class is 44 kg and the average weight of the girls is 40 kg. Find the average weight ofthe boys.

Solution. Here: n Number of students in a class = 50

1n = Number of boys in a class = 40; n2 = Number of girls in a class = 10

1X = Average weight of boys =.?; 2X = Average weight of girls = 40 kg.

Using2 21 2

1 2

n X + n XX =n + n , we get :

44 = 140X +10× 40

40 +10 50 44 = 40 1X + 400 1X = 45.

Example 20. The mean annual salary of all employees in a company is Rs. 25,000. The meansalary of male and female employees is Rs.27, 000 and Rs. 17,000 respectively. Find the percentageof males and females employed by the company.

Solution. The combined mean is given by:

1 21 212

1 2

N X +N XX =N +N ..........(1)

Here 12X = 25,000, X1 = 27,000, 2X = 17,000.

Let N1 + N2 = 100, where N1 denote males and N2 denote females.

N2 =1000 — N1

Substituting the values in the formula (1), we get:

25000 = N1(27,000)+17000(100 - N1)

100

2500000 = 27000N1 + 1700000 - 17000N1 10000N1 = 2500000 - 1700000 = 800000

or N1 = 80. N2 =100 - 80 = 20.

Hence, percentage of males = 80 and females = 20.


MEDIAN

Median is defined as the middle most or the central value of the variable in a set of bservations,when the observations are arranged either in ascending or in descending order of eir magnitudes.It divides the arranged series in two equal parts. Median is a position average, where as thearithmetic mean is a calculated average. When a series consists of an even number f terms, medianis the arithmetic mean of the two central items. It is generally denoted by M or Md.

CALCULATION OF MEDIAN.

When the Data is Ungrouped

Arrange the n values of the given variable in ascending (or descending) order of magnitudes.

Case I. When n is odd. In this case n +1

2 th term is the median

Median : Md or M = n +1

2 th term.

Case II. When n is even. In this case, there are two middle terms (n/2)th and (n/2 +1 )th. Thedian is the average of these two terms, i.e.,

Median : Md or M = (n/2)th term + [(n/2) + 1]th term

2Example 21. The number of runs scored by 11 players of a cricket team of a school are 5, 19, 42,11, 50, 30, 21, 0, 52, 36, 27.

Find the median.

Solution. Let us arrange the values in ascending order

0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52. ... (i)

Median : M =n +1

2

th value 11+1

2

= th value = 6th value.

Now, the 6th value in the data is 27.

Median = 27 runs.

Example 22. The weights (in kilogram) of 15 students are as follows:


31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36. 44, 45, 42, 30.

Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.

Solution. Weights of 15 students in ascending order is

27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45.

Median = thn +1

2

item = 15 +1

2 = 8th item = 35 kg.

When 44 is replaced by 46 and 27 by 25, then the new data in ascending order is

25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 46, 45.

New median = 8th item = 35 kg..

Example 23. Find the median of the following items:

6, 10, 4, 3, 9, 11, 22, 18.

Solution. Let us arrange the items in ascending order:

3, 4, 6, 9, 10, 11, 18, 22.

In this data, the number of items is n = 8, which is even.

Median : M = average of n nth and +1 th terms2 2

= Average of 82

th and 182

th terms

= average of 4th and 5th terms

9 +10 19=2 2 = 9.5.

Example 24. The following table represents the marks obtained by a batch of 12 students incertain class tests in Statistics and Physics.

Sr.No. 1 2 3 4 5 6 7 8 9 10 11 12

Marks (Statistics) : 53 54 32 30 60 46 28 25 48 72 33 65

Marks (Physics ) : 55 41 48 49 27 25 23 20 28 60 43 67

Indicate in which subject is the level of achievement higher?

Solution. The level of achievement is higher in that subject for which the median marks aremore.


Let us arrange the marks in the two subjects in ascending order,

Sr.No. 1 2 3 4 5 6 7 8 9 10 11 12

Marks (Statistics) : 25 28 30 32 33 46 48 53 54 60 65 72

Marks (Physics ) : 20 23 25 27 28 41 43 48 49 55 60 67

Since the number of students is 12, the marks of the middle-most student are the mean of themarks of 6th and 7th students.

Median marks in Statistics = 46 + 48

2 = 47.

Median marks in Physics = 41+ 43

2 = 42.

Here the median marks in Statistics are greater than the median iaarks in Physics, therefore,the level of achievement of the students is higher in Statistics.

Calculation of Median for Grouped Data

CASE I. When the series is discrete.

In this case, the values of the variable are arranged in ascending or descending order of magnitudes.A table is prepared showing the corresponding frequencies and cumulative frequencies.

Median : M = n +1

2

th value

Example 25. Calculate median for the following data:

No. of Students : 6 4 16 7 8 2

Marks : 20 9 25 50 40 80

Solution. Arranging the marks in ascending order and preparing the following table:

Table : CompulSion of Median

Marks Frequency Cumulative Frequency

9 4 4

20 6 10

25 16 26

40 8 34

50 7 41

80 2

n = f = 43


Here n = 43. Median : M = n +1

2

th value = 43 +1

2

th value = 22nd value.

The above table shows that all items from 11 to 26 have their values 25. Since 22nd item lies inthis interval, therefore, its value is 25.

Hence, Median = 25 marks.

Example 26. Find the median of the following frequency distribution:

x : 5 7 9 12 14 17 19 21

y : 6 5 3 6 5 3 2 4

Solution.

x : 5 7 9 12 14 17 19 21

f : 6 5 3 6 5 3 2 4

c.f. : 6 11 14 20 25 28 30 34

Median = Average of n2 and

n +12

items. ( Here n = 34 is an even number)

= Average of 17th and 18th items =12 +12

2 = 12.

The above table shows that all items between 12 to 14 have their values 9. Since 12th item lies inthis interval therefore its valus is 9.

Hence, Median = 9.

Example 27. The median of the observations 8, 11, 13, 15, x + 1, x + 3, 30, 35, 40, 43 arranged inascending order is 22. Find x.

Solution. The observations arranged in ascending order are:

8, 11, 13, 15, x + 1, x + 3,30, 35, 40, 43.

Total number of observations = 10 (even)

Median =

n nth item + +1 th item5th item + 6th item2 2 =

2 2

22 = (x +1)+ (x + 3)

2 22 = 2x + 4

2 22x + 4 x = 20.

CASE II. When the series is continuous.


In this case the data is given in the form of a frequency table with class-interval, etc., and thefollowing formula is used to calculate the Median.

Median class =

thn2 observations

Interpretation Formula

eM = L1 + 2 1L -LF (m - c) m =

thN2

observations

M = L2 - 1 2L -L (m- c)F (descending)

Median : (n/2) - CM = L + i,

f where

L = lower limit of the class in which the median lies,

n = total number of frequencies, Le., n = ff = frequency of the class in which the median lies,

C = cumulative frequency of the class preceding the median class,

i = width of the class-interval of the class in which the median lies.

Example 28. The following table gives the weekly expenditure of 100 families. Find the medianweekly expenditure.

WeeklyExpenditure : 0—10 10—20 20—30 39—40 40—50

(in ` )

Number of Families : 14 23 27 21 15

Solution. Let us prepare a table which gives the frequencies and cumulative frequencies.

Table : Computation of Median

Weekly Expenditure Number offamilies Cumulative

(in `) (frequency) f frequency

0—10 14 14

10—20 23 37

20—30 27 64

30—40 21 85

40 - 50 15 100

Here n = f = 100


Median = n2

th value = 100

2

th value = 50th value.

Median class = 20 — 30.

Here n2 = 50, L =20, C = 37, i = 10.

Median = L + (n/2) - C

f x i = 20 + 50 - 37

27 10 = 20 + 1327 x 10

= 20 + 4.815 = 24.815.

Hence, median = 24.815.

Example 29. Calculate the mean and median for the following data:

Height (in cm) No. of boys Height (in cm) No. of boys

135 - 140 4 155 - 160 24

140 -145 9 160 -165 10

145 -150 /8 165 -170 5

150 -155 28 170 - 175 2

Solution. Let us prepare the following table by taking 157.5 assumed mean.

Table: Computation of Median

Class xi fi c.f. ' ii

x -157.5d =5

fi'id

135-140 137.5 4 4 - 4 - 16

140-145 142.5 9 13 - 3 - 27

145-150 147.5 18 31 - 2 - 36

150-155 152.5 28 59 - 1 - 28

155-160 157.5 24 83 0. 0

160-165 162.5 10 93 .1 10

165-170 167.5 5 98 2 10

170-175 172.5 2 100 3 6

if =100 - 81

Now Arithmetic Mean =

n'

i ii=1

n

ii=1

fd(-81)a + ×i =157.5 + ×5100f


= 157.5 - 405100 = 157.5 - 4.05 = 153.45. ( a = 157.5, i = 5)

For Median : N2 =

1002 = 50.

Median calss : 150 - 155, L = 150; C = 31, i =5.

Median =

N - C2L + ×i

f

= 150 +

50 - 3128 × 5 = 150 +

9528

= 150 + 3.39 = 153.39.

PROPERTIES OF MEDIAN

1. If the two variables X and Y are related by Y = a + bx, where a and b are constants. Alsothe value of the median of X is given, then Ymd = a + bXmd.

2. For a given set of observations, the sum of deviations is minimum when the deviations andtaken from the median i.e., if A is the median then iX A is minimum.

Example 30. If x andy are related by 7x + 2y = 15 and median of x = 3, then find the median of y.

7x + 2y = 5 y = (-7/2)x + 15

ymd = (-7/2) xmd + 5 = (-7/2) x 3 + 15 = 4.5.

MODE

Mode is that value in a series which occurs most frequently. In a frequency distribution mode isthat variate which has the maximum frequency. In other words, mode represents that valuewhich is most frequent or typical or predominant. Mode is also known as Norm.

For example, in the series 6, 5, 3, 4, 3, 7, 8, 5, 9, 5, 4; we notice that 5 occurs most frequently,therefore, 5 is the mode.

Example 31. A shoe shop in Delhi had sold 100 pairs of shoes of a particular brand on a certainday with the following distribution:

Size of Shoe : 4 5 6 7 8 9 10

No. of Paris : 10 15 20 35 16 3 1

Find the mode of the distribution.


Solution. Let us prepare the table showing the frequency.

Size of Shoe : 4 5 6 7 8 9 10

No. of Paris : 10 15 20 35 16 3 1

In the above table, we notice that the size 7 has the maximum frequency, viz., 35. Therefore, 7 isthe mode of the distribution.

TYPES OF MODEL SERIES

A series of observations may have one or more modes.

Unimodal series. The series of observations which contains only one mode, is calledunimodal series.

Bimodal series. The series of observations which contains two modes is called abimodal series. In this series, the two modes are of same value of greatest density.

Trimodal series. The series of observations which contains three modes is called atrimodal series. In this series, the three modes are of same value of greatest density andhighest concentration of observations.

Ill-defined Mode. If a series of observations has more than one mode then the mode issaid to be ill-defined.

COMPUTATION OF MODE

Simple Series

In the case of simple series, the value which is repeated maximum number of times is the mode ofthe series.

Example 32. In Rajdhani Rubber Industry, Tilak Nagar, New Delhi seven labourers are receivingthe daily wages of ` 5, 6, 6 8, 8, 8 and 10. Find the modal wage.

Solution. In the series 5, 6, 6, 8, 8, 8, 10; since 8 occurs thrice and no other item occurs three timesor more than three times and hence the modal wage is ` 8.

Discrete frequency distribution series

In the case of discrete frequency distribution, mode is the value of the variable correspondingto the maximum frequency.

Example 33. A set of numbers consists offour 4 ‘s, five 5 ‘s, six 6 ‘s and nine 9’s. What is the mode?

Solution. Let us prepare the following frequency table.

Size of item : 4 5 6 9

Frequency : 4 5 6 9

Since 9 has the maximum frequency, viz., 9, therefore, 9 is the mode.

Computation of Mode in a Continuous Frequency Distribution


(i) Modal class. A class having maximum frequancy. It is that class in a grouped frequencydistribution in which the mode lies. The modal class can be determined either by inspection orwith the help of grouping table. After finding the modal class, we calculate the mode by thefollowing formula :

Mode = l + m 1

m 1 2

f - f × i,2f - f - f

where l = the lower limit of the modal class

i = the width of the modal class

f1 = the frequency of the class preceding the modal class

fm = the frequency of the modal class

f2 = the frequency of the class succeeding the modal class.

Sometimes, it so happened that the above formula fails to give the mode. In this case, the modalvalue lies in a class other than the one containing maximum frequency. In such cases, we takethe help of the following formula:

Mode = 1

1 2 m 21 2

Δl + × i,whereΔ = fm - f1,Δ = f - f ,Δ + Δ

where l, f1, f2, fm, and l have usual meanings.

The precedure of finding the mode by the above method is called Method of Interpolation.

Example 34. Find the mode for the following data:

Marks : 1—5 6—10 11—15 16—20 21—25

No. of Students : 7 10 16 32 24

Solution. From the above table, it is clear that the maximum frequency is 32 and it lies in the class16 20. Thus, the modal class is 16— 20.

Here l = 16, fm = 32, f1 = 16, f2 = 24, i = 5.

Mode = l + m 1

m 2 1

f - f i2f - f - f

= 16 + 32 -16 ×5

64 - 24 -16

= 16 + 1624 × 5 = 16 +

103 = 16 + 3.33 = 19.33

Example 35. Calculate Median and Mode for the following distribution:


Production perday (in Tons) : 21—22 23—24 25—26 27—28 29—30

No.of Days : 7 13 22 10 8

Solution. We shall first convert the given date into a continuous series.

Table: Calculations of Median and Mode

Class-boundaries No. of days (f) Cumulative

Frequency (cf)

20.5 - 22.5 7 7

22.5 - 24.5 13 20

24.5 - 26.5 22 42

26.5-28.5 10 52

28.5 - 30.5 8 60

Total N = f = 60

Since 12 N = 30 Median lies in the class 24.5 - 26.5.

Median = 24.5 + 32 - 20

22 x 2 = 24.5 + 0.91 = 25.41 tonnes.

Since the maximum frequency is 20 so the modal class is 24.5 — 20.5.

Mode = m 1

m 2 1

f - f 22 -13l+ ×i = 24.5 + ×2 = 25.36 tonnes.2f - f - f 2×22 -13 -10

Property of Mode

1. If X and Y are two variates related by: Y = a + bx, where a and b are real constants andvariates also the value of mode of X is given, then the mode of Y is given by :Ymo = a + b x mode .

EMPIRICAL RELATION BETWEEN MEAN, MEDIAN AND MODE

A distribution in which mean, median and mode coincide is called asymmetrical distribution. Ifthe distribution is moderately asymmetrical, then mean, median and mode are connected by theformula:

Mode = 3 Median - 2 Mean

or Median =13 (Mode + 2 Mean)


Example 36. If the value of mode and mean is 60 and 66 respectively, find the value of median.

Solution. We know that :

Median =13 (Mode + 2 Mean) =

13 (60 + 2 x 66) = 64.

GEOMETRIC MEAN

Geometric Mean. If x1, x2, x3,............. xn are n values of a variate x, none of them being negative,then the geometric mean G is defined as:

G = (x1 . x2 . x3 ........ xn)1/n. ... (1)

In particular, the geometric mean of 3, 9 and 27 is = (3. 9 . 27)1/3 = 9.

The difficulty of calculating the nth root is overpowered with the help of logarithms. Now takinglogarithms of both sides of (1), we get:

log G = log (x1 . x2 . x3...........xn)1/n = 1n log (x1 . x2 .............. xn)

= 1 2 nlog x + log x + ...........+ log xn

G = antilog 1 2 nlog x + log x + ...........+ log x

n

.

In the case of a frequency distribution, the geometric mean of n values x1, x2, ..... xn of a variate xoccurring with frequency f1,f2, .....,fn respectively is given by

31 2 n1/nff f f

1 2 3 nG = x .x .x .........x

log G = 1 1 2 2 n nf log x + f log x +..........f log xn

or G = antilog

n

i ii=1

f log x.

n

Thus, Geometric mean is the anti log of weighted mean of the different values of log whoseweights are their respective frequencies fi.

In the case of continuous or grouped frequency distribution, the values of the variate x are takento be the values corresponding to the mid-points of the class-intervals.

Properties of Geometric Mean

1. The logarithm of Gfor a set of observations x, x2, ...; xn is the arithmetic mean of the112 STATISTICS (CPT)

logarithm of the observations, i.e., log G = 1 2 nlog x + log x +....+ log xn

2. If all the observations of a variable assumes the same constant value K, i.e., x1 = x2 = ....xn =K, then the Geometric mean these observations is also K.

3. The G.M. of the product of two variables is the product of their G.M. i.e., if Z = XY then

G.M. of Z = (G.M. of X) x (G.M. of Y).

4. The G.M. of the ratio of two variables is the ratio of the G.M. of these two variables i.e., ifZ = X/Y. Then G.M. of Z = (G.M. of X)/(G.M. of Y).

Example 37. Compute the geometric mean for the following data:

10, 110, 120, 50, 52, 80, 37, 60.

Solution. Let us prepare the following table:

Size of item (x) log x Value of log x

10 log 10 1.000

110 log 110 2.0414

120 log 120 2.0792

50 log 50 1.6990

52 log 52 1.7160

80 log 80 1.9031

37 log 37 1.5682

60 log 60 1.7782

n = 8 log x = 13.7851

Now log G = 1n

13.7851 log x = 8 = 1.723.

G = antilog 1.723 = 52.84.

Example 38. Calculate the geometric mean for the following data:

x : 12 13 14 15 16 17

f : 5 4 4 3 2 1

Solution. Let us prepare the following table in order to calculate thç geometric mean for the givendata:

x log x f f log x


12 1.0792 5 5.3960

13 1.1139 4 4.4556

14 1.1461 4 4.5844

15 1.1761 3 3.5283

16 1.2041 2 2.4092

17 1.2304 1 1.2304

19n f f logx = 21.6029Here n = 19

log G = f log xn

= 21.6029

19 = 1.137

G = Antilog (1.137) = 13.71

Example 39. Find the geometric mean for the following data:

Marks : 0-10 10—20 20—30 30—40 40—50

No. of Students : 4 8 10 6 7

Solution. Let us prepare the following table in order to calculate the geometric mean for the givendata:

Marks Mid-value Frequency log x f x log x

(x) (f)

0 — 10 5 4 0.6990 2.7960

10—20 15 8 1.1761 9.4088

20—30 25 10 1.3979 13.9790

30—40 35 6 1.5441 9.2646

40—SO 45 7 1.6532 11.5724

Total 35f n f log x = 47.0208HARMONIC MEAN (none of them being non zero)

Definition. The harmonic mean of n items x1, x2, x3 ........xn , is defined as;


Harmonic Mean = 1 2 3 n

n1 1 1 1+ + +.........+x x x x

= n

1xi

i = 1

n

xi

For example, the harmonic mean of 2, 4 and 5 is =

3 60 3.161 1 1 19+ +2 4 5

Harmonic Mean of Frequency Distribution Data. Let x1, x2, x3 .......... , xn be n items which occurwith frequencies f1, f2, f3, ......... fn respectively. Then their Harmonic Mean is given by:

Harmonic Mean =

1 2 3 n

31 1 n

1 2 3 n

f + f + f + ..... fff f f+ + +.........+

x x x x

= 1i

ii

f.

fx

Properties of Harmonic Mean

1. If all the observations of a variable takes the same constant value K, then their H.M. is alsoK.

2. If there are two groups with n1 and n2 observations and H1 and H2 as their respective

H.M., then their combined H.M. = 1 2

1 1 2 2

n +n(n /H )+ (n /H ) .

WEIGHTED AVERAGE

When the observations under consideration have a hierarchical order of importance, we takerecourse to computing weighted average, which could be either weighted A.M. or weighted G.M.or weighted HM.

Let x1, x2, x3, be n observations with their respective weights, w1, w2, ...wn, then

Weighted A.M. = 1 1

1

w x;

w

Weighted G.M. = Ante-log 1 1

1

w log xw

; Weighted H.M. =

1

1

1

wwx

.

Example. 40. Find the H.M of 4. 6 and 10.

Solution. H. M. = 1n

T=1 i

n

x =

3(1/4)+ (1/6)+ (1/10) =

30.25 + 0.17 + 0.10 = 5.77


Example 41. The interest paid on each of three different sums of money yielding 3%, 4% and 5%simple interest p.a. respectively is the same. What is the average yield per cent on the total suminvested.

Solution :3 833 3 3×60 180H.M. = = = = . %1 1 1 20 +15 +12 47 47+ +

3 4 5 60

Example 42. A train travels first 300 kilometres at an average rate of 30 k.p.h. and further travelsthe same distance at an average rate of 40 k.p.h. What is the average speed over the wholedistance?

Solution. The average speed of train over the whole distance shall be the weighted harmonicmean of the speeds of 30 kilometres per hour over 300 kilometres and 40 kilometres per hour over300 kilometres.

Here the weights w1 = 300, w2 = 300.

We know that : 1

H.M. = (w/x)

w .

or1

average speed = 300 300+30 40

1

(300 + 300) .

or Average speed = 600

10 + 7.5 = 60017.5 = 34.286 k.p.h.

Example 43. Point out the mistake or ambiguity in the following statement, “A person goes X toY on cycle at 20 k.p.h. and returns at 24 k.p.h. His average speed was 22 kph”.

Solution. It is a question on rate where the variable is time but the constant is distance. We shallapply harmonic mean in cakulating the average speed.

Average speed = A.M. of 20 and 24 = 20 + 24

2 = 22 k.p.h.

So the average speed would not be 22 k.p.h. which is based on arithmetic mean. The appropriateaverage speed is the harmonic mean of 20 and 24.

Here1x =

120 +

124 =

11120


Harmonic Mean = N(1/x) = 2

12011 =

24011 = 21.82 k.p.h.

Example 44. If two grades of oranges sell at 10 for Re. 1 and 20 for Re. 1, respectively. Calculatethe average price per orange, stating your assumptions explicitly.

Solution. It is a question based on rate. Here the variable is orange and constant is rupee. Soharmonic mean is applicable in calculating the average rate.

Here 1x

= 1

10 + 1

20 = 320

N = number of items = 2.

Harmonic Mean = N 20= 2×(1/x) 3 = 13.33

Average rate of orange sold per Re. = 13.33.

Average price per orange = 1

13.33 = 7.5 paise.

Example 45. Find the H.M. for the following data:

x : 2 4 8 16

y : 2 3 3 2

Solution. H.M. i i

N(f /x ) =

10(2/2)+ (3/4)+ (3/8)+ (2/16) [Here N = if ]

= 10

1+ 0.75 + 0.37 + 0.125 = 10

2.25 = 4.44

Example 46. Find the weighted AM and weighted NM of first n natural weights being equal tothe squares of the corresponding numbers.

Solution. As given,

x : 1 2 3 ... n

w : 12 22 32 ... n2


Weighted A.M. = i i

i

w xw

=

2 2 2 2

2 2 2 2

1×1 + 2×2 + 3×3 +....n×n1 + 2 + 3 +......+n

= 3 2 3 3

2 2 2 2

1 + 2 + 3 +....n1 + 2 +3 +....+n

= 2[n(n+1)/2]

{n(n+1)(2n+1)/6 = 3n(n +1)2(2n +1)

Weighted H.M. =

i

i

i

wwx

=

2 2 2 2

2 2 2

1 + 2 + 3 +.....n1 2 3 n2+ + +......1 2 3 n

= 2 2 2 21 + 2 + 3 +.....n1+ 2 +3 + .....+n

=

n(n +1)(2n +1) /6

n(n +1) /2 = (2n +1)

3 .

RELATION BETWEEN ARITHMETIC MEAN, GEOMETRIC MEAN AND HARMONICMEAN

The arithmetic mean (A.M.), Geometric mean (G.M.) and Harmonic Mean (H.M.) for a given setof observations of a series are related as:

A.M. > G.M. > H.M.

Theorem : Given two positive numbers a and b, prove that A.H. = G2. Does the result hold for anyset of observations ?

Proof. For two positive numbers a and b, we have

A.M. = a + b

2 ; G.M. ab ; H.M. = 2

(1/a) + (1/b) = 2aba + b

(A.M) x (H.M.) = 2a b

x 2aba + b = ab = (G.M)2.

This result holds for only two positive observations or if the observations are in arithmeticalprogression.

Example 47. If the arithmetic mean of two numbers is 10 and their geometric mean is 8, find theirharmonic mean. Find the numbers also.

Solution. We are given that: A.M. = 10, G.M. = 8.

Also we know that (A.M.) (H.M.) = (G.M.)2

or (10) (H.M.) = 82 = 64 or H.M. = 6410 = 6.4.

Let the two numbers be a and b.


Now A.M. = a + b

2 = 10 or a +b = 20.

G.M. = ab = 8 or ab = 64.

But (a -b)2 = (a + b)2 - 4ab = 202 - 4 x 64 = 144.

(a -b) = 144 = 12.

Again, a + b = 20 and a - b = 12 implies that

2a = 32 or a = 16 and b = a - 12 = 16 - 12 = 4.

Hence, the numbers are 16, 4.

GENERAL REVIEW OF DIFFERENT MEASURES OF CENTRAL TENDENCY

The best measure of central tendency is the Arithmetic Mean (A.M.). It is rigidly defined and isbased on all the observations. It is easy to comprehend, simple to calculate and amenable tomathematical prbperties. However, A.M. has one drawback in the sense that it is very muchaffected by sampling fluctuations. In case of frequency distribution, mean cannot be calculatedfor open end classification.

Median, like A.M., is also rigidly defined and easy to comprehend and compute. But medianis not based on all the observations. It does not allow itself to mathematical treatment. However,median is not much affected by sampling fluctuations. It is the most appropriate measure ofcentral tendency for an open-end classifications.

Mode is the most popular measure of central tendency but there are cases when mode remainsundefined or the distribution has more than one mode. Unlike mean, it has no mathematicalproperty. Mode is also affected by sampling fluctuations.

G.M. and H.M. possess some mathematical properties. They are rigidly defined and based onall the observations. But they are difficult to comprehend and compute and, as such, have limitedapplications for the computation of average rates and ratios and such like things.

PARTITION VALUES

Median divides an arrayed series in ascending or descending series into two equal parts.When we are required to divide an arrayed series into more than two equal parts, the dividingplaces are known as partition values.

We know that one point divide a series into two equal parts called halves. Similarly, three


points (3) divides the arrayed series into four parts called quartiles, nine points (9) divide it intoten parts called deciles and ninety-nine (99) points divide it into one hundred parts called percentile.

Quartiles, Deciles and Percentiles are called Partition values.

For getting partition values, the most important rule is that the values must be arranged inascending order only. In case of finding the median, we could arrange the data either in ascendingor descending order but here there is no choice — only ascending order is possible for calculatingpartition values.

DIFFERENCE BETWEEN AVERAGES AND PARTITION VALUES

An average (mean, median, mode, G.M., H.M.) is the representative of whole series, but quartiles,deciles, percentiles are averages of parts of the distribution (series). For example, second quartile(D2) is the average of the second half of the series, first decile (D1) is the average of first tenobservations of the series and 35th percentile (P35) is the average of first 35 observations of theseries.

Thus, quartiles, deciles, percentiles are not averages like, mean, median and mode. Partition valueshelp us in understanding how various values are scattered around median.

Thus, partition values are used to study the scatteredness of the values of the variable in relationto the median. Therefore, the special use of partition values is to study dispersion of items inrelation to the median, i.e., it helps us in understanding the composition of a series.

QUARTILES

Quartiles. Quartiles are those values of variate which divides the series, when arrangedin ascending order, into four equal parts.

Each portion contains equal number of items. The first, second and third points are termed as firstquartile (Q1). second quartile (better named as median) and third quartile (Q3). The first quartile(Q1) or lower quartile, has 25% of the items of the distribution below it and 75% of the items aregreater than it Q2 (Median), the second quartile or median has 50% of the observations above itand 50% of the observations below it The upper quartile or third quartile (Q3) has 75% of theitems of the distribution below it and 25% of the items are above it

Note. It must be noted that Q1 <Q2 < Q3.

Computation of Quartiles

CASE I. Computation of Quartile for Individual Series


Let x1, x2, x3 ,..........., xn be n values of a variate X. We can compute quartiles of these values bythe following method.

METHOD

Step I. Arrange the given data in ascending order of magnitude.

Step II. Find the total number of observations. Let it be N.

Step III. Calculate the three quartiles Q1, Q2, Q3 by the following formulae.

Q1 = Value of thN+1

4

observation in the arrayed series

Q2 = Value of thN+1

2

observation in the arrayed series if n is odd.

Q2 = Mean of the values of thN

2

and thN +1

2

observation if n is even.

Q3 = Value of 4

thN+13

observation in the arrayed series .

Example 48. Compute Q1, Q2 and Q3 of the following data:

13, 14, 7, 12. 17, 8, 10, 6, 15, 18, 21, 20

Solution. Arranging the given data in ascending order, we get:

6,7, 8, 9, 10, 12, 13, 14, 15, 17, 18, 20, 21

Here n = 13.

Q1 = Value of thN+1

4

term = Value of th13 +1

4

, i.e., 3.5th term ( n = 13)

= Value of 3rd term + 12 (value of 4th term - 3rd term)

= 8 + 12 (9-8) = 8 +

12 = 8.5.


Q2 = Median = Value of thN+1

2

term

= Value of th13 +1

2

, i.e., 7th term = 13.

Q3 = Value of thN+13

4

= Value of th13 +13

4

, i.e., 10.5th term

= value of 10th term + 12 (Value of 11th term - value of 10th term)

= 17 + 12 (18-17) = 17 + 0.5 = 17.5.

Hence, Q1 = 8.5, Q2 = 13 and Q3 = 17.5.

CASE II. Computation of Quartiles for a Discrete Frequency Distribution

Let the variate X takes n values x1, x2, x3 ,........., xn with frequencies f1, f2, f3. .......... fn respectively.Then we calculate quartiles Q1,Q2, Q3 by the following method.

METHOD

Step I. Compute the cumulative frequencies.

Step II. Find N = 1

n

ii

f

Step III. Calculate the quartiles as under:

Q1 = Find the cumulative frequency just greater than N4 . Determine the

corresponding value of the variable X This value is the lower quartile Q1.

Q2 = Find the cumulative frequency just greater than N2 . Determine the

corresponding value of the variable X. This value is the middlequartile Q2, i.e., the median.

Q3 = Find the cumulative frequency just greater than 3N4 . Determine the

value of the variable X This value is the third quartile Q2.

Example 49. The following data relate to the sizes of shoes sold at a store during a given weekFind upper and lower quartile from the given data.


Find Q1and Q3 of the following series:

Size of shoes Frequency Size of shoes Frequency

4.0 10 6.0 40 4.5 18 6.5 15 5.0 22 7.0 10 5.5 25 7.5

8.0 7

Solution. Let us arrange the values of the variate in ascending order and prepare the cumulativefrequency table.

Table: Calculation of Q1 and Q3.

Size of shoes Frequency c.f

4.0 10 10

4.5 18 28

5.0 22 50

5.5 25 75

6.0 40 115

6.5 15 130

7.0 10 140

7.5 8 148

8.0 7 155

Here N= 155.

Lower Quartile Q1 = Size of (N+1)

4 th item = Size of 155 +1

4th item

= Size of 39th item = 5 [ 39 lies in the row with c.f.50]

Upper Quartile Q3 = Size of 3(N+1)

4 th item = Size of 3 155 +1

4th item

= Size of 1117th item = 6.5 [ 117 lies in the row with c.f.130]

Hence, Q1 = 5 and Q3 = 6.5.

CASE Ill. Computation of Quartiles for a Frequency Distribution with Class-Intervals

The following method is used to compute quartiles Q1, Q2, Q3 for a continuous distribution.

METHOD

Step I. Compute the cumulative frequency of the given distribution. Let N = 1

n

ii

f


Step II. Compute 4iN

, where i = 1 for lower quartile Q1, i = 2 for middle quartile Q2

(or median), i = 3 for upper quartile Q3.

Step III. Find the cumulative frequency just greater than 4iN

and the correspondingclass.

This class is called the quartile class.

Step IV. Use the following formula to calculate Q1, Q2, or Q3.

i(i×N/4) - CQ = L + ×h,i =1,2,3.

fwhere L = Lower limit of the class in which a particular quartile lies,

f = Frequency of the class-interval in which a particular quartile lies

h = Width of the class-interval of the class in which a particular quartile lies,

C = Cumulative frequency of the class preceding the class in which the quartilelies.

Example 50. Calculate median and the three quartiles from the following data:

Marks: 0—10 10—20 20—30 30—40 40—50 50—60 60—70 70—80

No. of

Students: 5 7 8 12 28 22 10 8

Solution.

Size of shoes Frequency c.f

0—10 5 5

10—20 7 12

20—30 8 20

30—40 12 32

40—50 28 60

50—60 22 82

60—70 10 92

70—80 8 100

Median = Size of n2

th item = 100

2 th item = Size of 50th item. [Here n = 100]

Now, the median lies in the group 40— 50 and L = 40; n2 = 50; C 32; h = 10; f = 28.


Median (Q2) = L + 2(n / ) Cf

h = 40 + 50 - 32

28 x 10

or Median = 40 + 18×10

28 = 40 + 6.43 = 46.43.

Again Q1 = size of n4

th item = 100

4 th, i.e. size of 25th item.

Q1 lies in the group 30 -40 and L = 30 - 40 and L = 30, C = 20, h =10, f =12

Q1 = L + (n/4) - C

f h = 30 + 25 - 20

12 10 = 34.17

Also Q3 = Size of 3n4

th item, i.e., size of 7th item, i.e.,

Q3 lies in the group 50 - 60 and L = 50, C = 60, h =10 and f =22.

Q3 = L + (3n/4) - C

f h = 50 + 75 - 60

22 10 = 50 + 1522 10 = 56.81.

Hence, Q1 = 34.17, Median = Q2 = 46.43 and Q3 = 56.81.

DECILES

The value of the variable which divides the series, when arranged in aséending order, into 10equal parts is called a dedile. The nine points which divide the given series (when arranged inascending order) into ten parts are called deciles.

Deciles are denoted by D1, D2, D3, ...., D9. The first decile, is the value of the variable such that itexceeds 10% of the observations and is exceeded by 90% of the observations. Similarly, D6, thesixth decile, has 60% observations before it and 40% observations after it.

The fifth decile, D5, is the median of the given data.

Computation of Deciles

CASE I. Computation of Deciles for Individual Series

In this case, the kth decile is given by

= Value of k n +110

th term, k = 1, 2, 3, 4 , ........... 9,

when the series is arranged in ascending order.


Example 51. Compute D3 of the following data.

13, 14, 7, 12, 9, 17, 8, 10, 6, 15, 18, 21, 20

Solution. Arranging the given data in ascending order, we get:

6, 7, 8, 9, 10, 12, 13, 14, 15, 17, 18, 20, 21.

Here n = 13.

D3 = Value of 3thn +1

10

term = Value of 3 3th13 +1

10

term

= Value of 4.2th term = Value of 4th term +15 (value of 5th term — value of 4th term)

= 9 +15 (10-9) = 9.2

Thus, 3rd decile = D3 = 9.2.

CASE II. Computation of Deciles for a Discrete Frequency Distribution

METHOD

Step I. Arrange the given data in ascending order.

Step II. Compute the cumulative frequencies

Step III. Find N = ifStep IV. Find

iN10 to compute D, the ith decile, i = 1, 2, 3 ........... 9.

Step V. Find the cumulative frequency just greater than iN10 .Then the corresponding

value of variable is the ith decile i = 1, 2, 3, ..............., 9.

Example 52. Compute D3 D6 and D7from the following data.

Marks : 10 20 30 40 50 80No. of Students : 4 7 15 18 7 2.

Solution. The cumulative frequency distribution is given below:

Marks No. of Students (f) Cumulative Frequency (c.f.)

10 4 4

20 7 11

30 15 26

40 18 44

50 7 51

80 2 53

iN = f = 53


Third Decile = D3, Here 3N10 =

3×5310 =

15910 = 15.9.

The cumulative frequency just greater than 3N10 = 15.9 is 26 and the corresponding value of the

variable is 30. Thus D3 = 30.

Sixth Decile = D6, Here 6N10 =

6×5310 = 31.8.


variable is 40. Thus = 40. Thus D6 = 40.

Seventh Decile = D7. Here 7N10 =

7×5310 = 37.1.


variable is 40. Thus D7 = 40.

Hence, D3 = 30, D6 = 40 and D7 = 40.

CASE III. Computation of Deciles for a Frequency Distribution with Class-Intervals.

METHOD

Step I. Compute the cumulative frequency table. Let N = if .Step II. Compute

iN10 to find Di, the ith decile i = 1, 2, 3, ........ 9.

Step III. Find the cumulative frequency just greater than iN10 and the corresponding class.

This class is 6th called the dedile class.

Step IV. Use the formula

i

iN - C10D = L + ×h.

f

where L = Lower limit of the decile class

C = Cumulative frequency of the class preceding the decile class

f = frequency of the ith decile class

h = width of the ith decile class.

Example 53. Compute Q3 and D7 for the following frequency distribution:


Marks : 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

No. of Students : 3 10 17 7 6 4 2 1

Solution. The cumulative frequency distribution is given by

Marks No. of Students (f) Cumulative Frequency (c.f)

0—10 3 3

10—20 10 13

20—30 17 30

30—40 7 37

40—50 6 43

50—60 4 47

60—70 2 49

70—80 1 50

N = if = 50

Third Quartile (Q3) : N= 50 3N4 =

3×504 = 37.5.

The cumulative frequency just greater than 3N4 = 37.5 is 43, so 40 - 50 is the upper quartile class

such that: L = 40, h = 10, f = 6, f = 37

3

3N - C 37.5 - 374Q = L + ×h = 40 + ×10 = 40.83f 6

Seventh Decile = D7. Here N = 50;7N 7 50= = 35.10 10

The cumulative frequency just greater than 35 is 37 and the corresponding class is 30 —40. Thus,the 7th decile class is 30— 40, such that L 30, h = 10,f = 7, C = 30.

7

7N - C10D = L + ×h

f =

35 - 3030 + ×10 = 27.14.7

Hence, Q3 = 40.83 marks and D7 = 24.14 marks.

PERCENTILES


The value of the variables which divides the series, when arranged in ascending order, into 100equal parts are called percentiles. There are 99 percentiles denoted by P1, P2, P3, P4, .............., P99

respectively.

The ninety-nine points which divide the given data, when arranged in ascending order, intohundred equal parts are called percentiles of the data.

P30 is the value of variable such that it exceeds 30% of the observations and is exceeded by 70%.Similarly, P50 is the median.

Computation of Percentiles

CASE I. Computation of Percentile of Individual Series.

In this case, the kth percentile is given byth

kn +1P = Value k100

term, when

arranged in ascending order, k = 1, 2, 3, ........., 99.

CASE II. Computation of Percentiles for Discrete Frequency Distribution

Step I. Arrange the given data in ascending order.

Step II. Compute the cumulative frequencies.

Step III. Find N = if .

Step IV. Compute 100i N

to compute Pi, the ith percentile, i = 1, 2, 3,....., 99.

Step V. Find the cumulative frequency just greater than iN

100 and the corresponding valueof the variable. This value is the Di, the ith percentile of the given data.

Example 54. Calculate D4 and P30 from the following data.

Marks : 20 25 30 35 45

No. of Students : 5 14 20 11 10.

Solution. We have

Table: Cumulative Frequency TableMarks Frequency (f) Cumulative Frequency (c.f)

20 5 525 14 1930 20 3935 11 5040 10 60

Total: N = f = 60


D4 = Value of th4N

10

item = Value of th4×60

10

item

= value of 24th item = 30. [ 19 < 24 < 39]

[The cumulative frequency just greater than 4N10 = 24 is 39 and the corresponding value of

marks is 30]

P30 = Value of th30N

100

item = Value of th30× 60

10

item

= Value of 18th item = 25. [ 5 < 18 < 19]

[The cumulative frequency just greater than 30N100 = 18 is 19 and the corresponding value of

marks is 25]

Case III. Computation of Percentiles for a Frequency Distribution with Class-Intervals. StepI. Compute the cumulative frequency table. Let N = if .

Step II. Compute i×N100 for Pi, the ith percentile, i = 1, 2, 3, ...... 99.

Step III. Find the cumulative frequency just greater than i×N100 and the corresponding

class.This class is called ith percentile class.

Step IV. Use the formula : i

i×N - C100P = L + ×h,

fwhere

L = lower limit of percentile class

C = cumulative frequency of the class precending the percentile class

f = frequency of the percentile class

h = width of the percentile class.

Example 55. Calculate Q1,Q3, D2, P5. P90 from the following data:

Marks : 0—10 10—20 20—30 40—60 60—80 80—100

Number of Students : 8 10 22 25 10 5

Solution. Let us prepare the following table showing the frequency and cumulative frequency.


Marks Number of Students (f) Cumulative Frequency (c.f)

0—10 8 8

10—20 10 18

26—40 22 40

40—60 25 65

60—80 10 75

80—200 5 80

Total = f = 80.Here n = 80.

Now Q1 =n4

th value =804


Here Q1 - class = 20 -40, L = 20, n4 = 20; C = 18, f = 22, i = 20

Q1 = L + (n/4) - C

f × i = 20 + 20 -18 × 20

22 = 20 + 1.8 = 21.8 marks.

Again Q3 =3×n

4

th value = 3× 80

4


Here Q3- class = 40 - 60, L =40, 3×n

4 = 60, C = 40, f = 25, i = 20.

Q3 = L + (3 n)/ 4 - C

f

× i = 40 + 60 - 40

25 ×20 = 40 + 16 = 56 marks.

Again D2 = 2×n10 th value =

2×8010 th value = 16th value.

Here D2 - class = 10 - 20; L = 10, 2×n10 = 16, C = 8, f = 10, i = 10.

D2 = L + (2 n)/ 10 - C

f

× i = 10 + 16 - 8

10 × 10 = 10 + 8 = 18 marks.

Again P5 = 5×n100

th value = 5 8100

= 4th value.


Here P5 - class = 0 - 10; L = 0, 5×n100 = 4, C = 0, f = 8, i =10.

P5 = L + (5 n)/ 100 - C

f

× i = 0 + 4 - 0 ×10

8 = 5 marks.

Again P90 = 90×n100

th value = 90×80

100 th value = 72nd value.

Here P90 - class = 60 -80; L = 60, 20×n100 = 72, C = 65, f = 10, i = 20

P90 - L + (90 n)/ 100 - C

f

i = 60 + 72 - 65

10 20 = 60 + 14 = 74 marks.

Example 56. Given below are the marks obtained by 50 students appearing for an admission test:

Marks : 0—10 10 - 20 20—30 30—40 40—50

No. of Students : 6 8 20 9 7

If cut off point was 34, find the percentage of students scoring more than 34 marks.

Solution. Suppose x % students scored 34 marks. Then, number of students scoring class 34

is x xth percentile 34 or Px = 34.

Table: The Cumulative Frequency

Marks Number of Students (f) Cumulative Frequency (c.f)

0—10 6 6

10—20 8 14

20—30 20 34

30—40 9 43

40—50 7 50

N = if = 50.Now, Px = 34 which lies in the interval 30— 40, so, the xth percentile class is 30—40 such that L =30, h = 10, f = 9, N = 50 and C = 34.


xN x ×50- C - 34

100 100 Px = L + ×h 34 = 30 + ×10f 9

4 × 9 = x - 342

× 10 5x = 376 x = 75.2.

Thus, 75.2% students scored upto 34 marks and hence the remaining 24.8% students scoredmore than 34 marks.

MEASURES OF DISPERSIONDISPERSION

The meaning of dispersion is scatteredness. It helps in finding out the variability of the data orscatteredness of individual items from an appropriate measure of central tendency in agiven distribution. In other words, the degree to which numerical data tend to spreadabout an average value is called the dispersion of the data.

These are two types of measures of dispersion:

1. Absolute measures of dispersion 2. Relative measures of dispersion

I. Absolute measures of dispersion. The absolute measures of dispersion are

(i) Range (ii) Quartile Deviation

(iii) Mean Deviation (iv) Standard Deviation

II. Relative measures of dispersion. Corresponding to these four absolute measures ofdispersion, we have the follOwing four relative measures of dispersions

(i) Coefficient of Range (ii) Coefficient of Quartile Deviation

(ii:) Coefficient of Mean Deviation

(iv) Coefficient of Variation or Coefficient of Standard Deviation

Difference between Absolute and Relative Measures of Dispersion

The following are the points of distinction between the absolute and relative measures ofdispersion

I. Absolute measures are dependent on the units of the variable under considerationwhereas the relative measures of dispersion are unit free.

II. The relative measures are used only for the purpose of comparison between two ormore series with varying size or the number of items or varying central values orvarying units of calculations.

III. Compared to absolute measures of dispersion, relative measures of dispersion aredifficult to compute and comprehend.


Characteristics for an Ideal Measure of Dispersion

An ideal measure of dispersion should be properly defined, and easy to comprehend, simple tocompute, based on all the observations and unaffected by sampling fluctuations. It shouldbe amenable to some desirable mathematical treatment

RANGE

It is the simplest measure of dispersion. It is the difference between the minimum andmaximum items of,the series.

For example, in the series 20, 21, 22, 25, 30, 32, 37, 47, 65 the range is 65 20 = 45. AbsoluteRange or Range = Xmax- Xmin or Range = L - S

where L is the largest value and S is the smallest value of the term.

Coefficient of Range or Relative Range = SLSL

valuesextremestwotheofSumRangeAbsolute

Range for Continuous Frequency Distribution

Range in continuous series is determined by the following two methods:

Method I.Find the differnce between the upper limit of the highest wage class and the lowerlimit of the lowest wage class.

Method II. Subtract the mid-point of the lowest wage class from the mid-point of the highestwage class.

Properties of Range

1. Range remains unaffected due to a change of origin but is affected in the same ratio due toa change in scale, i.e., itfor any two constants a and b, the two variables x andy are relatedbyy = a + bx, then the range ofy is given by: Ry = xb R ,

2. If all the observations assumed by a variable are constant, say then the range is zero.

Example 1. Find the range and coefficient of range of the weights of 10 students from thefollowing data:

41 20 15 65 73 84 53 35 71 55

Solution. Arranging the data in the ascending order, we get:

15 20 35 41 53 55 65 71 73 84

Here largest value: L = 84, Smallest value: S = 15

Range = L - S = 84 - 15 = 69


Coefficient of Range = 9669

15841584

SLSL

= 0.696

Example 2. Find the range and the coefficient of range of the marks obtained by 100 studentsgiven below.’

Marks No. of Students Marks No. of Students

0—10 13 40—50 11

10—20 8 50—60 23

20—30 7 60—70 18

30—40 10 70—80 10

Solution. Here L = 80, S = 0

Range = L - S = 80 - 0 = 80

Coefficient of Range = 18080

SLSL

Example 3. If the relationship between x andy is given by 3x + 2y = 13 and the range of x is 12,what would be the range of y.

Solution. 3x + 2y = 13 y = (- 3/2) x + (13/3)

We know that if y = a + bx, then range of y = Ry= | b | x Rx

Ry= | -3/2 | Rx = (3/2) x 12 = 18.

QUARTILE DEVIATION OR SEMI-INTER QUARTILE RANGE

Quartile Deviation is a measure of dispersion based on the Upper Quartile (Q3) and LowerQuartile (Q1) of a series. It is half of the difference between the upper quartile and thelower quartile. Since the difference is the range between these two quartiles so it is alsocalled inter-quartile range. The half ofthis range is semi4nter-quartile range. The quartiledeviation is also known as Semi-inter-quartile Range.

Quartile Deviation =2

QQ 13

where Q1 = first or lower quartile; Q3 = third orupper quartile.

In most cases, the central 50% observations of a series tend to be fairly typical. Thus quartiledeviation can be used as a suitable measure of dispersion in such cases. It can also becomputed from a frequency distribution with open-end classes at both ends. It is importantto note that range cannot be computed from a frequency distribution with open end classesat both ends, so, quartile deviation is a more suitable measure of dispersion than range.


Coefficient of Quartile Deviation :13

13QQQQ

Coefficient of Q.T.

There is another relative measures of dispersion where the coefficient of respective absolutemeasure is multiplied by 100 to convert the figures into percentages or in relation to 100.

Coefficient of Quartile Variation = Coefficient of Quartile Deviation x 100

100xQQQQ

13

13

PROPERTIES OF QUARTILE DEVIATION

1. It is the best measure of dispersion for open-end distribution.

2. It is less affected due to sampling fluctuations.

3. Quartile deviation remains unaffected due to a change of origin but it is affected in the.same ratio due to change in scale, i.e., if the two variables x and y are related as y = a + bx,where a and b are constants, then

Q.D. of y = b x ( Q.D. of x )

Example 4. Calculate the quartile deviation and coefficient of quartile deviation and thecoefficient of quartile variation from the following data

Age in years : 20 30 40 50 60 7080

of Members : 3 61 132 153 140 51 3

Solution. We have the following table

Age in years: (x) No. of Members: (f) Cumulative Frequency: (c.f)

20 3 3

30, 61 64

40 132 196

50 153 349

60 140 489

70 51 540

80 3 543

Here N= 543.


1N 1 543 1Q Value of th item 136th tiem 40 years.

4 4

.years60tiemth408136x3itemth4

1N3ofValueQ3

Quartile Deviation = 2

40602

QQ 13

= 10 years.

Coefficient of Q.D. = 2.010020

40604060

QQQQ

13

13

Coefficient of Quartile Variation =3

3 1

1

Q Q 100 0.2 100 20%.Q Q

Example 5. From the following data, calculate the quartiledeviation and the coefficient of quartile variation

Wages : 0—10 10—20 20—30 30—40 40—50 50—60 60—70 70—80

No. of workers : 20 45 85 160 70 55 35 30

Solution. We have the following table:

Wages: (x) No. of Workers: (f) Cumulative Frequency: (c.f)

0—10 20 20

10—20 45 65

20—30 T 85 150

30—40 160 310

40—50 70 380

50—60 55 435

60—70 35 470

70—80 30 500

Here, N = Total number of workers = 500.

COMPUTATION of Q1 and Q3

For Q1, we have = 4N

= 125, cumulative frequency just greater than 125 is 150, and therefore,it lies in the class 20 - 30, whose frequencyf= 85. Also C = 65 and i = 10

1(N / 4) C 125 65Q L x i 20 x 10 20 7.06 27.06

f 85


Similarly, for Q3, we have: 3754N3 , L = 40, C= 310, f= 70 and i = 10.

29.4929.94010x70

31037540ixf

C)4/N3(LQ3

Quartile Deviation = 115.11223.22

206.2729.49

2QQ 13

Coefficient of Quartile Variation = 3

3 1

1

Q Q 100Q Q

= 49.29 27.06 22.2349.29 27.06 76.35

x 100 = 29.12%.

Example 6. Calculate the appropriate measure of dispersion from the following data:

Income : Less than 35 35—37 38—48 41—43 over 43

No. of Persons: 14 62 99 18 7

Solution. Since the given data has frequency distribution with open end classes at the both ends,so quartile deviation is the appropriate measure of dispersion. Making the given series as inclusiveseries, we have the following table

Class-Interval Frequency: (f) Cumulative Frequency: (c.f)

Less than 34.5 14 14

34.5- 37.5 62 76

37.5-40.5 99 175

40.5—43.5 18 193

Over 43.5 7 200

Here N= 200, 4N

= 4200

= 50, 4N3

= 150.

Since, the cumulative frequenc-y just greater than 50 is 76, so Q1 lies in the class 34.5 37.5.Similarly, cumulative frequency just greater than 150 is 175, so Q3 lies in the class 37.5 40.5.

Now, Q1 = L + ixf

C)4/N( = 34.5 + 3x

621450

= 34.5 + 1.74 = Rs. 36.24.

Similarly, for Q3, we have


L=37.5, 4N3

=150, C=.76, i = 3, f = 99.

Q3 = L + ixf

C)4/N3( = 37.5 + 3x

9976150

= 37.5 + 2.24 = 39.74.

Quartile Deviation = 25.3

224.3674.39

2QQ 13

= Rs. 1.75.

Example 7. If the quartile deviation of x is 8 and 3x + 6y = 15, what is the quartile deviation of

Solution. 3x + 6y = 15 y = (15/6) (3/6) x.

Q.D. ofy =| -3/6 | x ( Q.D of x ) = (1/2) x 8 = 4

MEAN DEVIATION OR AVERAGE DEVIATIONMean deviation of a set of observations of a series is the arithmetic mean of all the deviations,(without their algebraic signs), taken from its central value (mean or median or mode). In otherwords, it is the average of the modulus of the deviations of the observations in a series takenfrom mean or median or mode. Mean deviation is one of the calculated measure in which all thevalues are considered in its calculations.

Coefficient of Mean Deviation = AAaboutdeviationMean

where we take A as mean or median or mode and accordingly, we get the mean deviationabout mean or median or mode.

Mean Deviation about A

Coefficient of Mean Variation = 100xA

AaboutdeviationMean

where A mean or median or mode.

Properties of Mean Deviation

1. If all the values taken by a variable x is a constant k, then mean deviation is zero.

2. Mean deviation takes minimum values when the deviations are taken from the median.

3. Mean deviation remains unchanged due to change of origin but changes in the same ratiodue to change in scale, i.e., if y = a + bx, a and b being constants, then

M.D.ofy = | b | x ( M.D of x )


Example 8. If a relationship between x andy is given by: 2x + 3y = 10 and M.D. of x is 15, whatis the M.D. of y

Solution.2x + 3y = 10 y = (—2/3) x + (10/3)

If y = a + bx, then M.D. of y l b | ( M.D. of x )

M.D. ofy = | -2/3 | x M.D. of x = (2/3) x 15 =10.

Mean Deviation for ungrouped data

While calculating the mean deviations, the algebraic sign of the deviation is always taken aspositive, because the sum of deviations with their algebraic signs, + and — ,from the arithmeticmean is always zero. We have the following rule

(i) Signs (plus and minus) of deviations are disregarded and absolute values of the deviationsare summed up. Symbolically, we use | Xi — X | which means the deviation of the ithobservation of x from the central value (which may be mean or median or mode) withpositive sign. Here the vertical lines stand for positive value.

Now add

up all n observations to get |XX| in

1i

.Then

Mean Deviation: MD x =

n

1ii |xx|

n1

, where x is the arithmetic mean.

Similarly, Mean deviation from the median for ungrouped data is denoted by MD med or d med

where

MDmed = |Mx|n1

di

(a) Mean deviation from the mode Z for ungrouped data

dv or MDv = Zx|n1

i

But in practice, we! generally use arithmetic mean as it is amenable to algebraic treatments.Median at times is not an actual quantity, while mode of a series may not exist, i.e., it may be illdefined.

Example 9. Calculate the mean deviation about the mean for the following series

15, 20, 17, 19, 21, 13, 12, 10, 17, 9, 12.

Solution. Here n = 11 and therefore


Mean = 111291710121321 19 17 20 15

= 11165

= 15 = M (say)

| d | | X M| X 15 = 0 + 5 + 2 + 4 + 6 + 2 + 3 + 5 + 2 + 6 + 3 = 38.

Mean deviation = N

|d| = 1138

= 3.455

Example 10. The marks obtained by 10 students in an examination were as follows:

70, 65, 68, 70, 75, 73, 80, 70, 83, 86.

Find the mean deviation about the mean and coefficient of variation.

Solution. Mean= 7410740

1086838075737070706865

nX

Mean Deviation about mean = n

|XX|

=6.5

1056

1012946114694

Coefficient of Variation = %57.7100X74

6.5100XMean

Deviation Mean

Mean Deviation for G!ouped data

Let x1,x2, x3,....., xn occur with frequencies f1,f2,f3......,fn respectively and let nf and

M can be either Mean or Median or Mode, then the mean deviation is given by the formula:

Mean Deviation =

f|Mx|f

= n

|d|f , where d = x M and f n. Example 11. Find the mean deviation from (i) mean and median for the following data:

Marks: 20 18 16 14 12 10 8 6

No. of Students: 2 4 9 18 27 25 14 1


Also calculate coefficient of variation.

Solution. (i) Mean deviation from mean.

Here f = 1+ 14 + 25 + 27 + 18 +9 +4 +2= 100.

xxf = 6 + 112 + 250 + 324 + 252 + 144 + 72 + 40 = 1200

Arithmetic mean: X =

fxf

= 1001200

= 12

Also f | d | = 6 + 56 + 50 + 0 + 36 + 36 + 24 + 16 = 224.

Mean deviation (about mean) = n

|d|f = 100224

= 224.

Coefficient of Variation = M.DX x 100 = 12

24.2x 100 = 18.67%.

(ii) Mean deviation about median. Let us prepare the following table in order to calculate the median:

Table: Computation of Mean Deviation about Median

Marks: (x) Frequency : f Cumulative frequency : c.f | d | = | x - 12 |

6 1 1 6

8 14 15 56

10 25 40 50

12 27 67 0

14 18 85 36

16 9 94 36

18 4 98 24

20 2 100 16

f =100 |d|f =224

Median = Average of 2n

th and

1

2n

item. = Average of 50th and 51st item = 12.

Mean Deviation (about median) = n

|d|f = 100224

= 2.24

Coefficient of Variation (mean deviation) = MedianD.M

x 100 = 1224.2

x 100 = 18.67%.


Example 12. Calculate the mean deviation about the mean for the following data:

x : 5 15 25 35 45 55 65

f : 8 12 10 8 3 2 7

Also find the coefficient of Mean Variation.

Lettheassumedmeanbe A=35. Let = d = iAx

= 1035x

Here fd = - 24 - 24 -10 + 0 + 3 + 4 + 21 = - 30

Mean : 2910x503035ix

ffd

Ax

)29x(f)xx(f 192 + 168 + 40 + 48 + 52 + 252 = 800

MeanDeviation = 16)800(501|xx|f

N1

.

Coefficient of Mean Variation = 2916100x

xD.M

x 100 = 55.17%

Example 13. Calculate the mean deviation from the mean for the following data:

Class lnterval: 0—4 4—8 8—12 12—16 16—20

Frequency: 4 6 8 5 2

Solution. Let us prepare the following table by assuming that the frequencies in each class arecentred at its mid-value.

Table: Computation of Mean Deviation

Class interval Mid-value: (x) f fx | d | = x 9.2 f I d I

0—4 2 4 8 7.2 28.8

4—8 6 6 36 3.2 19.2

8—12 10 8 80 0.8 6.4

12—16 14 5 70 4.8 24.0

16—20 18 2 36 8.8 17.6

25f 230fx 0.96|d|f


Now Arithmetic Mean: x = 2.925

230fxf

Mean Deviation: = 84.32596

f|d|f

STANDARD DEVIATION

Standard deviation is the most important and conmionly used measure of dispersion. It measuresthe absolute dispersion or variability of a distribution. A small standard deviation means a highdegree of uniformity in the observations as well as homogeneity of the series, it is extremelyuseful in judging the representativeness of the mean.

Standard deviation is the positive square root of the average of squared deviations taken fromarithmetic mean. It is, generally, denoted by the Greek alphabet a or by S.D. or s.d. Let x be arandom variate which takes on n values, viz., x1,x2, x3...... , xn, then the standard deviation ofthese n observations is given by

Standard Deviation:

,n

)xx( 2 where

nx

x = is the mean of these observation

ALTERNATIVELY

22

nx

nx

Also n

)xx( 2 = 2

2x

nx

nx

x

But the formula

22

nx

nx


is used when the items are very small. On the other hand the relevance of this method isparticularly useful when computers are used whereby the values, even when they are of highmagnitude, can be used directly for calculating a.

Standard deviation is also khown as Root Mean Square. Deviation.

Variance

The variance is the square of standard deviation and is denoted by 2

Coefficient of Standard Deviation = x

where x is the arithmetic mean of the given series. It is a relative measure of standarddeviation.

Coefficient of Variation

Coefficient of Variation. It is a relative measure of dispersion. It is, generally, denoted by C.V.and is given by the formula:

Coefficient of Variation : C.V. = x

100,

where a is the standard deviation and x is the mean of the given series.

It is important to note that the coefficient of variation is always a percentage.

The coefficient of variation has a great practical significance and is the best measure for comparingthe variability of the two series. The series or group fol which the coefficient of variation isgreater is said to be more variable (or less consistent). On the other hand, the series for whichthe variation is less is said to be less variable (or more consistent).

Properties of Standard Deviation

1. If all the observations assumed by a variable are constant, i.e., equal, then the S.D. is zero.This means that if all the values taken by a variable x is a constant k, then the standarddeviation is zero.

2. S.D. remains unaffected due to a change of origin but is affected in the same ratio due to achange of scale, i.e., if there are two variables x andy related asy = a + bx for any twoconstants a and b, then S.D. of y is given by y = | b | x ,

Example 14. If the two variables x andy are related as : 2x + 3y = 10 and the standard deviationof x is 15 then what is the standard deviation of y.


Solution. 2x + 3y = 10 y = 10 + (— 2/3) x.

Also, S.D.of y = | b | x (S.D.of x ) ; where y = a + bx.

S.D. of y = | - 2/3 | x (S.D. of x) (2/3) x 15 = 10.

Example 15. If AM and the coefficient of variation are 10 and 50 respectively, what is thevariance of (15 - 2x) ?

Solution. C.V.(x) = .5100x10

50100xx x

xx

Variance of x = 252x

S.D.of(15—2x) = | -2 | (S.D of x ) = 2 x 5 = 10 = Sy ( say )

Variance of (15 — 2x) = Sy2 = 102 = 100.

Example 16. Find the standard deviation of 3, 4, 5, 6.

Solution. Here n=4, x = 3 + 4 + 5 + 6 = 18.2x = 32 + 42 + 52 + 62 = 9 + 16 + 25 + 36 = 86.

2 22x x 86 18n n 4 4

12.125.125.205.21)5.4(5.21 2 nearly.

CALCULATION OP STANDARD DEVIATION - INDIVIDUAL OBSERVATIONS

When the data under consideration consists of individual observations, the standard deviationmay be computed by any of the following two methods

(a) By taking deviations of the items from the actual mean.

(b) By taking deviations of the items from an assumed mean.

When the Deviations are Taken from the Actual Arithmetic Mean

This method is known as Direct Method.

Direct Method

In case of simple series, the standard deviation can be obtained by the formula :

n)xx( 2

i or

nd2

, where d = xi - x


and xi = value of the variable or observation, x = arithmetic mean, and

n = total number of observations.

Example 17. Find the standard deviation of 16, 13, 17, 22.

Solution. Here A.M. = x = 422171316

= 468

=17. Here d = x - x = x - 17.

Let us prepare the following table in order to calculate the standard deviation.

x : 16 13 17 22

d : -1 -4 0 5

d2 : 1 16 0 25

d2 = 1 + 16 + 25 = 42.

Now = n

)xx( 2 =

nd2

= 4

42= 3.24.

When the Deviations are Taken from the Assumed Mean

This method is also known as Short-cut Method.

Short-cut Method

This method is applied to calculate standard deviation, when the mean of the data comes out tobe a fraction. In that case, it is very difficult and tedious to find the deviations of all observationsfrom the actual mean by the above method. The formula used is:

22

nd

nd

where d = x - A, A = assumed mean, n = total number of observations.

Example 18. Find the standard deviation of the following data :

48, 43, 65, 57, 31, 60, 37 48, 59, 78.

Solution. Let us prepare the following table in order to calculate the value of S.D. by assumingA = 50. Also, d = x - A = x - 50 and n = 10.

x : 48 43 65 57 31 60 37 48 59 78

d : -2 -7 15 7 -19 10 -13 -2 9 28

d2 : 4 49 225 49 361 100 169 4 81 784


d = 26, d2 = 1826.

Here 6.52102650

ndAx

,

which is a fraction. Let us apply the short-cut formula in order to calculate S.D.

S.D. =

22

nd

nd

2

1026

101826

84.17576.660.182 = 13.26.

CALCULATION OF STANDARD DEViATION DISCRETE SERIES OR GROUPED DATA

The standard deviation of a discrete series or grouped data can be calculated by any one of thefollowing three methods.

(a) Actual Mean Method or Direct Method

(b) Assumed Mean Method or Short-cut Method

(c) Step Deviation Method

Actual Mean Method or Direct Method

The standard deviation for the discrete series is given by the formula :

n)xx(f 2

where x is the arithmetic mean, x is the size of item,f is the corresponding frequency and

n = f.

However, in practice, this method is rarely used because if the arithmetic mean is a fraction,the calculations take a lot of time and are cumbersome.

Assumed Mean Method or Short-cut Method

In this method we use the following formula to calculate the standard deviation a:

22

nfd

nfd

,


where A = is the assumed mean, d = x - A, and n = f.

Example 19. Find the standard deviation from the following data:

Size ofthe item : 10 11 12 13 14 15 16

Frequency : 2 7 11 15 10 4 1.

Also find the coefficient of variation.

Solution.

Table: Computation of Standard Deviation

Size of the Frequency: (f) dx - A, fd d2 fd2

item : (x) A = 13

10 2 -3 - 6 9 18

11 7 -2 -14 4 28

12 11 -1 -11 1 11

13 15 0 .0 0 0

14 10 1 10 1 10

15 4 . 2 8 4 16

16 1 3 3 9 9

Total n = f = 50 fd = -10 fd2 = 92

Now Mean : x = A + nfd

= 13+ 50)10(

= 12.8.

Here x = 12.8, is a fraction.

S.D. = 22

nd

nd

=

2

5010

5092

= 04.084.1 = 80.1 = 1.342.

Coefficient of Variation = 8.12342.1100x

x

= x 100 = 10.4.

Step Deviation Method

In this method we divide the deviations by a common class interval and use the followingformula for computing standard deviation:

22fd fd in n


where i = common class interval, d = iAx

, A is assumed mean,f is the respectivefrequency.

Example 20. Daily sales (recorded in rupees) of a retail shop are given below :

Daily sales in ` 102 106 110 114 118 122 126

(X-Mid-point of interval) :

No.ofdays 3 9 25 35 17 10 1

(Frequency-f) :

Calculate the mean and the standard deviation of the above data. Also calculate the coefficient ofvariation.

Solution. Let us take the assumed mean A = 114.

Then d = (x - A) = (x - 114) / 4, where i = 4.

Table : Computation of Mean and Standard Deviation

No.of days : x f d = 4114x

fd fd 2

102 3 -3 -9 27

106 9 -2 -18 36

110 25 -1 -25 25

114 35 0 0 0

118 17 1 17 17

122 10 20 40

126 1 3 3 9

N = 100 fd= -12 fd = 154

Arithmetic Mean : x = fdA x i

N

Here A = 114 ; fd = -12 ; n = 100 ; i = 4

x = 114 - 10012

x 4 = 114 - 0.48 = 113.52.


22

nfd

nfd

x i =

22

nfd

nfd

x 4 = 2)12.0(54.1 x 4

= 0144.054.1 x 4 = 1.235 x 4 = 4.94.

Coefficient of Variation: C.V. = x

x 100 = 52.11394.4

x 100 = 4.35%.

CALCULATION OF STANDARD DEVIATION CONTINUOUS SERIESThe standard deviation of a continuous series can be calculated by any one of the methodsdiscussed for discrete frequency distribution. However, in practice only Step DeviationMethod is mostly used. In this method the formula used is:

22

nfd

nfd

x i,

where d = iAm

, i = class interval (or the common factor in case the class intervals areunequal), rn is the mid-value of the interval, A is the assumed mean.

Example 21. Find the standard deviation for the following distribution :Marks 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 -80No. of Students 5 12 15 20 10 4 2Solution. Let us prepare the following table in order to calculate the standard deviation, by

assuming A = 45.

Table : Computation of Standard Deviation

Marks (Class No. of Students Mid-value d = 1045m

fd fd2

interval) (f) (m)10 - 20 5 15 -3 -15 4520 - 30 12 25 -2 -24 4830 - 40 15 35 -1 -15 1540 - 50 20 45 0 0 050 - 60 10 55 1 10 1060 - 70 4 65 2 8 1670 - 80 2 75 3 6 18Total f= n = 68 fd = -30 fd2 = 152

22 2fd fd 152 30i 10n n 68 86


Example 22. The arithmetic mean and the standard deviation of a set of 9 items are 43 and 5respectively. Than item of value 63 is added to the set, find the mean and standard deviation ofall the 10 items.

Solution. Here it is given that :

FIRST CASE. n = 9, x = 43, = 5.

But x = nx

x = n x = 9 x 43 = 387.

Also 2 =

22

nx

nx

=

nx2

- 2)x(

or 2x = n( 2 + 2)x( ) = 9 [25+(43)2] = 9 (25 + 1849) = 16866.

SECOND CASE.

If a new item 63 is added then the new number of terms becomes 10.

New x = (old x) + 63 = 387 + 63 = 450.

New Mean : x = 10xNew

= 10450

= 45.

x2 = x2 + (63)2 = 16866 + 3969 = 20835.

New S. D. = 22

x10x

= 24510

20835 = 20255.2083 = 5.58 = 7.65.

Example 23. You are provided with the following raw data of two variables x and y.

x = 235, y = 250, x2 = 6750, y2 = 6840.

Ten pairs of values are included in the survey. Calculate their standard deviations.

Solution.

x = 22

nx

nx

=

2

10235

106750

= 25.552675 = 75.122 =11.08.


22y n

yny

=

2

10250

106840

= 625684 = 59 = 7.68.

Example 24. A student obtained the mean and standard deviation of /00 observations as 40and 5.1 respectively. It was later discovered that he had wrongly copied down anobservation 50 instead of 40. Calculate the correct mean and standard deviation.

Solution. Here N X = 100, x = 40, = 5.1

Also X = Nx 40 = 100

x

or x = 40 x 100 = 4000 (Uncorrected).

Correct x = uncorrected x - wrong item + correct item

= 4000 - 50 + 40 = 3990.

Corrected X = NxCorrect

= 1003990

= 39.9.

Uncorrected x2 = N ( 2 + X 2) = 100 (5.1)2 + (40)2] = 1,62601.

Correct x2 = 162601 - 502 + 402 = 161701.

Corrected = 2

2Corrected x (Corrected X)N

= 2)9.39(100

161701 = 01.159201.1617 = 25 = 5.

Example 25. The quarterly profits (in lakhs of `) of three companies are given below. Calculatetheir standard deviation and coefficient of variation which company is more solid ?

Quarters Company A Company B

Jan-March 9.5 15.5

April-June 13.7 21.2

July-September 10.4 23.4

October-December 8.6 18.8

Solution.


Company A

Here x = 9.5 + 13.7 + 10.4 + 8.6 = 42.2. x A = nx

= 42.42

= 10.55

x2 = (9.5)2 + (13.7)2 + (10.4)2 + (8.6)2 = 90.25 + 187.69 + 108.16 + 73.96 = 460.06.

Now A = 22

nx

nx

=

2

42.42

406.460

= 3.111115 = 7.3 = 1.92.

C.V. (A) = AA

x

x 100 = 55.1092.1

x 100 = 18.23%

Company B

x = 15.5 + 21.2 + 23.4 + 18.8 = 78.9. x B = 49.78

= 19.72.

x2 = (15.5)2 + (21.2)2 + (23.4)2 + (18.8)2 = 240.25 + 449.44 + 547.56 + 353.44 = 1590.69.

2

B 49.78

469.1590

= 877.38867.397 = 8.8 = 2.96.

C.V. (B) = BB

x

x 100 = 72.1996.2

x 100 = 15%.

Since, C.V. (B) is less than the C.V. (A), so the company B is more solid in profits.

Example 26. For a set of 100 observations, taking assumed mean as 4. the sum of the deviationsis -11 cm. and the sum of the squares of these deviations is 257 cm2. Find the coefficient ofvariation.

Solution. In order to find the coefficient of variation, we require mean and standard deviations.They are calculated as follows :

x = A + nfd

= 4 + 10011

= 3.89.

Also = 22

nd

nd

=

2

10011

100257

= 0012.057.2 = 558.2 = 1.6.

Coefficient of Variation = x

x 100 = 89.36.1

x 100 = 41.13%.


Example 27. Comment on the statement ‘after settlement the average weekly wage in a factoryhas increased from ` 8 to ` 12 and standard deviation has increased from 2 to 2.5. Afterthe settlement, the wage has become higher and more unVorm

Solution. It is clear that after the settlement, the weekly wage has increased from ` 8, ` 12, sothe total wages received per week by all the workers have increased. Thus the wages havebecome higher.

C.V. before settlement = BeforeBefore

x

x 100 = 82

x 100 = 25%.

C.V. after settlement = AfterAfter

x

x 100 = 125.2

x 100 = 20.83%.

Thus after the settlement C.V. is less, so the variation is also less.

Example 28. If n = 10, x = 12, x2 = 1530, find the coefficient of variation.

Solution. We know that :

= 22

)x(nx

= 2)12(10

1530 = 9 = 3.

Coefficient of variation = x

x 100 = 12100x3

= 25%.

Example 29. The arithmetic mean of the runs scored by three batsman, Amit, Sumeet and Kapilin the series are 50, 48 and 12 respectively. The standard deviations of their runs are respectively15, 12 and 2. Who is the most consistent of the three? If one of the three is to be selected, who willbe selected?

Solution. Let x 1, x 2, x 3 be the means and 1, 2, 3 be the standard deviations of the runsscored by Amit, Sumeet and Kapil. Here x 1 = 50, x 2 = 48, x 3 = 12; 1 = 15, 2 = 12, 3 = 2.

Coefficient of variation of runs scored by Amit = 11

x

x 100 = 50100x15

= 30%

Coefficient of variation of runs scored by Sumeet = 22

x

x 100 = 48100x12

= 25%

Coefficient of variation of runs scored by Kapil = 33

x

x 100 = 122

x 100 = 16.67%


The most consistent among the three is one whose C.V. is least. Here C.V. of Kapil is least,so he is most consistent. He is to be selected.

COMBINED STANDARD

It is possible to find out the standard deviation of a composite series from the standarddeviations of its component parts. Let A1 and A2 be two groups having N1 and N2 itemsrespectively. Let their means be X 1 and X 2 and their respective standard deviations be 1

and 2. Then the combined standard deviation 12 of two groups A1 and A2 is given bythe formula :

12 = 21

22

212

122

212

1NN

dNdNNN

=

21

22

22

212

12

1NN

)d(N)d(N

where d1 = ( X 1 - X ), d2 = ( X 2 - X ),

X =21

2211NN

XNXN

their combined mean.

Similarly, for three groups A, B, C with means X 1, X 2, X 3, and S.D. 1, 2, 3, havingnumber of items N1, N2,N3, their combined Standard Deviation is given by :

321

32

32

322

22

212

12

1123 NNN

)d(N)d(N)d(N

where d1 = X 1 - X ,d2 = X 2 - X ,d3 = X 3 - X , X is the combined mean of A, B and C.

Example 30. The mean and standard deviation of the marks obtained by two groups of students,consisting of 50 each, are given below. Calculate the mean and standard deviation of themarks obtained by all the in 100 students:

Group Mean Standard Deviation

1 60 8

2 55 7


Solution. Here

For Group X 1 : = 60 1= 8 n1 = 50

For Group X 2 : = 55 2 = 7 n2 = 50

Combined Mean : x 12 = 21

21nn

nxnx

= 5050

50555060

= 100

5750 = 57.5.

Combined Standard Deviation : 12 = 21

222

112

222

112

nnndndnn

[ d1= x 12 - x 1 = 57.5 - 60 and d2 = x 12 - x 2 = 57.5 - 55 = 2.5 ]

505050)5.2(50)5.2(507508 2222

12

= 100

5.3125.31224503200 = 75.62 = 7.92.

Example 31. The means of two samples of sizes 500 and 600 were respectively 186 and 175.The corresponding standard deviations were respectively 9 and 10. The variable studied washeight in centimetres. Obtain the mean and variance of the combined sample.

Solution. Here n1 = 500 x = 186 1 = 9

n2 = 600 x 2 =175 2 = 10.

Combined mean : X 12 = 21

2211nn

xnxn

X 12 = 600500600175500186

= 180 cms.

Combined S.D. : 12 = 21

22

212

122

212

1nn

dndnnn

where d1 = x - x 1 = 180 - 186 = -6 ; d2 = x - x 2 = 180 - 175 = 5.


Combined Standard Deviation: 12 = 600500

)5(600)6(500106009500 2222

Variance : 212 = 1100

)25100(600)3681(500 = 1100

133500 = 121.36.

Example 32. After settlement the average weekly wage in afactory has increased from 8000 to12000 and the standard deviation had increased from ` 100 to ` 150. After settlement the wagehas become higher and more un~form. Do you agree?

Solution. Mean before settlement = ` 8000.

Standard deviation before settlement = ` 100

Mean after settlement = ` 12000.

Standard deviation after settlement = ` 150.

C.V. (Before settlement) = X

x 100 = 8000100

x 100 = 1.25%.

C.V. (After settlement) = X

x 100 = 12000150

x 100 = 1.25%.

Hence, C.V. (Before settlement) = C.V. (After settlement) there is no “more uniformity” in theweekly wages.

The statement is not correct, i.e., we do not agree as there is no change in the wages.

* * *


CLASS WORK

1. Mean is a measure of :

(a) location (central value) (b) dispersion

(c) correlation (d) none of the above

2. Which of the following is a measure of central value ?

(a) Median (b) Variance (c) Standard deviation

(d) (b) and (c) both

3. Which of the followings represents median ?

(a) First quartile (b) Fiftieth percentile (c) Sixth decile(d) None of the above

4. The mean of the following observations is

10, 19, 22, 16, 15, 18, 20, 18, 14, 18, 23.

(a) 17.55 (b) 18 (c) 15 (d) 16.5

5. If a constant value 40 is subtracted from each observation of a set, the mean of the set is

(a) increased by 40 (b) decreased by 40 (c) is not affected(d) zero

6. If a constant 25 is added to each observation of a set, the mean is :

(a) increased by 25 (b) decreased by 25

(c) 25 times the original mean (d) not affected

7. If each observation of a set is multiplied by 5, the mean of the new set of observations:

(a) remains the same (b) is 5 times the original mean

(c) is (1/5) times the original mean (d) is increased by 5

8. If each value of a series is multiplied by 7, the median of the coded values is :

(a) not affected (b) 7 times the original median value

(c) (1/7)th of the original median value (d) increased by 7

9. If each value of a series is multiplied by 8, the mode of the coded values is :

(a) not affected (b) (1/8)th of the original modal value

(c) 8 times of the original modal value (d) none of them

10. If each observation of a set is divided by 5, then the mean of new values :

(a) is 5 times the original mean (b) is decreased by 5

(c) is (1/5)th of the original mean (d) remains the same


11. The median of the following observations is

10, 19, 22, 16, 15, 18, 20, 18, 14, 18, 23.

(a) 17.55 (b) 18 (c) 15 (d) 16.5

12. The mode of the following observations is

10, 19, 22, 16, 15, 18, 20, 18, 14, 18, 23.

(a) 17.55 (b) 18 (c) 15 (d) 16.5

13. The first quartile of the following observations is

10, 19, 22, 16, 15, 18, 20, 18, 14, 18, 23.

(a) 17.55 (b) 18 (c) 15 (d) 20

14. The third quartile of the following observations is

10, 19, 22, 16, 15, 18, 20, 18, 14, 18, 23.

(a) 17.55 (b) 18 (c) 15 (d) 20

15. The second quartile of the following observations is

10, 19, 22, 16, 15, 18, 20, 18, 14, 18, 23.

(a) 17.55 (b) 18 (c) 15 (d) 20

16. The mean of the following observations is

10, 8, –9, –12, 15, 0, 23, –3, –2, –13, –24, 28, 35, 42

(a) 4 (b) –2 (c) 7 (d) –9.75

17. The median of the following observations is

10, 8, –9, –12, 15, 0, 23, –3, –2, –13, –24, 28, 35, 42

(a) 4 (b) –2 (c) 7 (d) –9.75

18. The mode of the following observations is

10, 8, –9, –12, 15, 0, 23, –3, –2, –13, –24, 28, 35, 42

(a) 4 (b) –2 (c) 7 (d) –9.75

19. The first quartile of the following observations is

10, 8, –9, –12, 15, 0, 23, –3, –2, –13, –24, 28, 35, 42

(a) 4 (b) –2.25 (c) 24.25 (d) –9.75

20. The third quartile of the following observations is

10, 8, –9, –12, 15, 0, 23, –3, –2, –13, –24, 28, 35, 42

(a) 14.25 (b) –2.25 (c) 24.25 (d) –9.75


21. The following distribution of ages (in complete years) is obtained for the students of highersecondary. The median of the distribution is

Age (in years) 15 16 17 18 19 20 21

Number of students 12 18 20 10 7 6 2

(a) 17.11 (b) 17 (c) 18 (d) none of them

22. The average of five numbers is 40 and the average of another six numbers is 50. The average ofall numbers taken together is :

(a) 44.44 (b) 45.00 (c) 45.45 (d) 90.00

23. If the two observations are 10 and 0, their arithmetic mean is

(a) 10 (b) 0 (c) 5

(d) none of the above

24. What percentage of value is greater than 3rd quartile?

(a) 75 per cent (b) 50 per cent (c) 25 per cent (d) 0 per cent

25. What percentage of values is less than 4th decile?

(a) 50% (b) 70% (c) 40% (d) none of the above

26. What percentage of values lies between 5th and 25th percentiles

(a) 20% (b) 30% (c) 75% (d) none of the above

27. If the A.M. of a set of observations is 9 and its G.M. is 6. Then the H.M. of the set ofobservations is :

(a) 4 (b) 6 (c) 3 (d) none of them

28. The A.M. of two numbers is 6.5 and their G.M. is 6. The two numbers are :

(a) 8,6 (b) 8, 5 (c) 7, 16 (d) 4, 9

29. If Md, Q, D and P stand for median, quartile, decile and percentile respectively, then which ofthe following relation between them is true ?

(a) Md = Q2 = D6 = P50 (b) Md = Q3 = D6 = P75

(c) Md = Q5 = D5 = P50 (d) Md = Q2 = D5 = P50

30. The following distribution of ages (in complete years) is obtained for the students of highersecondary. The mean of the distribution is

Age (in years) 15 16 17 18 19 20 21


(a) 17.11 (b) 17 (c) 18 (d) none of them


31. The following distribution of ages (in complete years) is obtained for the students of highersecondary. The mode of the distribution is

Age (in years) 15 16 17 18 19 20 21


(a) 16 (b) 17 (c) 18 (d) none of them

32. The following distribution of ages (in complete years) is obtained for the students of highersecondary. The first quartile of the distribution is.

Age (in years) 15 16 17 18 19 20 21


(a) 19 (b) 17 (c) 18 (d) none of them

33. The first quartile of the following frequency distribution is

Observation 8 7 6 5 4 3 2 1

Frequency 10 8 12 23 30 18 19 15

(a) 4 (b) 7 (c) 5 (d) 2

34. The second quartile of the following frequency distribution is


Frequency 10 8 12 23 30 18 19 15

(a) 4 (b) 7 (c) 5 (d) 2

35. The third quartile of the following frequency distribution is


Frequency 10 8 12 23 30 18 19 15

(a) 4 (b) 7 (c) 5 (d) 2

36. Which of the following relation is true between 3rd decile and 30th percentile ?

(a) D7 = P70 (b) D7 = P30 (c) D3 = P30 (d) D3 = P70

37. Which of the deciles are less than first quartile ?

(a) D1 and D3 (b) D1 and D3

(c) D1 and D2 (d) None of the above

38. The mean of seven observations is 18. A new observation 16 is added. The mean of eightobservations is :

(a) 22 (b) 17 (c) 18 (d) none of them

39. If the sum of n observations is 630 and their mean is 21, then the value of n is :

(a) 21 (b) 30 (c) 15 (d) 20


40. If the two observations are 10 and 0 then their geometric mean is :

(a) 10 (b) 0 (c) 5 (d) 8

41. If the two observations are 7 and –7, their geometric mean is :

(a) 7 (b) –7 (c) 0 (d) None of the above

42. If the two observations are 10 and –10, their arithmetic mean is :

(a) 10 (b) 20 (c) 0 (d) none of the above

43. Can a quartile, a decile and a percentile be the median ?

(a) Only quartile but not decile

(b) Quartile and decile but not percentile and percentile

(c) Decile and percentile but

(d) Quartile, decile and percentile, all the not quartile three

44. When all the observations are same, then the relation between A.M., G.M. and H.M. is :

(a) A.M. = G.M. = H.M. (b) A.M. = G.M. > H.M.

(c) A.M. > G.M. > H.M. (d) A.M. < G.M. < H.M.

45. The average of n natural numbers is

(a) (n + 1)/2 (b) (2n + 1)/2 (c) n(n + 1)/2 (d) n (2n + 1)/2

46. Mode is that value in a frequency distribution which possesses:

(a) minimum frequency (b) maximum frequency

(c) frequency one (d) none of the above

47. The mean of the following frequency distribution is

Class 0–6 6–12 12–18 18–24 24–30 30–36 36–42 42–48 48–54

Frequency 13 25 57 79 105 79 57 25 13

(a) 27 (b) 28 (c) 25 (d) none of them

48. The median of the following frequency distribution is

Class 0–6 6–12 12–18 18–24 24–30 30–36 36–42 42–48 48–54

Frequency 13 25 57 79 105 79 57 25 13

(a) 27 (b) 28 (c) 25 (d) none of them

49. The mode of the following frequency distribution is

Class 0–6 6–12 12–18 18–24 24–30 30–36 36–42 42–48 48–54

Frequency 13 25 57 79 105 79 57 25 13

(a) 27 (b) 28 (c) 25 (d) none of them

50. The combined mean of the following data isMEASURES OF CENTRAL TENDENCY OR AVERAGES 163

Group A B C

Number of observations 400 250 350

Mean 52 44 60

(a) 52.8 (b) 44 (c) 52 (d) none of them

51. The average marks of 120 students of a class is 48.2. If the average marks of 50 girls of that classis 54, the approximate average marks of the boys is

(a) 48 (b) 58 (c) 44 (d) none of them

52. The approximate weighted mean of the following data isxi 56 40 68 55 37w i 3 2 2 4 1

(a) 55.5 (b) 58.3 (c) 56.4 (d) 53.4

53. The 8th decile of the following observations is29, 18, 15, 30, 42, 35, 34, 28, 45, 34

(a) 44.6 (b) 40.6 (c) 34.15 (d) 38.15

54. The 65th percentile of the following observations is29, 18, 15, 30, 42, 35, 34, 28, 45, 34

(a) 44.6 (b) 40.6 (c) 34.15 (d) 38.15

55. The median for the following distribution isx: 2 3 4 5 6 7 8 9 10 11f: 3 6 9 18 20 14 10 10 7 2

(a) 6 (b) 5 (c) 8 (d) 9

56. The Q1 for the following distribution isx: 2 3 4 5 6 7 8 9 10 11f: 3 6 9 18 20 14 10 10 7 2

(a) 6 (b) 5 (c) 8 (d) 9

57. The Q3 for the following distribution is

x : 2 3 4 5 6 7 8 9 10 11

f : 3 6 9 18 20 14 10 10 7 2

(a) 6 (b) 5 (c) 8 (d) 9

58. The D9 for the following distribution is

x: 2 3 4 5 6 7 8 9 10 11

f: 3 6 9 18 20 14 10 10 7 2

(a) 6 (b) 5 (c) 8 (d) 9


59. The P82 for the following distribution is

x : 2 3 4 5 6 7 8 9 10 11

f : 3 6 9 18 20 14 10 10 7 2

(a) 6 (b) 5 (c) 8 (d) 9

60. Which of the following relations among the location parameters does not hold ?

(a) Q2 = median (b) P50 = median

(c) D5 = median (d) D4 = median

61. If the group data has open end classes, one cannot calculate:

(a) median (b) mode (c) mean (d) quartiles

62. Geometric mean of two observations can be calculated only if :

(a) both the observations are positive (b) one of the two observations is zero

(c) one of them is negative (d) both of them are zero

63. Geometric mean is a good measure of central value if the data are :

(a) categorical (b) on ordinal scale

(c) in ratios or proportions (d) none of the above

64. Harmonic mean is better than other means if the data are for :

(a) speed or rates (b) heights or lengths

(c) binary values like O and l (d) ratios or proportions

65. The geometric mean of 3, 7, 11, 15, 24, 28, 30, 0 is

(a) 6 (b) 0 (c) 8 (d) 9

66. The Harmonic mean of 5, 6, 10, 15, 20, 24 is

(a) 6.9 (b) 9.6 (c) 8.6 (d) 9.8

67. A cyclist travels 20 kms at a speed of 16 kms. per hour, 30 kms at a speed of 12 kms per hourand 40 kms at a speed at 10 kms per hour. His average speed is

(a) 16 (b) 15.5 (c) 11.6 (d) 19.5

68. The mean of a distribution is 22.2 and its mode is 23.3. The median is

(a) 26.5 (b) 22.6 (c) 28.5 (d) 26.6

69. If Z – M = 4 and Z + M = 60, then M is

(a) 26 (b) 25 (c) 28 (d) 32

70. If Z – M = 4 and Z + M = 60, then Z is

(a) 32 (b) 25 (c) 28 (d) 26


71. If Z – M = 4 and Z + M = 60, then mean is

(a) 32 (b) 25 (c) 28 (d) 26

72. The mean of 4, 3, 9, x, 6 is 5.4 then x is

(a) 3 (b) 5 (c) 8 (d) 6

73. The mean of 6 observations is 15. Four of the observations are 10, 18, 23 and 9. of the remainingtwo observations one is four times the other, They are(a) 32 ,8 (b) 24,6 (c) 28,7 (d) 26,6.5

74. The mean of 10 observations is 48.4. If by mistake one observation was taken as 54 instead of45. The correct mean is(a) 32.5 (b) 25.5 (c) 47.5 (d) 26.5

75. The mean of 15 observations is 22. One observation was taken as –12 instead of 12, the correctmean is(a) 32.5 (b) 25.6 (c) 28.6 (d) 23.6

76. The value of the variable corresponding to the highest point of a frequency distribution curverepresents :

(a) mean (b) median

(c) mode (d) none of the above

77. A frequency distribution having two modes is said to be :

(a) unimodal (b) bimodal

(c) trimodal (d) without mode

78. If modal value is not clear in a distribution, it can be ascertained by the method of:

(a) grouping (b) guessing

(c) summarizing (d) trial and error

79. Shoe size of most of the people in a city is No.7. Which measure of central value does it represent ?

(a) mean (b) median

(c) eighth decile (d) mode

80. In a discrete series having (2n + 1) observations, median is :

(a) nth observation (b) (n + 1)th observation

(c) [(n + 2)/2]th observations (d) [(2n +1)/2]th observation

81. The median of the variate values 7, 3, 8, 16, 18, 25,11 is :

(a) 8 (b) 18 (c) 11 (d) 16

82. To find the median , it is necessary to arrange the data in :

(a) ascending order (b) descending order

(c) ascending or descending order (d) any of them


83. For a grouped data, the formula for median is based on :

(a) interpolation method (b) extrapolation method

(c) trial and error method (d) iterative method

84. Which of the measure of central tendency is not affected by extreme values?

(a) mode (b) median

(c) sixth decile (d) all the above

85. The mean of 25 observations is 73.408. If one observation 64 is removed, the revised mean is

(a) 72.8 (b) 73.8 (c) 80.8 (d) 76.8

86. In a group, 2 students spend ` 8 daily, 3 students spend ` 10 daily and 5 students spend ` 6daily. The average spending of all 10 students is

(a) 7.6 (b) 5.8 (c) 8.5 (d) 6.7

87 The mean of 20 observations of a series is 18.7. If from each of the observations 3 is subtractedand then multiplied by 2, the mean of the new observations is

(a) 32.4 (b) 30.4 (c) 31.4 (d) none of them

88. The average marks of 50 students of a class is 54. If 20 students of the class have average marks60, the average marks of the remaining 30 students is

(a) 30 (b) 40 (c) 60 (d) 50

89. The correct relationship between A.M. G.M. and H.M. is :

(a) A.M. = G.M. = H.M (b) G.M > A.M. > H.M

(c) H.M. > G.M. > A.M. (d) A.M. > G.M. > H.M.

90. Extreme value have no effect on :

(a) average (b) median

(c) geometric mean (d) harmonic mean

91. Average of 12 members = 11.0. Average of the first six members = 10.5 average of the last sixmembers =

(a) 10.5 (b) 11.5 (c) 11.0 (d) 10.0

92. The average of the 7 number 5,9,14, x, 5, 6, 9 is 9. The missing number x is :

(a) 13 (b) 14 (c) 15 (d) none of them

93. The second of the two samples has 50 items with mean 15. If the whole group has 150 itemswith mean l6, the mean of the first sample is

(a) 18 (b) 15 (c) 16.5 (d) none of the above


94. For a group of 100 candidates, the mean was found to be 40. Later on it was discovered that avalue 45 was misread as 54. The correct mean is :

(a) 40 (b) 39.8 (c) 39.9 (d) 39.91

95. The average runs of a cricketer in four innings is 32. How many runs he should score in the fifthinning so that his average score becomes 50 runs?

(a) 132 (b) 122 (c) 128 (d) 126

96. The combined mean of two groups is 57. The mean of the first group is 52 and the mean of thesecond group is 60. The proportion of observations in the two groups is

(a) 3:5 (b) 2:5 (c) 2:8 (d) 5:2

97. The mean of first 10 natural numbers.

(a) 6.5 (b) 2.5 (c) 4.5 (d) 5.5

98. Given mean = 70.2 and mode = 70.5, find median using empirical relationship among them.

(a) 70.3 (b) 70.5 (c) 70.6 (d) 70.4

99. The middle value of an ordered series is called :

(a) 2nd quartile (b) 5th decile

(c) 50th percentile (d) all the above

100. The variate values which divide a series (frequency distribution) into four equal parts are called :

(a) quintiles (b) quartiles (c) Octiles (d) percentiles

101. The variate values which divide a series (frequency distribution) into 100 equal parts are called :

(a) quintiles (b) quartiles (c) deciles (d) percentiles

102. The variate values which divide a series (frequency distribution) into ten equal parts are called:

(a) quartiles (b) deciles (c) Octiles (d) percentiles

103. The number of partition values in case of quartiles is :

(a) 4 (b) 3 (c) 2 (d) 1

104. The first quartile divides a frequency distribution in the ratio :

(a) 4 : 1 (b) 1 : 4 (c) 3 : 1 (d) 1 : 3

105. The first quartile is also known as :

(a) median (b) lower quartile

(c) mode (d) third decile

106. The third quartile is also called :

(a) lower quartile (b) median (c) mode (d) upper quartile


107. If for an individual series, assumed mean, A = 25 dx = –21 for dx = X – A and N = 7, then themean of the series is :

(a) 20 (b) 21 (c) 22 (d) 6.57

108. If we plot the more than type and less than type frequency distributions of the same set of data,their graphs intersect at the point which is known as:

(a) median (b) mode

(c) mean (d) none of the above

109. Mean of a set of values is based on :

(a) all values (b) 50 percent values

(c) first and last value (d) maximum & minimum value

110. Which average is most affected by extreme values?

(a) Geometric mean (b) Harmonic mean

(c) Arithmetic mean (d) Trimmed mean

111. Harmonic mean gives more weightage to

(a) small values (b) large values

(c) positive values (d) negative values

112. For further algebraic treatment geometric mean is

(a) suitable (b) not suitable

(c) sometime suitable (d) none of the above

113. For further algebraic treatment harmonic mean is :

(a) suitable (b) not suitable

(c) sometime suitable (d) none of the above

114. In a class test, 40 students out of 50 passed with mean marks 6.0 and the overall average ofclass marks was 5.5. The average marks of students who failed were:

(a) 2.5 (b) 3.0 (c) 4.8 (d) 3.5

115. Sum of the deviations about mean is :

(a) Zero (b) minimum (c) maximum (d) One

116. Sum of the absolute deviations about median is

(a) Zero (b) minimum (c) maximum (d) One

117. Sum of square of the deviations about mean is:

(a) Zero (b) minimum

(c) maximum (d) none of the above


118. The suitable measure of central tendency for qualitative data is:

(a) mode (b) arithmetic mean

(c) geometric mean (d) median

119. In a frequency distribution with open ends, one can not find out :

(a) mean (b) median (c) mode (d) all the above

120. The geometric mean of 3, 6, 24 and 48 is

(a) 32 (b) 12 (c) 14 (d) 24

121. The mean annual salary of all employees in a company is ` 25,000. The mean salaries of maleand female employees are ` 27,000 and ` 17,000 respectively, the percentage of males andfemales employed by the company is

(a) 80,20 (b) 20,80 (c) 30,70 (d) 70,30

122. From which average sum of deviation is zero?

(a) median (b) mean (c) mode (d) none of them

123. In a moderately skewed distribution the arithmetic mean is 10 units and the mode is 7 units, themedian is

(a) 9 (b) 5 (c) 8 (d) 6

124. If for a series the arithmetic mean is 25 and the harmonic mean is 9, what is the geometricmean?

(a) 25 (b) 15 (c) 28 (d) 16

125. Which of the following is a unit less measure of dispersion?

(a) Quartile deviation (b) Mean deviation

(c) Coefficient of variation (d) Range

126. Which one of the given measures of dispersion is considered best?

(a) Standard deviation (b) Range

(c) Mean deviation (d) Coefficient of variation

127. For comparison of two different series, the best measure of dispersion is

(a) Range (b) Mean deviation

(c) Standard deviation (d) Coefficient of variation

128. Correct formula for mean deviation from a constant A of a series in which the variate valuesx1, x2, …..., xn have frequencies f1, f2, ..…, fn respectively is:

(a) 1i i

if x A

N (b) 1

1i

if x A

N

(c) 11 | |i

if x A

N (d) 1

1 | |ii

f x AN

where i = 1,2,…, k and ii

f N170 STATISTICS (CPT)

129. The correct relation between variance and standard deviation (S.D) of a variable X is.

(a) S.D. = Var (b) S.D. = [Var (X)]1/2

(c) S.D. = [Var (X)]2 (d) None of the above

130. Formula for coefficient of variation is :

(a) C.V. = 100

. .mean S D (b) C.V. =. .

meanS D x 100

(c) C.V. = . .

100mean S D

(d) C.V. = . .S D

mean x 100

131. Formula for coefficient of range of the set of observations X1, X2, …, Xn is :

(a) Xmin – Xmax (b) Xmax – Xmin

(c)X XX X

max min

max min

(d)X XX X

max min

max min

132. Coefficient of quartile deviation is given by the formula:

(a)Q QQ Q

3 1

1 3

(b)

Q QQ Q

3 1

1 3

(c)Q QQ Q

3 13 1

(d)Q QQ Q

3 13 1

133. The range from the following observations is

10, 18, 20, 28, 15, 17, 22, 25, 29, 32, 34

(a) 24 (b) 20 (c) 30 (d) 42

134. The co–efficient of range from the following observations is

10, 18, 20, 28, 15, 17, 22, 25, 29, 32, 34

(a) 0.6 (b) 0.5 (c) 0.55 (d) none of them

135. The quartile deviation from the following observations is

10, 18, 20, 28, 15, 17, 22, 25, 29, 32, 34

(a) 8 (b) 6 (c) 10 (d) 5

136. The co–efficient of quartile deviation from the following observations is

10, 18, 20, 28, 15, 17, 22, 25, 29, 32, 34

(a) 0.26 (b) 0.16 (c) 0.06 (d) 0.36


137. The co– efficient of Q.D of 63, 65, 68, 75, 76, 77, 78, 81, 83, 70 is

(a) 0.018 (b) 0.028 (c) 0.08 (d) 0.008

138. The co–efficient of mean deviation of 15, 17, 22, 18, 19, 11, 13, 18, 20, 17 is

(a) 0.44 (b) 0.34 (c) 0.24 (d) 0.14

139. The variance of 1, 2, 4, 5 and x is 2, then x is

(a) 3 (b) 6 (c) 1 (d) 2

140. Semi inter-quartile deviation is given by the formula:

(a) Q3 – Q1 (b)Q Q3 1

2(c) (Q3 + Q1)/2 (d) (Q3–Q1)/4

141. Mean deviation is minimum when deviations are taken from:

(a) mean (b) median (c) mode (d) zero

142. Sum of squares of the deviations is minimum when deviations are taken from:

(a) Mean (b) median (c) mode (d) zero

143. If a constant value 15 is subtracted from each observation of a set, the variance is:

(a) reduced by 15 (b) reduced by 225

(c) unaltered (d) increased by 225

144. If each observation of a set is divided by 10, the S.D. of the new observations is:

(a) 10 times of S.D. of original obs. (b)1

100 th of S.D. of original obs.

(c) not changed (d)1

10 th of S.D. of original obs.

145. Which of the following formula for standard deviation of a frequency distribution is not correct?

(a) s = 1 2N

( ) f x xi i (b) s = 1 2 2N f x xi i

(c)

2

i i2i i

i

f x1 f xN N

(d) 2

i i2i i

i

f x1 f xN N

146. The measure of dispersion which ignores signs of the deviations from a central value is:

(a) Range (b) quartile deviation

(c) standard deviation (d) mean deviation

147. Which measure of dispersion is least affected by extreme values ?


(c) Standard deviation (d) Quartile deviation172 STATISTICS (CPT)

148. Which measure of dispersion is most affected by extreme values ?

(a) Standard deviation (b) Mean deviation

(c) Range (d) Quartile deviation

149. Which measure of desperision ensures highest degree of reliability?


(c) Standard deviation (d) Quartile deviation

150. Which measure of disperision ensures lowest degree of reliability?


(c) Quartile deviation (d) standard deviation

151. A set of values is said to be relatively uniform if it has:

(a) high dispersion (b) zero dispersion

(c) little dispersion (d) negative dispersion

152. Range of a set of values is 60 and maximum value in the series is 80. The minimum value of theseries is :

(a) 140 (b) 20 (c) 70 (d) none of the above

153. If the minimum value in a set is 15 and its range is 55, the maximum value of the set is

(a) 40 (b) 60 (c) 80 (d) 70

154. If the values of a set are measured in kgs., the unit of variance will be:

(a) kg2 (b) kg (c) kg3 (d) no unit

155. Which measure of dispersion has a different unit other than the unit of measurement of values:


(c) Standard deviation (d) Variance

156. If Q3 + Q1 = 108.5, Q3 – Q1 = 74; then Q.D. is

(a) 33 (b) 3.7 (c) 37 (d) 0.07

157. If Q3 + Q1 = 108.5, Q3 – Q1 = 74; then co–efficient of Q.D is

(a) 33 (b) 3.7 (c) 37 (d) 0.68

158. In a distribution 25% of the observations are less than 46 and 25% of the observations are morethan 54. The quartile deviation of the distribution is

(a) 3 (b) 7 (c) 4 (d) 6

159. The co-efficient of mean deviation of first five natural numbers is

(a) 0.3 (b) 0.4 (c) 0.5 (d) 0.7


160. The mean deviation of 3, 4, 5, 3 is

(a) 0.75 (b) 0.85 (c) 0.65 (d) 0.55

161. The mean and S.D. of 13, 13, 13, 13, 13 are respectively

(a) 0,13 (b) 13, 13 (c) 13,0 (d) 0,0

162. The average of the sum of squares of the deviations about mean is called:

(a) Variance (b) absolute deviation

(c) standard deviation (d) mean deviation

163. Quartile deviation is equal to:

(a) Semi–interquartile range (b) double the interquartile range

(c) Interquartile range (d) none of the above

164. Which measure of dispersion can be calculated in case of open end intervals ?

(a) Range (b) Standard deviation

(c) Quartile deviation (d) Coefficient of variation

165. If each value of a series is multiplied by 15, the coefficient of variation will be increased by

(a) 5% (b) 0% (c) 15% (d) 10%

166. The C.V. of 4, 6, 8, 10, 12 is

(a) 33 (b) 35.33% (c) 42 (d) 46

167. The sum of 10 observations is 110 and the sum of squares of observations is 2900, their standarddeviation is

(a) 13 (b) 8 (c) 10 (d) 9

168. The sum of 25 observations is 400 and the sum of squares of observations is 8900 find C.V.

(a) 33.5 (b) 60.5 (c) 62.5 (d) 70.5

169. The C.V of a distribution is 80% and the mean of the distribution is 40, the S.D. of the distributionis

(a) 33 (b) 32 (c) 35 (d) 0.30

170. The mean of 100 observations is 18.4 and sum of squares of deviations from mean is 1444, theco-efficient of variation is

(a) 30.6 (b) 35.6 (c) 20.6 (d) 10.6

171. The standard deviation of 5 items is found to be 15. What will be the standard deviation if thevalues of all the items are increased by 5?

(a) 15 (b) 20 (c) 10 (d) none of them


172. What will be the relative range, if the spread of items in a given distribution lies from 100 and180 kg ?

(a) 0.29 (b) 0.4 (c) 0.3 (d) 0.55

173. The standard deviation of a set of 50 items is 8, what is the standard deviation, if each item ismultiplied by 2?

(a) 32 (b) 8 (c) 4 (d) 16

174. If each value of a series is multiplied by a constant , the coefficient of variation as compared tooriginal value is:

(a) increased (b) unaltered (c) decreased (d) zero

175. If each value of a set is divided by a constant ‘d’, the coefficient of variation will be:

(a) more than original value (b) less than original value

(c) same as original value (d) none of the above

176. The relation between variance and standard deviation is:

(a) variance is the square root of (b) square of the standard deviation

standard deviation is equal to variance

(c) variance is equal to standard (d) standard deviation is the square

deviation of the variance

177. Which of the following statements is true of a measure of dispersion ?

(a) Mean deviation does not follow (b) Range is a crudest measure

algebraic rule

(c) Coefficient of variation is a (d) All the above statements

relative measure

*****************


ANSWER KEYS


HOME WORK - 1

1. Measures of central tendency for a given set of observations measures

(a) The scatterness of the observations (b) The central location of the observations

(c) Both (i) and (ii) (d) None of these.

2. While computing the AM from a grouped frequency distribution, we assume that

(a) The classes are of equal length (b) The classes have equal frequency

(c) All the values of a class are equal to (d) None of these.

the mid-value of that class

3. Which of the following statements is wrong?

(i) Mean is rigidly defined

(ii) Mean is not affected due to sampling fluctuations

(iii) Mean has some mathematical properties (iv) All these

4. Which of the following statements is true?

(a) Usually mean is the best measure

(b) Usually median is the best measure of central tendencyof central tendency

(c) Usually mode is the best measure of central tendency

(d) Normally, GM is the best measure of central tendency

5. For open-end classification, which of the following is the best measure of central tendency?

(a) AM (b) GM (c) Median (d) Mode

6. The presence of extreme observations does not affect

(a) AM (b) Median (c) Mode (d) Any of these.

7. In case of an even number of observations which of the following is median?

(a) Any of the two middle-most value

(b) The simple average of these two middle values

(c) The weighted average of these two middle values

(d) Any of these


8. The most commonly used measure of central tendency is

(a) AM (b) Median (c) Mode

(d) Both GM and HM.

9. Which one of the following is not uniquely defined?

(a) Mean (b) Median

(c) Mode (d) All of these measures

10. Which of the following measure of the central tendency is difficult to compute?

(a) Mean (b) Median

(c) Mode (d) GM

11. Which measure(s) of central tendency is(are) considered for finding the average rates?

(a) AM (b) GM

(c) HM (d) Both (ii) and(iii)

12. For a moderately skewed distribution, which of he following relationship holds?

(a) Mean — Mode = 3 (Mean — Median) (b) Median — Mode = 3 (Mean — Median)

(c) Mean — Median = 3 (Mean — Mode) (d) Mean — Median = 3 (Median — Mode)

13. Weighted averages are considered when

(a) The data are not classified (b) The data are put in the form of groupedfrequency distribution

(c) All the observations are not of (d) Both (i) and (iii)

equal importance

14. Which of the following results hold for a set of distinct positive observations?

(a) AM > GM > HM (b) HM > GM > AM

(c) AM > GM > HM (d) GM > AM > HM

15. When a firm registers both profits and losses, which of the following measure of central tendencycannot be considered?

(a) AM (b) GM

(c) Median (d) Mode

16. Quartiles are the values dividing a given set of observations into

(a) Two equal parts (b) Four equal parts

(c) Five equal parts (d) None of these.


17. Quartiles can be determined graphically using

(a) Histogram (b) Frequency Polygon

(c) Ogive (d) Pie chart.

18. Which of the following measure(s) possesses (possess) mathematical properties?

(a) AM (b) GM

(c) HM (d) All of these

19. Which of the following measure(s) satisfies (satisfy) a linear relationship between two variables?

(a) Mean (b) Median (c) Mode (d) All of these

20. Which of he following measures of central tendency is based on only fifty percent of the centralvalues?

(a) Mean (b) Medium (c) Mode (d) Both (i) and(ii)

21. If there are 3 observations 15, 20, 25 then the sum of deviation of the observations from theirAM is

(a) 0 (b) 5 (c) —5 (d) None of these

22. What is the median for the following observations?

5, 8, 6, 9, 11, 4.

(a) 6 (b) 7 (c) 8 (d) None of these

23. What is the modal value for the numbers 5, 8, 6, 4, 10, 15, 18, 10?


24. What is the GM for the numbers 8, 24 and 40?

(a) 24 (b) 12 (c) 38 15 (d) 10

25. The harmonic mean for the numbers 2, 3, 5 is

(a) 2.00 (b) 3.33 (c) 2.90 (d) 3 3026. If the AM and GM for two numbers are 6.50 and 6 respectively then the two numbers are

(a) 6 and 7 (b) 9 and 4 (c) 10 and 3 (d) 8 and 5.

27. If the AM and HM for two numbers are 5 and 3.2 respectively then the GM will be

(a) 16.00 (b) 4.10 (c) 4.05 (d) 4.00.

28. What is the value of the first quartile for observations 15, 18, 10, 20, 23, 28, 12, 16?

(a) 17 (b) 16 (c) 12.75 (d) 12


29. The third decile for the numbers 15, 10, 20, 25, 18, 11 9, 12 is

(a) 13 (b) 10.7 (c) 11 (d) 11.50

30. If there are two groups containing 30 and 20 observations and having 50 and 60 as arithmeticmeans, then the combined arithmetic mean is

(a) 55 (b) 56 (c) 54 (d) 52.

31. The average salary of a group of unskilled workers is ` 10000 and that of a group of skilledworkers is ` 15,000. If the combined salary is ` 12000, then what is the percentage of skilledworkers?

(a) 40% (b) 50% (c) 60% (d)none of these

32. If there are two groups with 75 and 65 as harmonic means and containing 15 and 13 observationthen the combined HM is given by

(a) 65 (b) 70.36 (c) 70 (d) 71

33. What is the HM of 1,½, 1/3, ...................... 1/n?

(a) n (b) 2n (c) 2

(n +1) (d) n (n +1)

234. An aeroplane flies from A to B at the rate of 500 km/hour and comes back from B to A at the

rate of 700 km/hour. The average speed of the aeroplane is

(a) 600 km. per hour (b) 583.33 km. per hour

(c) 100 35 km. per hour (d) 620 km. per hour.

35. If a variable assumes the values 1, 2, 3...5 with frequencies as 1, 2, 3. ..5, then what is the AM?

(a)113 (b) 5 (c) 4 (d) 4.50

36. Two variables x and y are given by y= 2x - 3. If the median of x is 20, what is the median of y?

(a) 20 (b) 40 (c) 37 (d) 35

37. If the relationship between two variables u and v are given by 2u + v + 7 = 0 and if the AM ofu is 10, then the AM of v is

(a) 17 (b) - 17 (c) - 27 (d) 27.

38. If x and y are related by x - y - 10 = 0 and mode of x is known to be 23, then the mode of y is

(a) 20 (b) 13 (c) 3 (d) 23.

39. If GM of x is 10 and GM of y is 15, then the GM of xy is

(a) 150 (b) Log 10 x Log 15 (c) Log 150 (d) None of these.


40. If the AM and GM for 10 observations are both 15, then the value of HM is

(a) Less than 15 (c) More than 15

(b) 15 (d) Cannot be determined.

41. Which of the following statements is correct?

(a) Two distributions may have identical measures of central tendency and dispersion.

(b) Two distributions may have the identical measures of central tendency but differentmeasures of dispersion.

(c) Two distributions may have the different measures of central tendency but

identical measures of dispersion.

(d) All the statements (a), (b) and (c).

42. Dispersion measures

(a) The scatterness of a set of observations

(b) The concentration of a set of observations

(c) Both (a) and (b) (d) Neither (a) and b).

43. When it comes to comparing two or more distributions we consider

(a) Absolute measures of dispersion (b) Relative measures of dispersion

(c) Both a) and b) (d) Either (a) or (b).

44. Which one is difficult to compute?

(a) Relative measures of dispersion (b) Absolute measures of dispersion

(c) Both (a) and (b) (d) Range

45. Which one is an absolute measure of dispersion?

(a) Range (b) Mean Deviation

(c) Standard Deviation (d) All these measures

46. Which measure of dispersion is the quickest to compute?

(a) Standard deviation (b) Quartile deviation

(c) Mean deviation (d) Range

47. Which measures of dispersions is not affected by the presence of extreme observations?


(c) Standard deviation (d) Quartile deviationMEASURES OF CENTRAL TENDENCY OR AVERAGES 181

48. Which measure of dispersion is based on the absolute deviations only?


(c) Quartile deviation (d) Range

49. Which measure is based on only the central fifty percent of the observations?

(a) Standard deviation (b) Quartile deviation

(c) Mean deviation (d) All these measures

50. Which measure of dispersion is based on all the observations?

(a) Mean deviation (b) Standard deviation

(c) Quartile deviation (d) (a) and (b) but not (c)

51. The appropriate measure of dispersions for open - end classification is


(c) Quartile deviation (d) All these measures.

52. The most commonly used measure of dispersion is

(a) Range (b) Standard deviation

(c) Coefficient of variation (d) Quartile deviation.

53. Which measure of dispersion has some desirable mathematical properties?


(c) Quartile deviation (d) All these measures

54. If the profits of a company remains the same for the last ten months, then the standard deviationof profits for these ten months would be ?

(a) Positive (b) Negative

(c) Zero (d) (a) or (c)

55. Which measure of dispersion is considered for finding a pooled measure of dispersion aftercombining several groups?


(c) Quartile deviation (d) Any of these

56. A shift of origin has no impact on


(c) Standard deviation (d) All these and quartile deviation.


57. The range of 15, 12, 10, 9, 17, 20 is

(a) 5 (b) 12 (c) 13 (d) 11

58. The standard deviation of, 10, 16, 10, 16, 10, 10, 16, 16 is

(a) 4 (b) 6 (c) 3 (d) 0

59. For any two numbers SD is always

(a) Twice the range (b) Half of the range

(c) Square of the range (d) None of these.

60. If all the observations are increased by 10, then

(a) SD would be increased by 10

(b) Mean deviation would be increased by 10

(c) Quartile deviation would be increased by 10

(d) All these three remain unchanged.

61. If all the observations are multiplied by 2, then (a) New SD would be also multiplied by 2 then

(a) New SD would be also multiplied by 2

(b) New SD would be half of the previous SD

(c) New SD would be increased by 2

(d) New SD would be decreased by 2

62. What is the coefficient of range for the following wages of 8 workers?

` 80, ` 65, ` 90, ` 60, ` 75, ` 70, ` 72, ` 85.

(a) ` 30 (b) ` 20 (c) 30 (d) 20

63. If Rx and Ry denote ranges of x and y respectively where; and yare related ,by 3x+2y+10=0,what would be the relation between x and y?

(a) Rx = Ry (b) 2 Rx = 3 Ry (c) 3 Rx = 2 Ry (d) Rx = 2 Ry

64. What is the coefficient of range for the following distribution?

Class Interval : 10-19 20-29 30-39 40-49 50-59

Frequency : 11 25 16 7 3

(a) 22 (b) 50 (c) 72.46 (d) 75.82

65. If the range of x is 2, what would be the range of -3x +50 ?

(a) 2 (b) 6 (c) -6 (d) 44


66. What is the value of mean deviation about mean for the following numbers ? 5, 8, 6, 3, 4.

(a) 5.20 (b) 7.20 (c) 1.44 (d) 2.23

67. What is the value of mean deviation about mean for the following observations?

50, 60, 50, .50, 60, 60, 60, 50, 50, 50, 60, 60, 60, 50.

(a) 5 (b) 7 (c) 35 (d) 10

68. The coefficient of mean deviation about mean for the first 9 natural numbers is

(a) 200/9 (b) 80 (c) 400/9 (d) 50.

69. If the relation between x and y is 5y-3x = 10 and the mean deviation about mean for x is 12,then the mean deviation of y about mean is

(a) 7.20 (b) 6.80 (c) 20 (d) 18.80.

70. If two variables x and yare related by 2x + 3y -7 =0 and the mean and mean deviation aboutmean of x are 1 and 0.3 respectively, then the coefficient of mean deviation of y about mean is

(a) -5 (b) 12 (c) 50 (d) 4.

71. The mean deviation about mode for the numbers 4/11, 6/11, 8/11, 9/11, 12/11, 8/11 is

(a) 8/11, (b) 1/6 (c) 6/11 (d) 5/11

72. What is the standard deviation of 5, 5, 9, 9, 9,10, 5, 10, 10?

(a) 4 (b) 42 / 3 (c) 4.50 (d) 8

73. If the mean and SD of x are a and b respectively, then the SD of x a

b

is ,

(a) -1 (b) 1 (c) ab (d) a/b.74. What is the coefficient of variation of the following numbers?

53, 52, 61, 60, 64.(a) 8.09 (b) 18.08 (c) 20.23 (d) 20.45

75. If the SD of x is 3, what is the variance of (5-2x)?

(a) 36 (b) 6 (c) 1 (d) 9

76. If x and y are related by 2x+3y+4 = 0 and SD of x is 6, then SD of y is

(a) 22 (b) 4 (c) 5 (d) 9

77. The quartiles of a variable are 45, 52 and 65 respectively. Its quartile deviation is

(a) 10 (b) 20 (c) 25 (d) 8.30


78. If x and y are related as 3x+4y = 20 and the quartile deviation of x is 12, then the quartiledeviation of y is

(a) 16 (b) 14 (c) 10 (d) 9

79. If the SD of the 1st n natural numbers is 2, then the value of n must be

(a) 2 (b) 7 (c) 6 (d) 5

80. If x and yare related by y = 2x+ 5 and the SD and AM of x are known to be 5 and 10 respectively,then the coefficient of variation is

(a) 25 (b) 30 (c) 40 (d) 20

81. The mean and SD for a,b and 2 are 3 and 23

respectively, the value of ab would be

(a) 5 (b) 6 (c) 12 (d) 3


ANSWER KEYS

1 16 31 46 61 76

2 17 32 47 62 77

3 18 33 48 63 78

4 19 34 49 64 79

5 20 35 50 65 80

6 21 36 51 66 81

7 22 37 52 67

8 23 38 53 68

9 24 39 54 69

10 25 40 55 70

11 26 41 56 71

12 27 42 57 72

13 28 43 58 73

14 29 44 59 74

15 30 45 60 75

HOME WORK - 2

1. Which of the following statements is true?

(a) Usually mean is the best measure of central tendency.

(b) Usually median is the best measure of central tendency.

(c) Usually mode is the best measure of central tendency.

(d) Normally, GM is the best measure of central tendency.

2. The mean salary for a group of 40 female workers is 5200 per month and that for a group of 60male workers is 6800 per month . What is the combined mean salary?

(a) 6500 (b) 6200 (c) 6160 (d) 6100

3. The standard deviation of, 10, 16, 10, 16, 10, 10, 16, 16 is

(a) 4 (b) 6 (c) 3 (d) 0

4. If there are 3 observations 15, 20, 25 then the sum of deviation of the observations from theirAM is

(a) 0 (b) 5 (c) –5 (d) None of these.

5. If the profits of a company remains the same for the last ten months, then the standard deviationof profits for these ten months would be ?

(a) Positive (b) Negative (c) Zero (d) (a) or (c)

6. Which measure of dispersion is based on all the observations?


(c) Quartile deviation (d) (a) and (b) but not (c)

7. Usually _________ is the best measure of central tendency.

(a) Median (b) Mode (c) Mean (d) G.M.

8. ________ are used for measuring central tendency , dispersion & skewness.

(a) Median (b) Deciles (c) Percentiles (d) Quartiles.

9. What is the value of the first quartile for observations 15, 18, 10, 20, 23, 28, 12, 16?

(a) 17 (b) 16 (c) 12.75 (d) 12


10. What is the coefficient of range for the following wages of 8 workers?

` 80, `65, ` 90, `60, ` 75, ` 70, ` 72, ` 85.

(a) ` 30 (b) ` 20 (c) 30 (d) 20

11. If the mean deviation of a normal variable is 16, what is its quartile deviation?

(a) 10.00 (b) 13.50 (c) 15.00 (d) 12.05

12. If S.D.= 20 and sample size is 100 then standard error of mean is

(a) 2 (b) 5 (c) (d) None of these.

13. For any two numbers SD is always

(a) Twice the range. (b) Half of the range.

(c) Square of the range. (d) None of these.

14. ______ is an absolute measure of dispersion.

(a) Range (b) Mean Deviation

(c) Standard Deviation (d) All these measures

15. What is the median for the following observations?

5, 8, 6, 9, 11, 4.

(a) 6 (b) 7 (c) 8 (d) None of these.

16. The third decile for the numbers 15, 10, 20, 25, 18, 11, 9, 12 is

(a) 13 (b) 10.70 (c) 11 (d) 11.50

17. If the range of x is 2, what would be the range of –3x +50 ?

(a) 2 (b) 6 (c) –6 (d) 44

18. What is the standard deviation of 5, 5, 9, 9, 9, 10, 5, 10, 10?

(a) 14 (b)423

(c) 4.50 (d) 8

19. If the quartile deviation of a normal curve is 4.05, then its mean deviation is

(a) 5.26 (b) 6.24 (c ) 4.24 (d) 4.80

20. In case of an even number of observations which of the following is median ?

(a) Any of the two middle-most value.

(b) The simple average of these two middle values.

(c) The weighted average of these two middle values.

(d) Any of these.MEASURES OF CENTRAL TENDENCY OR AVERAGES 187

21. If all the observations are increased by 10, then

(a) SD would be increased by 10.

(b) Mean deviation would be increased by 10.

(c) Quartile deviation would be increased by 10.

(d) All these three remain unchanged.

22. Mode of 0,3,5,6,7,9,12,0,2 is

(a) 6 (b) 0 (c) 3 (d) 5

23. ______ always lies in between the arithmetic mean & mode.

(a) G.M. (b) H.M. (c) Median (d) G.M. and H.M.

24. The harmonic mean for the numbers 2, 3, 5 is

(a) 2.00 (b) 3.33 (c) 2.90 (d) –3 30

25. The coefficient of mean deviation about mean for the first 9 natural numbers is

(a) 200/9 (b) 80 (c) 400/9 (d) 50

26. Which of the following measures of central tendency is based on only fifty percent of the centralvalues?

(a) Mean (b) Median (c) Mode (d) Both (i) and (ii)

27. Difference between the maximum & minimum value of a given data is called______ .

(a) Width (b) Size (c) Range (d) Class

28. ______ is used when distribution pattern has to be studied at varying levels.

(a) A.M (b) Median (c) G.M (d) Mode

29. ______ is extremely sensitive to the size of the sample.

(a) Range (b) Mean (c) Median (d) Mode

30. If there are two groups containing 30 and 20 observations and having 50 and 60 as arithmeticmeans, then the combined arithmetic mean is

(a) 55 (b) 56 (c) 54 (d) 52

31. The algebraic sum of deviations of a set of observations from their AM is

(a) Negative. (b) Positive. (c) Zero. (d) None of these.

32. For a set of observations, the sum of absolute deviations is ______ when the deviations aretaken from the median.

(a) Zero (b) Maximum

(c) Minimum (d) None of these


33. When we want to divide the given set of observations into two equal parts, we consider ______.

(a) Mean (b) Median (c) Mode (d) None of these

34. The mean weight for a group of 40 female students is 42 kg and that for a group of 60 malestudents is 52 kg. What is the combined mean weight?

(a) 46 (b) 47 (c) 48 (d) 49

35. The wages of 8 workers expressed in rupees are 42,45,49,38,56,54,55,47. Find median wage.

(a) 47 (b) 48 (c) 49 (d) 50

36. The variables x and y are related by 5x+6y=70 and median of x is 8. What is the median of y?

(a) 4 (b) 4.5 (c) 5 (d) 5.5

37. If the relationship between x and y is given by 4x+5y=10 and the range of x is 15, what wouldbe the range of y?

(a) 10 (b) 11 (c) 12 (d) 13

38. First quartile is the value for which one fourth of the observations are __________Q1 and theremaining three–fourths observations are _____________Q1.

(a) Less than or equal to, More than or equal to

(b) More than or equal to, Less than or equal to

(c) Less than, More than

(d) More than, Less than

39. 50% of actual values will be below & 50% of values will be above ________.

(a) Mode (b) Median (c) Mean (d) Q1

40. Find D6 for the following observations.

7,9,5,4,10,15,14,18,6,20

(a) 11.40 (b) 12.40 (c) 13.40 (d) 13.80

41. For a moderately skewed distribution of marks in statistics for a group of 100 students, themean mark and median mark were found to be 50 and 40. What is the modal mark?

(a) 15 (b) 20 (c) 25 (d) 30

42. Following are the marks of 10 students :

82, 79, 56, 79, 85, 95, 55, 72, 70, 66 .

Find coefficient of range.

(a) 25.66 (b) 26.67 (c) 27.66 (d) 28.67


43. The value of middlemost item when they are arranged in order of magnitude is called

(a) Standard deviation (b) Mean (c) Mode (d) Median

44. The value of deciles divides the total no. of observations into ________ equal parts


45. Coefficient of variation of two series are 60% and 80% respectively. Their standard deviationare 20 and 16 respectively what are their A.M.

(a) 15 and 20 (b) 33.3 and 20

(c) 33.3 and 15 (d) 12 and 16

46. The height of 8 boys in a class (in cumulative) are 135, 138, 160, 141, 155, 146, 158, 149 Find61st percentile.

(a) 139.81 (b) 151.91 (c) 153.98 (d) None of these

47. What is the modal value for the numbers 4, 3, 8, 15, 4, 3, 6, 3, 15, 3, 4.


48. A graphical representation of __________ can be prepared in 2 different ways.

(a) Median (b) Mode (c) Mean (d) None

49. The mean weight of 150 students in a class is 60 kg. The mean weight of the boys is 70 kg, whilethat of girls is 55 kg find the number of boys and that of the girls in the class.

(a) 50 boys and 100 girls (b) 100 boys and 50 girls

(c) 75 boys and 75 girls (d) None of these

50. The class having maximum frequency is called

(a) Modal class (b) Median class

(c) Mean Class (d) None of these

51. Theoretically, A.M. is the best average in the construction of index nos. but in practice, mostlythe G.M. is used

(a) False (b) True (c) Both (d) None of these

52. In __________________ the quantities are in ratios

(a) A.M. (b) G.M. (c) H.M. (d) None of these

53. If each item is reduced by 12 A.M. is

(a) Reduced by 12 (b) Increased by 12

(c) Unchanged (d) None of these


54. There are 5 bags of wheat weighing on an average 102 kgs and another 8 bags weighing 98 kgson an average. What is combined mean of 13 bags.


55. The standard deviation of 25, 32, 43, 53, 62, 59, 48, 31, 24, 33 is


56. Which result is true?

(a) HM GM AM (b) HM GM AM

(c) HM < GM < AM (d) GM > AM > HM

57. Median is effected by extreme valves

(a) True (b) False (c) Both (d) None of these

58. Wages of 8 workers expressed in Rs. as follows – 80, 96, 51, 72, 67, 50, 70. find Coefficient ofrange.

(a) 30.5 (b) 35.3 (c) 31.00 (d) 31.51

59. If Arithmetic Mean and coefficient of variation of x are 5 and 20 respectively. What is thevariance of (15-2x)?

(a) 16 (b) 2 (c) 4 (d) 32

60. More laborious numerical calculation involves in A.M. than GM.

(a) True (b) Fals (c) Both (d) None of these

61. Which measures of dispersion is the quickest to compute


(c) Quartile deviation (d) Range

62. 2nd decile is greater than 1st decile


63. The mean height of 8 student is 152 cm. Two more students of heights 143 cm and 156 cm jointhe group. New mean height is equal to

(a) 153 (b) 152.5 (c) 151.5 (d) 151

64. If the first Quartile is 104 and quartile deviation is 18, the third quartile will be

(a) 140 (b) 116 (c) 20 (d) 0

65. If the quartile deviation of x is 6 and 4x + 8y = 20, What is the quartile deviation of y

(a) 3 (b) 4 (c) 5 (d) 1


66. The presence of extreme observations does not affect

(a) AM (b) Median (c) Mode (d) Any of these

67. S.D. of first n natural numbers is

(a) (b)

(c) (d) None of these

68. For any two numbers range is always

(a) Twice the SD (b) Half the SD

(c) Square the SD (d) None of these

69. G.M. is defined only when

(a) All observation have the positive sign and none is zero.

(b) All observation have the different sign and none is zero

(c) All observation have same sign and one is zero

(d) All observation have the different sign and one is zero

70. The standard deviation is required to determine sample size for

(a) Estimating a mean (b) Estimating a proportion

(c) Both (d) None of these

71. The standard deviation of 100 and 150 items are 40,6 respectively; if mean of 250 item is 44,mean of 100 and 150 item are 50 and 5, then find S.D. for 250 items.

(a) 7.46 (b) 7.64 (c) 6.74 (d) 4.67

72. If the median of 5, 9, 11, 3, 4, x, 8 is 6, the value of x is equal to

(a) 6 (b) 5 (c) 4 (d) 3

73. The average marks scored by 50 students in a class were calculated to be 38. Later it was found,that marks of two students were wrongly copied as 34 and 23 instead of 43 and 32. Findcorrect average marks.

(a) 37.36 (b) 39.00 (c) 38.36

(d) None of these

74. For open end classification, which of the following is the best measure of central tendency

(a AM (b) GM (c) Median (d) Mode


75. Businessman uses to find out the operation cost, profit per unit of article, output per man etc.

(a) AM (b) GM (c) Median (d) Mode

76. The marks obtained by10 students in an examinations were as follows:

70, 65, 68, 70, 75, 73, 80, 70, 83, 86. Find mean deviation about the mean

(a) 5.3 (b) 5.4 (c) 5.5 (d) 5.6

77. Mean may lead to fallacious conditions in the absence of original observations


78. Quartile deviation can be affected by

(a) Poisson distribution (b) Binomial distribution

(c) Sampling fluctuations (d) None of these

79. Which measure of dispersion has some desirable mathematical properties.


(c) Quartile deviation (d) All these measure

80. If the same quantity is multiplied to all the values the mean shall ______ by the same amount.

(a) Add (b) Subtract (c) Multiply (d) Divide

81. The mean of numbers 1, 7, 5, 3, 4, 4 is m. The numbers 3, 2, 4, 2, 3, 3, P have mean m- 1 andmedian 1. Then mean of P and 1 is equal to _______

(a) 4.0 (b) 2.5 (c) 4.5 (d) 3.5

82. For the numbers 1, 2, 3, 4, 5, 6, 7 standard deviation is:


83. The combined mean of three groups is 12 and the combined mean of first two groups is 3. If thefirst, second and third groups have 2, 3 and 5 items respectively, then the mean of third groupis

(a) 10 (b) 21 (c) 12 (d) 13

84. Mode is

(a) Least frequent value (b) Middle most value

(c) Most frequent value (d) None of these

85. The sum of deviations of the given values from their ………… is always 0.

(a) Arithmetic Mean (b) G.M. (c) H.M. (d) Median


86. The sum of squares of the deviations of the given values from their ………… is minimum.

(a) Arithmetic Mean (b) Median (c) Mode (d) None of these

87. Which is greatly affected by the extreme values?

(a) Arithmetic Mean (b) Median (c) Mode (d) None of these

88. Which is not amenable to further algebraic treatment?

(a) Arithmetic Mean (b) Median (c) Mode (d) Both (d) (b) and (c)

89. In a frequency distribution, the mid value of a class is 15 and the class interval length is 4. Thelower limit of the class is

(a) 10 (b) 12 (c) 13 (d) 14

90. The range of 10 observations is 20. If each item is increased by 15, then the range of new seriesis


91. In grouped frequency distribution, if the Class interval Gap is unequal; then which dispersionis more appropriate?

(a) Q.D. (b) Range

(c) Mean deviation (d) Standard Deviation.

92. If the Standard Deviation of 10 observations is 4 and if each item is divided by – 2 then StandardDeviation of new series is

(a) 2 (b) –2 (c) 4 (d) None of these

93. Find the average of first 25 multiples of 5.


94. If a, b, c, d, e are five consecutive odd integers, then their average is

(a) a + 5 (b)

(c) 5(a + b + c + d + e) (d) a + 4

95. The variance of 10 observations is 4, then their Standard Deviation is +2 or –2. This is

(a) True (b) False

(c) Either (a) or (b) (d) None of these

96. The average of 2 numbers is 20 and their Standard Deviation is 5. Find the two numbers.

(a) 15, 25 (b) 30, 40

(c) 10, 15 (d) None of these194 STATISTICS (CPT)

97. In series of 5 observations, the values of mean and variance are 4.4. and 8.24 respectively. Ifthree observations are 1, 2 and 6 then the value of other two observations are

(a) 3, 2 (b) 4, 9 (c) 10, 4 (d) None of these98. For individual series, the rank of the median is

(a) (b)


99. Find the rank of the Median in the given series 3, 2, 4, 6, 5, 7(a) 3.5 (b) 4 (c) 4.5 (d) None of these

100. The mean of the values of 1, 2, 3 ........, n with respective frequencies x, 2x, 3x, ........ nx is

(a) (b)

(c) (d)

101. The mean of four observations is 10 and when a constant a is added to each observation, themean becomes 13. The value of a is

(a) 2 (b) - 3 (c) 3 (d) None of these

102. The Standard Deviation of a set of 50 items is 10. Find the Standard Deviation if every item isincreased by 5.


103. Find the coefficient of variation if the sum of squared deviations taken from mean 40 of 10observations is 360.


104. If the coefficient of mean deviation is 0.44 and the mean deviation from mean is 5.77; then themean is –

(a) 14 (b) 13.11 (c) 16 (d) None of these

105. The Standard Deviation of two values is equal to half their difference. This statement is

(a) True (b) false

(c) cannot say (d) None of theseMEASURES OF CENTRAL TENDENCY OR AVERAGES 195

106. The weighted arithmetic mean of first n natural numbers whose weights are equal to thecorresponding numbers is equal to:

(a) (b)


107. The mean weight of 15 persons is 110 kg. The mean weight of 5 of them is 100 and another 5 is125 kgs. What is the mean weight of the remainder?

(a) 110 kgs. (b) 105 kgs. (c) 100 kgs. (d) None of these

108. The sum of diviations of certain number of items measured from 2.5 is 50 and the sum ofdeviations of the same series measured from 3.5 is –50. Find the number of observations andtheir mean?

(a) 100, 3 (b) 200, 6 (c) 100, 4 (d) None of these

109. In which Central value arranging is required.

(a) Mean (b) G.M. (c) Median (d) H.M.

110. The average of n numbers is x. If each of the numbers is multiplied by (n+1); then the averageof new set of numbers is

(a) x (b)

(c) (n + 1).x (d) None of these

111. The average weight of 8 person increases by 1.5 kg, if a person weighing 65 kg replaced by anew person, what would be the weight of the new person?(a) 76 kg (b) 80 kg (c) 77 kg (d) None of these

112. The average of marks obtained by 120 students in a certain examination is 34. If the averagemarks of passed students is 39 and that of the failed students is 15; what is the number ofstudents who passed in the examination?(a) 100 (b) 150 (c) 200 (d) None of these

113. The average of 17 numbers is 45. The average of first 9 of these numbers is 51 and the last 9 ofthese numbers is 36. Find the 9th number?(a) 5 (b) 14 (c) 18 (d) None of these

114. The average of 11 results is 30, that of the first five is 25 and that of the last five is 28. Find thevalue of the 6th number?(a) 60 (b) 65 (c) 75 (d) None of these

115. If SD = 40 and the sample size is 150 the standard error of mean is _______

(a) 2 (b) 5 (c) 0.5 (d) None of these



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MEASURES OF CENTRAL TENDENCY OR AVERAGES · MEASURES OF CENTRAL TENDENCY OR AVERAGES MEASURES OF CENTRAL TENDENCY OR LOCATION A single expression, representing the whole group, is