class x english maths chapter06

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This unit facilitates you in,

explaining the meaning of standard

deviation.

deriving formula to find the standard

deviation.

calculating variance and standard

deviation of ungrouped and grouped data

by direct, actual mean, assumed mean

and step deviation method.

explaining the meaning and use of

coefficient of variation.

calculating coefficient of variation and

interpret the result.

constructing pie-charts.

interpreting pie-charts.

Meaning of standard deviation

Formula to find standard

deviation.

Standard deviation of un-

grouped and grouped data by

direct method

actual mean method.

assumed mean method

step deviation method

Coefficient of variation

Pie charts.

6 Statistics

Statistical thinking will one day be as necessary

for efficient citizenship, as the ability to read

and write.

-H.G. Wells

Karl pearson

(1857-1936, England)

Karl Pearson, British statistician,

is a leading founder of modern field of statistics. He established

the discipline of mathematical sta-

tistics. The term 'standard devia-

tion' was first used by karl

pearson in 1894 as a replacement

of the term 'mean error' used by

Carl Guass.



124 UNIT-6

In our previous classes, we have already learnt that in any distribution of scores (data),the tendency of scores is to cluster or concentrate around the central part of thedistribution. The measures of central tendency are mean, median and mode. We havealso learnt how the knowledge of measures of central tendency cannot give a completeidea about the distribution. For example, consider the number of runs scored by twocircketers in a series.

A 59, 65, 73, 61, 67

B 83, 120, 40, 22, 60

The two sets of scores have the same mean 65. i.e. = 65.

In set A, all the scores are nearer to the mean 65, where as in set B, the scores are

widely scattered about the mean 65.

Thus, the measures of central tendency may not give a definite picture about thedistribution. We need to have a measure which tells us how the scores are distributed

or dispersed around the mean.

One of the measures which gives an idea about the scatterdness of the data of thedistribution is dispersion. It measures the degree and direction of variation of the scoresin the distribution. Out of the four measures of dispersion, we have already learnt aboutrange, quartile deviation and mean deviation, and also their limitations.

It is known that range depends only on the two extreme scores and the quartiledeviation takes into account only 50% of the scores. Mean deviation is considered as abetter measure than range and quartile deviation. In mean deviation, we have observed

that the deviation of scores from the mean may be positive or negative. But to get the realpicture we ignore the negative sign and take only the absolute values of the deviations.

Is there any other mathematical way to take only positive values for deviation fromthe mean?

Consider the example discussed above.

Scores ( ) 59 65 73 61 67

Deviation ( ) 6 0 8 4 2

x A

d x x

Instead of taking absolute values of d = –6 as |d| = +6 and of d = –4 as |d| = +4, wecan also square them and get the positive values.

i.e., d2 = (–6)2 = +36

d2 = (–4)2 = +16

Let us discuss another case where squaring the deviation is helpful to measuredispersion. Consider the set of data having a uniform distribution.

12 13 14 15 16

2 1 0 1 2

x

d Mean = x = 14

We observe that in both the above cases the average of deviations becomes zero.

Hence, further calculation of measure of dispersion will also be zero. Does this say that



Statistics 125

no score has deviated from the mean? This is not true. Therefore, there is a need to avoidthe negative sign of scores. This is done by squaring the deviations.

This is a better way to measure dispersion. Here the difference between each score

and the mean is squared before averaging them.

This measure of dispersion is known as Variance. The variance is always positive.

If we take the square root of the positive value of the variance, we get a positive and a

negative value. The positive square root of the variance is known as standard deviation.

Hence, standard deviation is defined as follows.

The square root of the average of the squared deviations from the arithmetic mean is

called standard deviation, or Root mean square deviation (RMS).

Stadard deviation is denoted by ''. It is read as 'sigma' sigma is a Greek letter. It

measures the spread of individual scores around the mean of the distribution. It is always

expressed in the same units as that of data. It shows how much variation is there from

the mean. A low standard deviation indicates that the scores tend to be very close to the

mean, where as a high standard deviation indicates that the data is spread very much

away from the mean.

Now, let us learn how to calculate the measure of dispersion, i.e., standard deviation.

Method of calculating standard deviation:

We know that the distribution of data may or may not be evenly spread, the scores

may be whole numbers or integers or decimals, and mean value may be either an integral

number or decimal number. Depending on the nature of data, we calculate standarddeviation () by different methods using specified formula.

Standard deviation for ungrouped data

Let us learn each method and understand the procedure of calculating standard

deviation. These methods are applicable for both ungrouped and grouped data. First, weshall take ungrouped data, understand the different methods and then extend them togrouped data.

Variance and standard deviation for ungrouped data are calculated by using thefollowing methods:

i) Direct method

ii) Actual mean method

iii) Assumed mean method

iv) Step deviation method.

(i) Standard deviation by direct method

Let us first derive the formula for calculating standard deviation by direct method.

By definition, standard deviation is the positive square root of the mean of thesquared deviations from the mean.

Let, x 1, x

2, x

3 ...........x

n be the given scores and x be their mean.

Deviations from the mean are1 2, ,...,

n x x x x x x .



126 UNIT-6

Sum of deviations = x x

Squared deviations from the mean are2 22

1 2, ,...n

x x x x x x

Average of squared deviations =

2

x x

n

Variance or 2 =

2

x x

n (Average of squared deviation

= variance and represented by 2)

Standard deviation = =

2

x x

n

(by definition)

=

22 2x xx x

n

=

22 2. .x x x x

n n n

=

22

2. .x x

x x n n

x x

n

We know that, 2 2 2 2 2.... ' ' times =x x x x n nx

=

222

2x nx

x n n

=2

2 2

2x

x x n

=2

2x x

n =

22x x

n n

x x

n

22

S.D orx x

= n n

S.D. is the square root of the difference between average of the squared values andthe square of the average of all the values.

This is the formula for calculating standard deviation by direct method.

We know that, standard deviation is the positive square root of variance and varianceis the square of standard deviation i.e., variance = ()2

By writing the formula for , we get variance

22 22 2

Variance = x x x x

n n n n

In both the formula for standard deviation and variance, observe that 'x ' stands forscores and 'n' for total number of scores. That means, we do not need deviations of scoresfrom the mean for calculation of variance or standard deviation.



Statistics 127

This method of calculating standard deviation is known as direct method.

Steps of direct method of calculating S.D ()

1. Find x 2

2. Find x and x 2

3. Substitute the values of x , x 2 and n in the formula,

variance = =

22x x

n n or, =

22x x

n n

Direct method is used when the squares of the scores can be easily obtained.

Let us consider an example to understand this method.

Example 1: The number of saplings planted by 8 students during a year are 2, 6, 12, 5,

9, 10, 7, 4. Calculate the standard deviation for the data.

Sol. Let us rewrite the scroes in ascending order and find the squares.

Note: In this method, arranging scores in an order is not very important. But it is apractice in statistics to arrange the scores before any calculation.

x 2 4 5 6 7 9 10 12 x = 55

x 2 4 16 25 36 49 81 100 144 x 2 = 455

Here n = 8, x = 55 and x 2 = 455

=

22x x

n n =

2455 55

8 8

= 256.88 (6.88) = 56.88 47.33 = 9.55 3.09

On an average, each score deviates from the mean by 3.09.

Limitations of direct method

In direct method, we need x , x 2 and n to calculate variance or standard deviation. Welater find x and x 2, substitute the values in the formula and calculate variance. Thefirst important calculation in this method is squaring the scores.

So, this method is more suitable when the scores are less in number and the scoresare small integers.

It is not very easy to square non-integers. So, direct method is not suitable when* the data is large

* the scores are not integers

Hence, we need alternate methods for calculating standard deviation in such cases

Also note that, in direct method we do not use deviations of scores to calculatestandard deviation.

Is there a method where standard deviation is calculated using deviation of scoresfrom the mean?

Let us now study the other methods, where deviation of the scores from the meanare used for calculation of standard deviation.

3.09

3 9.55

3 9

609 5500

5481

19



128 UNIT-6

(ii) Actual mean method.

Steps of actual mean method of calculating

1. Find actual mean (arithmetic mean) x

2. Find the deviation of each score x from the actual mean d = x x

3. Square the deviations d 2 =2

x x

4. Find the sum of squared deviations d 2

5. Find the average of squared deviations (This is variance)2

d

n

6. Find the square root of the average of squared deviations2

d

n

we get standard deviation.

Example 2: The number of children born in 10 different hospitals during a month are

9, 12, 15, 18, 20, 22, 23, 24, 26, 31 Caclulate the standard deviation

Sol. x d x x d2

9 –11 121

12 –8 64

15 –5 25

18 –2 4

20 0 0

22 2 4

23 3 9

24 4 16

26 6 36

31 11 121

x = 200 d 2 = 400

Arithmetic mean =200

2010

x

n

We know that

=2

d

n =

40040 6.32

10

In the above example, which is easier?

* finding the square of scores (x )

* finding the square of deviations (d )

Why is actual mean method more suitable than direct method? Discuss in groups.



Statistics 129

Limitations of acutal mean method This method seems to be suitable as long as the mean value is an integer. If mean

is an integer, deviations are also integers and hence squaring them is easy.

Suppose that the mean value x of a distribution is not an integer, but a decimal

number. Then all the deviations x x will become decimal numbers and squaring the

decimal numbers may be a tedious task.

In such cases, where mean is not an integer we follow another alternate methodknown as assumed mean method to calculate the standard deviation.

Here, the actual mean is not calculated. Instead we choose a score (A) from thedistribution which is supposed to be close to the mean, such that the difference (x – A) areall small numbers, possibly integers.

(iii) Assumed mean method

In this method, we assume some score of the distibution to be the mean (prefferably the scores at the centre), and then find the deviations of each score from the assumedmean.

Steps of assumed mean method of calculating 1. Assume a score as the mean A

2. Find the deviation of each score from the d = x – Aassumed mean.

3. Square the deviations. d 2 or (x – A)2


5. Find the sum of deviation d

6. Use the formula22

d d

n n

7. Substitute the values d 2 and d .

8. Calculate standard deviation .

Example 3: The number of sick people who were treated as outpatients in a hospitalon each day during a week are given below:

50, 56, 59, 60, 63, 67, 68 Caclulate the standard deviation.

Sol. For the given scores, let us take A = 60 as the assumed mean.

From this assumed mean, let us find the deviation of each score. i.e., d = x – A

x d = x – A d 2

50 –10 100

56 –4 16

59 –1 1

60 0 0

63 +3 9+

67 +7 49

68 +8 64

n = 7 d = +3 d 2 = 239

22d d

n n

2

239 3

7 7

34.14 (0.18)

33.96

5.83

1 5

1 8



130 UNIT-6

In this example,

* Suppose you were asked to follow the direct method, what difficulty you wouldhave faced?

* Suppose you were asked to follow the actual mean method, what would havehappened?

The actual mean is 60.42. This is not an integer, and hence all the deviationswill be decimal numbers. What diffculty you would have faced?

Find the standard deviation by using both these two methods and discuss theanswers in groups. Compare these two methods with the assumed mean method.

Why is assumed mean method much easier compared to actual mean methodand direct method in this type of situation.

When is assumed mean method more suitable? Discuss in groups.

(iv) Step deviation method

Example 4: The number of books issued in a school library during the first ten days of the month are as follows:

20, 30, 40, 60, 80, 90, 110, 120, 130, 140 Calculate the standard deviation

Sol. For the given scores, let us take the value A = 90 as the assumed mean.

Observe the scores, Do you find any speciality?

Note that all the scores have a common factor 10. They are all multiples of 10.

Hence, divide the difference of the score and the assumed mean by common factor10 to find the deviation. This makes the calculations much easier.

i.e., d =x A

c , this deviation is called step deviation.

So, when the data is very large in size and has a common factor, we choose an

assumed mean A, calculated d by usingx A

d C

where C is the common factor of

the scores and find S.D. using the formula =

22d d C

n n

This method is known as step deviation method.

In example 4, which is easier?

* Finding the squares of d before taking out the common factor C.

* Finding the squares of d after taking out the common factor C.

How does removing the common factor of deviations useful in calculating thestandard deviation.



Statistics 131

Steps of step deviation method of calculating standard deviation.1. Assume a score as the mean A

2. Find the difference of each score and the d = x – A

assumed mean

3. Identify the common factor of the scores C

4. Divide the difference by common factor, to get dx A

d C

5. Square the deviations d 2


7. Use the formula and find the standard deviation

22d d C

n n

x d = x – A Step deviation d 2

x A d

c

20 –70 –7 49

30 –60 –6 36

40 –50 –5 25

60 –30 –3 9

80 –10 –1 1

90 0 0 0

110 20 2 4

120 30 3 9

130 40 4 16

140 50 5 25

n = 10 d = –8 d 2 = 174

=

22d d

n n C =

2174 8

10 1010 =

17.4 0.64× 10

= 16.76 10 = 4.09 × 10 = 40.9

So for we have learnt that variance and standard deviation can be found by any oneof the four method.

As expected, the different methods should not give different answers for for thesame data. Hence, we can follow any one of these methods suitable to the data.

2 2

1 4



132 UNIT-6

Now, let us consolidate the details of all the four methods in the table,

Note: Assumed mean method and step deviation method are just simplified formsof direct method.

Now, let us take a set of scores and calculate the variance and standard deviationby all the four methods.

Example 5: Calculate variance and standard deviation for the first eight even naturalnumbers.

Sol. The first eight even natural numbers are

2, 4, 6, 8, 10, 12, 14, 16(i) By direct method

x 2 4 6 8 10 12 14 16 x = 72

x 2 4 16 36 64 100 144 196 256 x 2 = 816

Variance =

22x x

n n =

2816 72

8 8 = 102 – 81

Variance = 21

=

22

Variance x x

n n = 21 = 4.58

(ii) By actual mean methodx d = x – x d 2

2 –7 494 –5 256 –3 98 –1 110 1 112 3 914 5 2516 7 49

x = 72 d 2 = 168

Method of calculating

Direct method

Actual meanmethod

Assumed mean

method

Step deviationmethod

When to use

When the scores are less in number andthe squares of scores are easily obtained.

When the mean is an integer (not afraction)

When the mean of the given data is not

an integer.When the scores are large in numberand have a common factor.

Formula

22x x

n n

2d

n

22d d

n n 22d d

C n n

2

2

Mean = =

729

8

9

168Variance=

8

21

21

4.58

x x n

x

d

n

d

n

4.58

4 21.00,00

4 16

85 500

5 425

908 7500

7264

236



Statistics 133

(iii) By assumed mean method

x d = x – A d 2

2 –6 36

4 –4 16

6 –2 4

8 0 0

10 2 4

12 4 16

14 6 36

16 8 64

n = 8 d = +8 d 2 = 176

= Variance 21 4.58

= 4.58

(iv) By step deviation method:

x step deviation d 2

d =x A

C

2 –3 9

4 –2 4

6 –1 1

8 0 0

10 1 1

12 2 4

14 3 9

16 4 16

d = +4 d 2 = 44

Note: While solving problems on standard deviation, unless the method is

mentioned, you can judge and select the most suitable method to calculatestandard deviation.

So far we have learnt to calculate variance and standard deviation for ungroupeddata. Do the same methods hold good for grouped data also?

Is there any change in the formulae to be used? Let us learn.

Standard deviation of grouped data

We know that, the grouped data may be arranged in an order as individual scores orwith class intervals. Each of the individual scores or class intervals has frequency (f).While calculating the mean, in order to find the sum of scores, we multiply the score ormid-point of class-interval with frequency. i.e., f x .

12

20

6

10

Let the assumed mean be 8.

Common factor of scores

C = 2

i.e., A = 8, Variance

=

222d d

C n n

=

2

244 42

8 8

= (5.5 – 0.25) × 4

= 5.25 × 4

Variance = 21

= Variance 21 4.58

Let the assumed mean be 8.

i.e., A = 8

Variance =

22d d

n n

=

2176 8

8 8

= 22 – 1

Variance = 21



134 UNIT-6

In the same way, while calculating the standard deviation for grouped data, in orderto find the sum of deviations, we have to multiply the deviation of each score or thedeviation of the mid-point of class-interval with frequency i.e.fd.

Hence, fd and fd2 are calculated

The formula for standard deviation of groped data are given in the table below.

Observe that all the four methods we have discussed for ungrouped data holds goodfor grouped data also, with modifications in the formulae

Method of Direct Actual Assumed Step deviation

calculating method mean method mean method method

Formula

22 fx fx

n n

2 fd

n

22 fd fd

n n

22 fd fd C

n n

Study the following examples.

Example 6: The rainfall recorded in various places of five districts for six days aregiven below:

Rainfall in mm 35 40 45 50 55

Number of places 6 8 12 5 9

Calculate the standard deviation

Sol. In this example, rainfall in mm represents the scores (x) and number of places thefrequency (f).

Let us calcualte standard deviation by all the four methods.

(i) By direct method.

x f fx x 2 fx 2

35 6 210 1225 7350

40 8 320 1600 12800

45 12 540 2025 24300

50 5 250 2500 12500

55 9 495 3025 27225

n = 40 fx = 1815 fx 2 = 84175

=

22 fx fx

n n =

284175 1815

40 40

= 2104.38 2058.9 = 45.48 = 6.7

= 6.7



Statistics 135

(ii) By actual mean method

x f fx d x x d 2 fd 2

35 6 210 –10.4 108.2 649.2

40 8 320 –5.4 29.2 233.6

45 12 540 –0.4 1.6 19.2

50 5 250 4.6 21.2 106.0

55 9 495 9.6 92.2 829.8

n = 40 fx = 1815 fd 2 = 1837.8

Mean

fx

x n

1815

40 = 45.38 = 45.4

2 fd

n

1837.845.95 6.7

40

(iii) By assumed mean method.

x f d = x – A fd d 2 fd 2

35 6 –10 –60 100 600

40 8 –5 –40 25 200

45 12 0 0 0 0

50 5 +5 +25 25 12555 9 +10 +90 100 900

n = 40 fd = +15 fd 2 = 1825

Let A = 45

22 fd fd

n n

21825 15

45.46 6.740 40

(iv) By step deviation method

x f Step deviation fd d 2 fd 2

x A d C

35 6 –2 –12 4 24

40 8 –1 –8 1 8

45 12 0 0 0 0

50 5 +1 +5 1 5

55 9 +2 +18 4 36

n=40 fd = +3 fd 2 = 73

100

115

20

23



136 UNIT-6

Let the assumed mean, A = 45.

Identify the common factor of the scores, C = 5

22 fd fd C

d n

273 3

5 1.82 5 6.740 40

Example 7: The time (in seconds) taken by a group of students to solve a problem inmathematics is given in the table below.

Calculate the standard deviation of the data.

C-I 0-10 10-20 20-30 30-40 40-50

f 7 10 15 8 10

Sol. Let us find S.D by assumed mean method. In this example, the scores are groupedin class-intervals. In such cases, we first find the mid-point of the class intervals andthen consider any one of them (preferrably the mid-point of C.I. with highest frequency)as the assumed mean. The following steps are followed to find S.D by assumed meanmethod.

Assume the mid-point of C.I with highest frequency as mean (A). i.e., A = 25

CI f Midpoint d = x – A d 2 fd fd 2

x = x – 25

0–10 7 5 –20 400 –140 2800

10–20 10 15 –10 100 –100 1000

20–30 15 25 0 0 0 0

30–40 8 35 +10 100 +80 800

40–50 10 45 +20 400 +200 4000

n=50 fd =+40 fd 2 = 8600

2 22 8600 40

50 50

fd fd

n n = 172 0.64 171.36 = 13.1

13.1 Therefore on an average, the time taken by each student deviates from the actualmean time by about 13 seconds.

Observe that the mid-points of C.I have a common factor 5. Hence, step-deviationmethod is a more convenient method. Let us try it.

Here, we have to find the deviations as25

5

x A x d

C . The remaining steps

are followed as same.

240

280



Statistics 137

x f x Step deviation d 2 fd fd 2

x A d

C

0–10 7 5 –4 16 –28 112

10–20 10 15 –2 4 –20 40

20–30 15 25 0 0 0 0

30–40 8 35 +2 4 +16 32

40–50 10 45 +4 16 +40 160

n=50 fd = +8 fd 2 = 344

2 22 344 85

50 50

fd fd C

n n = 6.88 0.03 5 6.85 5 13.1

It is interesting to know that step deviation method can be carried out faster in aslightly different way when the grouped data is given in class intervals eitherarranged in ascending or decending order.

In this case, we simply write o against the assumed mean in the column d andthen write –1, –2..... above o and +1, +2 ...... below o.

You can check that these values will be nothing butx A

d i

, where i is the size of

the class intervals.

The remaining steps of finding d2, fd, fd2, fd and fd2 are followed as usual and S.Dis calculated using the same formula with slight modification,

22 fd fd i

n n .

Now let us calculate by this process.

x f x x A

d i

d 2 fd fd 2

0–10 7 5 –2 4 –14 28

10–20 10 15 –1 1 –10 10

20–30 15 25 0 0 0 0

30–40 8 35 +1 1 +8 8

40–50 10 45 +2 4 +20 40

n=50 fd = +4 fd 2 = 86

22 86 410

50 50

fd fd i

n n = 21.72 (0.08) 10 = 13.1

48

56

24

28



138 UNIT-6

ILLUSTRATIVE E XAMPLES

Example 1: The score points of 10 students in a quiz competition out of 25 totalpoints is given below. Calculate the mean, variance and standard deviation.Also interpret the results.

14, 16, 21, 9, 16, 17, 14, 12, 11, 20

Sol. Scores Deviation d 2

d x x

14 –1 1

16 +1 1

21 +6 36

9 –6 36

16 +1 1

17 +2 4

14 –1 1

12 –3 9

11 –4 16

20 +5 25

x = 150 d 2 = 130

Example 2: The number of accidents that happened during the 12 month of a year ina city were recorded as : 43, 41, 58, 55, 57, 42, 50, 47, 48, 58, 50, 58. Find themean and standard deviation.

Sol. * Let us rewrite the scores in an order.* The number of scores, n = 12 Scores x Deviation d 2

d = x – A

41 –9 81

42 –8 64

43 –7 49

47 –3 9

48 –2 4

48 –2 4

50 0 0

50 0 0

55 +5 25

57 +7 49

58 +8 64

58 +8 64

x = 597 d = –3 d 2 = 413

The number of scores, n = 10

(i) Finding the mean

Mean =150

1510

x x

n

Mean = 15x

(ii) Variance =2 130

1310

d

n

Variance = 13

(iii) S.D = Variance

= 13 = 3.6

S.D of the scores = 3.6

This means, each score on an averagedeviates from the mean by 3.6

(i) Mean =597

49.812

x x

n

x = 49.8

(ii) Since the mean value is not an

integer, let us find the S.D by assumed mean method.

Let the assumed mean A = 50

=

22d d

n n =

2413 3

12 12

34.4 0.06 34.34

= 5.86

On an average, each score deviatesfrom the mean value 49.8 by 5.86

31

28



Statistics 139

Example 3: Find the variance and standard deviation of the following scores : 68, 72,80, 84, 92, 100.

Sol. We observe that all the scores have 4 as common factor.

Let us take 84 as assumed mean i.e., A = 84

Score Deviation d 2

x 84

4

x d

68 –4 16

72 –3 9

80 –1 184 0 0

92 +2 4

100 +4 16

x = 496 d = –2 d 2 = 46

Mean =496

82.66

x

n

Example 4: The marks obtained by 60 students in a test are given as follows:

Marks 5-15 15-25 25-35 35-45 45-55 55-65

No. of students 8 12 20 10 7 3

Calculate the mean and standard deviation of the distribution. Also interpret theresults.

Sol. The mid-points of class intervals have 10 as a common factor and the mean is not aninteger.

Let us follow step deviation method.

C-I f x fx Step deviation d 2 fd fd 2

x A d

C

5-15 8 10 80 –2 4 –16 32

15-25 12 20 240 –1 1 –12 12

25-35 20 30 600 0 0 0 0

35-45 10 40 400 +1 1 +10 10

45-55 7 50 350 +2 4 +14 28

55-65 3 60 180 +3 9 +9 27

n = 60 fx= 1850 fd = +5 fd 2 = 109

Since, the data has a common factor and themean value (82.6) is not an integer, let us usestep deviation method to find 2 and .

d =84

4

x A x

C

i.e.,68 84 16

4

4 4 and so on.

(i) Variance =

2 22 46 2

6 6

d d

n n = 7.7 – 0.1

2 = 7.6

(ii) = Variance 7.6 = 2.76

8

6



140 UNIT-6

(i) Mean =1850

30.860

fx

n

(ii) S.D =

2 22 109 510

60 60

fd fd C

n n = 1.81 10 1.34 10

= 13.4

This means, on an average each score deviates from the mean value 30.8 by 13.4

Example 5: The mean of 30 scores is 18 and their standard deviation is 3. Find thesum of all the scores and also the sum of the squares of all the scores.

Sol. The number of scores = n = 30

The mean of scores = x = 18

We know that, sum of scores = x = 18 × 30 = 540

x x

n

Sum of all scores = 540

Standard deviation of 30 scores = = 3

2 =

22x x

n n

2218

30

x = 9

2

30

x = 9 + 324

x 2 = 333 × 30 = 9990

sum of squares of all the scores = 9990

E XERCISE 6.1

1. Calculate the standard deviation of the following data.

x 3 8 13 18 23

f 7 10 15 10 8

2. The number of books bought by 200 students in a book exhibition is given below.

No. of books 0 1 2 3 4No. of students 35 64 68 18 15

Find the variance and standard variation.

3. The daily minimum temperature recorded (in degrees F) at a place on a hill stationduring a week was as follows:

Monday Tuesday Wednesday Thursday Friday Saturday Sunday

4.1 3.2 2.1 1.8 1.6 2.2 1.4

Calculate the variance and standard deviation. Interpret the results.



Statistics 141

4. The marks scored by 60 students in a science test are given below.

Marks (x ) 10 20 30 40 50 60

No. of students( f ) 8 12 20 10 7 3

Calculate the variance and standard deviation.

5. The daily wages of 40 workers of a factory are given in the following table.

Wages in ` 30-34 34-38 38-42 42-46 46-50 50-54

No. of workers 4 7 9 11 6 3

Calculate (i) Mean (ii) Variance and (iii) Standard deviation of wages and interpretthe findings.

6. Calculate (i) Arithmetic mean and (ii) Standard deviation for the following frequency distribution.

C.I 20-25 25-30 30-35 35-40 40-45 45-50

f 8 3 15 12 8 4

7. Find the variance and standard deviation for the following distribution.

C.I 3.5-4.5 4.5-5.5 5.5-6.5 6.5-7.5 7.5-8.5

f 9 14 22 11 17

8. Mean of 100 items is 48 and their standard deviation is 10. Find the sum of all theitems and the sum of the squares of all the items.

9. In a study of diabetic patients in a village, the following observations were noted:

Age in years 10-20 20-30 30-40 40-50 50-60 60-70

No. of patients 2 5 12 19 9 3

Calculate the mean and standard deviation. Also interpret the results.

10. A group of 45 house owners contributed money towards the development of greenenvironment in a locality. The amount of money collected per month is shown below.

Amount ( ` ) 20-40 40-60 60-80 80-100 100-120

No. of house 2 7 12 19 5owners

Calculate the variance and standard deviations. interpret the results.

Coefficient of variation

Two garment factories A and B manufacture readymade shirts. The number of shirtsstiched on six consecutive days by the two factories are given in the table below.

A 85 102 37 60 72 106

B 44 60 55 70 63 56

On an average which factory prepares more number of shirts?



142 UNIT-6

Let us find the average number of shirts.

Factory A : Mean =462

776

x

n Factory B : Mean =

34858

6

x

n

We observe that the average number of shirts that A prepares is more than B.

Suppose a textile whole sale shop owner wants to place orders with a factory fordaily supply of shirts. The owner prefers consistency in supply, i.e., without much variability.

Which fact among A and B should the owner select?

It is a known fact that consistency in production and supply is always better thanmere higher output.

In our daily life we come across many such situations where consistency inperformance and production is preferred just quantity in large number.

For example,

* Selecting a batsman or bowler for cricket match

* Selecting a student for national level quiz competition.

* Selecting an athelete for olympics.

* Choosing a vegetable or a flower vendor.

How to calculate the consistency or variability?

In statistics, we calculate coefficient of variation (C.V) for this purpose.

The coefficient of variation is a relative measure of dispersion. It is based onthe arithmetic mean and standard deviation of a frequency distribution. It is alsocalled as a relative standard deviation.

Coefficient of variation is defined as: C.V =Standard deviation

100Arithmetic mean

C.V = 100x

It can be observed that coefficient of variation.

* is expressed as a percentage.

* is independent of units.

* is determined by mean x and standard deviation ().

* determines consistency or variability of the distribution.

But, how to decide the consistency or variability using coefficient of variation?

How to calculate C.V?

Let us again consider the example of factories A and B.

We know their mean value, let us now find S.D. and then calculate C.V.



Statistics 143

By comparing (1) and (2), we observe that the coefficient of variation for B is less

than the coefficient of variation for A.

We say, factory B is more consistent than factory A in preparing the shirts.

From the above discussion we can conclude that the coefficient of variation helps

us to comapre the consistency of two or more collections of data.

is more

is less

the given data is

consistent.less

the given data is consistent.more

Factory A

Score x d = x – A d 2

37 –48 2304

60 –25 625

72 –13 169

85 0 0

102 +17 289

106 +21 441

n=6 d = –48 d 2

=3828Let A = 85

=

22d d

n n

=

23828 48

6 6

= 638 64

= 574

= 23.96

C.V = 100x

=23.96

10077

C.V = 31.12

The coefficient of variation for the

number of shirts prepared by factory A

is 31.12 ......(1)

Factory B

Score x d = x – A d 2

44 –16 256

55 –5 25

56 –4 16

60 0 0

63 +3 9

70 +10 100

n=6 d = –12 d 2

=406Let A = 60

=

22d d

n n

=

2406 12

6 6

= 67.67 4

= 63.67

= 7.98

C.V = 100x

=7.98

10058

C.V = 13.76

The coefficient of variation for the number

of shirts prepared by factory B is 13.76.

.....(2)

86

38

25

13

When the coefficent

of variation



144 UNIT-6

ILLUSTRATIVE E XAMPLES

Example 1 : The total runs scored by two cricket players Arun and Bharath in 15matches are 1050 and 900 with standard deviation 4.2 and 3.0 respectively. Who isbetter run getter? Who is more consistent in performance?

Sol. Number of matches played n = 15

Average score of Arun1050

7015

x Average score of Bharath900

6015

x

Let us calculate C.V using x and S.D.

Player Mean S.D(s) C.V = 100x

Arun 70 4.24.2

100 6.070

Bharath 60 3.03.0

100 5.060

(i) The average score of Arun is greater than average score of Bharath, hence

Arun is a better run getter.

(ii) The coefficient of variation of Bharath is less than coefficient of variation of

Arun, hence Bharath is more consistent.

Example 2 : Calculate the standard deviation and coefficient of variation for the followingdistribution.

Marks (x ) 10 20 30 40 50

Number of students ( f ) 4 3 6 5 2

Solution :

Scores Frequency fx Deviation from d 2 fd 2

x f mean d x x

10 4 40 –19 361 144420 3 60 –9 81 243

30 6 180 +1 1 6

40 5 200 +11 121 605

50 2 100 +21 441 882

n = 20 fx = 580 fd 2 = 3180



Statistics 145

(i) Arithmetic mean =580

2920

fx x

n Mean = 29x

(ii) =2 3180

159 12.6120

fd

n = 12.61

(iii) C.V =12.61

100 10029x

=1261

43.4829

C.V = 43.48

E XERCISE 6.2

1. Calculate the coefficient of variation of the following data : 40, 36, 64, 48, 52.

2. The scores of a batsman in 8 innings are given as: 48, 40, 36, 35, 46, 42, 36, 37.Calculate (i) Mean (ii) Standard deviation (iii) Coefficient of variation.

3. If the coefficient of variation of a collection of data is 45 and its standard deviationis 2.5, then find the mean.

4. A group of 100 candidates attending a physical test for recruitment have theiraverage height as 163.8 cm with coefficent of variation 3.2. What is the standarddeivation of their heights?

5. If n = 10, 12x and x 2 = 1530, then calculate the coefficient of variation.

6. The coefficient of variations of two series are 58 and 69. Their standard deviations

are 21.2 and 51.6. What are their arithmetic means?

7. Batsman A gets an average of 64 runs per innings with standard deviation of 18runs, while batsman B get an average score of 43 runs with standard deviation of 9runs in an equal number of innings. Discuss the efficiency and consistency of boththe batsmen.

8. In two construction companies A and B, the average weekly wages in rupees andthe standard deviations are as follows:

Company Average of wages S.D of wages

(in ` ) in ( ` )

A 3450 6.21

B 2850 4.56

9. Calculate (i) Arithmetic mean (ii) Standard deviation and (iii) coefficient of variationfor the following frequency distribution.

Class interval Frequency

30-35 5

35-40 10

40-45 16

45-50 15

50-55 4

Determine which factory has greater variability inindividual wages?



146 UNIT-6

10. Marks obtained in the test by students of two sections A and B are given below.

Marks No. of students No. of students

25-30 5 5

30-35 10 12

35-40 25 20

40-45 8 8

45-50 2 5

Determine (i) Which section's performance is better?

(ii) Which section's performance is more variable?Pie charts

We have learnt that numerical data can be represented pictorially or through graphs.Recall that statistical data can be presented using pictographs, line graphs, bar graphs,histogram and pie charts.

In pie charts, the data is distributed over a circle. Hence, a pie chart is also calledCircle graph. A pie chart is a way of showing how something is shared or divided.

In a pie chart, if the circle represents the whole data then, each part of the datamakes a certain angle at the centre of the circle. So, each part of the data is representedby the sectors formed by drawing the central angle. Hence pie charts are also calledsector graphs.

Pie charts are useful in analysing statistical data regarding money, poll information,budget, weather, population etc.

Study the following examples.

Example 1 : There are 36 students in a class.

The following table shows how they usually come to school :

Walk Bicycle Bus Car School Van

12 8 3 4 9

Let us learn to represent this data by a pie- chart.

The whole group of 36 students is to be represented by a circle.We know, that the centre of the circle is 360°.

Thus, 36 students is represented by 360°. Each category of students are in proportionto the central angle formed by them.

If the whole group of 36 students correspond to 360° then each student corresponds

to360

36 = 10°

12 students who come by walk correspond to 12 × 10° = 120°. This means, 12students who walk to school is represented by 120° at the centre of the circle.



Statistics 147

Similarly, the number of students who come to school by different modes and the centralangle they represent is given in the following table.

Students who come Number of Central angle

to school by students

Walk 1212

36 × 360º =120°

Bicycle 88

36 × 360º = 80°

Bus 33

36 ×360º = 30°

Car 44

36 ×360º = 40°

School van 99

36 ×360º = 90°

36 360°

By constructing the above calculated central angles in a circle, we can representthe data in a pie - chart.

The steps for constructing pie-chart are:

Step 1 : Take the total number of components (students). 12 + 8 + 3 + 4 + 9 = 36

Step 2 : Find the central angle for each component. [as shown in the above table]

Step 3 : Draw a circle of suitable radius

Step 4 : Construct the central angles at the centre of the circle, using protractor.Step 5 : Mark the angles and write the component in each sector.

Example 2 : The four important types of trees found in one square kilometer of aforest area are given in the table. Draw a pie-chart.

Teak wood Rose wood Devadaru Eucalyptus

360 300 285 135

Sol: Let us follow the steps of constructing a pie chart.

Step 1 : Take the total number of trees 360 + 300 + 285 + 135 = 1080

Step 2 : Find the central angle for each type of tree

Type of tree Number of trees Central angle

Teak wood 360360

1080 × 360º = 120°

Rose wood 300300

1080× 360º = 100°

Devadaru 285285

1080× 360º = 95°

Eucalyptus 135135

1080 × 360º = 45°

1080 360°

R o s e w o o d

T e a k

w o o d

E u c a l y

p t u s

D e v a

d a r u

120° 100°

45° 95°

S c h o o l V a n

30° Bus

40°

Car80°

Bicycle

Walk

120°

90°



148 UNIT-6

Step 3 : Draw a circle of suitable radius.

Step 4 : Construct the central angles as calculated in the table.

Step 5 : Mark the angles and write the type of trees in each sector.

So far we have learnt to construct a pie chart to represent given numerical data.

Now let us learn to read a given pie chart and interpret the data.

Reading and interpreting pie charts

Example 3 : The pie chart given below shows the expenditure of a family on variousitems and its savings during a year.

Study the pie chart and answer the following questions

(i) If the total annual income of the family is ` 75,000, what is the expenditure on

children education?

(ii) What amount of income was spent on clothing?

(iii) How much of family's income is saved?

(iv) How much is the expenditure on food more than that on housing?

(v) What is the difference in the expenses on housing and transport?

Sol.

(i) Given, 12% of income is spent on children's education,

Total income of the family = ` 75,000.

Expenditure on children education =12

100 × 75,000 = ` 9,000

Out of ` 75,000, expenditure on children's education is ` 9,000

(ii) Given, 10% of income is spent on clothing

Expenditure on clothing =10

100× 75000 = 7500

Amount spent on clothing = ` 7500

(iii ) Given, 15% of income is saved

Amount saved =

15

100 × 75,000 = 11,250

Out of an income of 75,000, the money saved is ` 11,250

(iv) Given, the expenditure on food is 23% and on housing is 15%

Expenditure on food =23

100× ` 75,000 = ` 17,250

Expenditure on housing =15

100 × ` 75,000 = ` 11,250

Expenditure on food - expenditure on housing = ` 17,250 – ` 11,250 = ` 6,000

Expenditure on food is ` 6,000 more than the expenditure on housing

S a v i n g s

15%O t h e r s

2 0 %

T r a n s p o r t

2 0 % 12%

Housing

15%

Clothing10%

Food 23%



Statistics 149

(v) Expenditure on housing = ` 11,250 (calculated)

Expenditure on transport =5

100× ` 75,000 = ` 3,750

Difference in the expenses on housing and transport is ` 11,250 – ` 3,750 = ` 7,500

Example : 4

A pie chart representing the population of four cities is shown below. Read thepie chart and find the population of S city.

Sol.

Let the central angle of the sector representing the city S be x °

Then, 120° + 96° + 60° + x ° = 360°

276° + x ° = 360°

x ° = 360° - 276° = 84°

A sector of 120° corresponds= 100 lakh

to the Population

A sector of 84° corresponds 5= 84 × lakh = 70 lakh

to the Population 6

The Population of the city S = 70 lakh

E XERCISE 6.3

I. Draw pie charts to represent the following data.

1. The number of students who are willing to join their favourite sports.

Name of Foot ball Tennis Volley ball Hockey Basket ball Other

the sport

Number of

students 35 14 10 6 5 2

2. The survey carried out in the class regarding places of visit for excursion and thenumber of students who opted each place.

Places Mysore Bijapur Gokarna Chitradurga

Number of

students 14 6 2 18

3. The number of books issued to students by a school library on each day of the week.

Day Monday Tueday Wednesday Thursday Friday Saturday

Number of

books issued 52 124 168 84 36 16

?

60°

120°

96°

R

SQ

P

100 lakh

Try:

Find the population of Q city and R city



150 UNIT-6

4. The ages of the drivers of cars involved in fatal accidents during a certain year.

Age of driver under 25 25 to 40 40 to 60 Above 60 (in years)

% of drivers 15% 60% 20% 5%

5. A survey was conducted to study the various brands of soaps used by people in avillage.

Brand of soap A B C Others

Percent of Villagers 50% 30% 15% 5%

II. Study the pie charts given below and answer the questions in each case.

1. The given pie chart shows the annual agricultural yield of a certain place. If thetotal production is 8100 tons,

Answer the following questions

(i) What is the yield in tons of rice, ragi,

sugar cane and others.

(ii) How much percent do the production

of ragi exceeds that of rice?

(ii i) If the yield of sugarcane on a different year was 2400tons, find the yield of rice?

2. A group of people were interviewed and asked which T.V. Channel they liked themost. The results are shown in the pie chart.

Answer the questions.

(i) What fraction of the people who wereinterviewed watched

(a) Channel 3 (b) Channel 5

(c) Channel 1 (d) Channel 2

(e) Channel 9

(ii) If there were 200 people, how many viewed each of the Channels.

3. A housewife was asked to estimate how she spenttime on a particular day from 5. 00 a.m. to 8. 00 pm.

The results are shown on the pie chart given below

(i) What fraction of the time did she spend on each of the activities?.

(ii) What percentage of time did she spend on (a) washing(b) cleaning (c) cooking.

Ragi100°

140°

Other

Rice40°

80° S u g a r c a n e

C h a n

n e l

1

C h a n n e l

9

90°

Channel

2

45°

Channel

3

90°

C h a n n e

l

5

27°

108°

C l e

a n i n

g

120°60°

W a s h i n g

60°72°

48°L u n c h

a n d

t e a b r e a k s

S h o p p i n g

Cooking



Statistics 151

4. Study the following pie charts and answer the questions given below them.

Percentage composition of Human body

Bones

S k i n

MusclesHor monesand Enzy mes

1 1 0

2

5

61

3

1

Water

70%

P roteins

Other dry elements

14%

16%

(i) What percent of the total weight of human body is equivalent to the weight of theproteins in skin in human body?

(ii) What will be the quantity of water in the body of a person weighing 50kg?

(ii i) What is the ratio of distribution of proteins in the muscles to that of the distributionof proteins in the bones?

(iv) In the human body, what part is made of bones orskin?

(v) What should be the central angle to represent the

distribution of proteins and other dry elements inthe human body?

5. The pie chart given below shows the percentagedistribution of the expenditure incurred inpublishing a book. Study the pie chart and answerthe questions.

(i) If the total expenditure incurred is ` 3,60,000 whatis the expenditure incurred for each of the items?

(ii) If for a certain quantity of books, the publisher hasto pay ` 30,600 as printing cost, then what is the amount of royalty to be paid forthese books?

(ii i) By what percentage is the royalty on the book less than paper cost?

(iv) By what percentage is the binding cost more than transport cost?

(v) If the transport cost incurred is ` 36,600, what will be the binding cost?

(vi) How much is the binding cost less than the printing cost, if the printing cost incurredis ` 52, 600?

(vii) If 5500 copies are published and the transportation cost on them amounts to ` 82, 500, then what should be the selling price of the book so that the publisher canearn a proft of 25%?

(vii i) The price of a book is marked 20% above the cost price. If the marked price of thebook is ` 180, then what is the cost of the paper used in a single copy of the book?

Binding 20%

Tr anspor t10%

Paper cost25%

Ro y alty 15%

P r inting cost20%

A d v e r t i -

s e m e n t

1 0 %



152 UNIT-6

ANSWERS

E XERCISE 6.1

1] 6.32 2] v = 1.23, = 1.109 3] v = 0.82, = 0.905 4] v = 180.98, = 13.45

5] = 5.789 6] Mean = 34.6, = 7.286 7] v = 1.72, = 1.314

8] x = 4800, x 2 = 2,40, 400 9] X = 42.4, S.D = 11.63 10] 20.39

E XERCISE 6.2

1] C.V = 20.39 2] (i) 40 (ii) = 4.60 (iii) C.V = 11.5 3] X = 5.55 4] = 5.2

5] C.V = 25 6] 36.55 and 74.78 7] A = 28.125, B = 20.93 8] A = 0.18, B = 0.16; A

9] X = 42.8, = 5.5, C.V. = 12.85 10] Sec A = 4.7, C.V = 12.8 Sec B = 5.4,

C.V = 14.7, Section B performance is better and section B is more variable.

E XERCISE 6.3

II 1] (i) Ragi = 2,250 tons, Sugarcane = 1,800 tons, Rice = 900 tons, Others = 3,150 tons

(ii) 150% (iii) 1,200

2] (i) (a)1

8 (b)

3

40 (c)

3

10 (d)

1

4 (e)

1

4 (ii) (a) 25 (b) 15 (c) 60 (d) 50 (e) 50

3] (i) Cleaning =1

3, Washing =

1

6, Lunch & tea =

10

75, Shopping =

1

6, Cooking =

1

5

(ii) (a) washing = 16.66%, (b) cleaning = 33.33%, (c) cooking = 20%

4] (i) 1.6% (ii) 35 kg (iii) 2:1 (iv)4

15(v) 108º

5] (i) Transport = ` 36,000 ; Paper cost = ` 90,000 ; Binding = ` 72,000; Royalty = ` 54,000;

Advt = ` 36,000; printing = ` 72,000 (ii) Royalty = ` 22,950 (iii) 10% (iv) 10%

(v) Binding = ` 73,200 (vi) 0 (vii) S.P = ` 187.50 (viii) ` 37.5

Statistics

Standard deviation

v

Variance2v

Coef?cient of variation

Pie charts

Ungrouped data Grouped data

22x x

n n

22 fx fx

n n

2d

n

2 fd

n

22 fd fd

n n

22 fd fd

n n

22 fd fd

C n n

Step deviationmethod

22 fd fd

C n n

. . 100C V x

Constructingpie charts

Interpretationof pie charts

C.V. is more C.V. is less

Data is lessconsistent

Data is moreconsistent

Coefficient of

class x english maths chapter06

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