sTATICS qUESTIONS

MB1115: Statistics

Practice Assignment 1A (Solutions)

1: Find the mean deviation about the mean for the data

2, 5, 6, 7, 9, 10, 11, 14.

If 1 2 3, , ,..., nx x x x are n observations, then the mean deviation about the mean is given by,

1

1M.D.

n

i

i

x x xN

, where x = mean of the data.

First, we will find the mean which is obtained by dividing the sum of all the observations by the total number of

observations.

Thus, we obtain mean as:

2 5 6 7 9 10 11 14

8x

64

8

8.

Now, ix x represents the deviation of each observation from the mean. Thus, the deviations are found as

follows:

1 2 8x x

6,

2 5 8x x

3,

3 6 8x x

2,

4 7 8x x

1,

5 9 8x x

1,

6 10 8x x

2,

7 11 8x x

3,

MB1115: Statistics


8 14 8x x

6.

Remember ix x represents the absolute value of a deviation. Therefore, the absolute values of the deviations

are:

6, 3, 2, 1, 1, 2, 3, 6.

The summation sign in the formula indicates that the sum of the absolute values of these deviations is to be

found.

Thus, we find the mean deviation about mean as:

8

1

1M.D. i

i

x x xN

1

6 3 2 1 1 2 3 68

24

8

3.

Thus, the mean deviation about the mean for the given data is 3.

MB1115: Statistics


2: Find the mean deviation about the mean for the data

28, 60, 38, 30, 32, 45, 53, 36, 44, 34.

If 1 2 3, , ,..., nx x x x are n observations, then the mean deviation about the mean is given by

1

1M.D.

n

i

i

x x xN


First, we will find the mean which is obtained by dividing the sum of all the observations by the total number of

observations.

Thus, we obtain mean as:

28 60 38 30 32 45 53 36 44 34

10x

400

10

40.

Now, ix x represents the deviation of each observation from the mean.

Thus, the deviations are found as follows:

1 28 40x x

12,

2 60 40x x

20,

3 38 40x x

2,

4 30 40x x

10,

5 32 40x x

8,

6 45 40x x

5,

7 53 40x x

MB1115: Statistics


13,

8 36 40x x

4,

9 44 40x x

4,

10 34 40x x

6.

Remember ix x represents the absolute value of a deviation. Therefore, the absolute values of the deviations

are:

12, 20, 2, 10, 8, 5, 13, 4, 4, 6.

The summation sign in the formula indicates that sum of the absolute values of these deviations is to be found.

Thus, we find the mean deviation about the mean as:

10

1

1M.D. i

i

x x xN

1

12 20 2 10 8 5 13 4 4 610

84

10

8.4.

Thus, the mean deviation about the mean for the given data is 8.4.

MB1115: Statistics


3: Find the mean deviation about the median for the data.

11, 15, 14, 12, 9, 11, 8, 14, 9, 16, 10, 15.

If 1 2 3, , ,..., nx x x x are n observations, then the mean deviation about the median is given by:

1

1M.D. M

n

i

i

M xN

, where M = median of the data.

First, we find the median.

To find the median, we arrange the data in ascending order, to get

8, 9, 9, 10, 11, 11, 12, 14, 14, 15, 15, 16.

The number of observations, N = 12.

Since the number of observations is even, to find the median, we use the formula

observation 1 observation2 2

M .2

th thN N

Thus, substituting N=12, in the formula, we get

12 12observation 1 observation

2 2M

2

th th

6 observation 7 observation

2

th th

11 12

2

11.5.

The deviation of each observation is obtained by subtracting the median from it. Therefore, the deviations are

1 8 11.5x M

3.5,

2 9 11.5x M

2.5,

3 9 11.5x M

2.5,

MB1115: Statistics


4 10 11.5x M

1.5,

5 11 11.5x M

0.5,

6 11 11.5x M

0.5,

7 12 11.5x M

0.5,

8 14 11.5x M

2.5,

9 14 11.5x M

2.5,

10 15 11.5x M

3.5,

11 15 11.5

3.5,

x M

12 16 11.5

4.5.

x M

Now,

10

1

1M.D. M.i

i

M xN


found.

Thus, we find the mean deviation about median as:

1

M.D. 3.5 2.5 2.5 1.5 0.5 0.5 0.5 2.5 2.5 3.5 3.5 4.512

M

28

12

2.33.

Thus, the mean deviation about the median for the given data is 2.33.

MB1115: Statistics


4: Find the mean deviation about the median for the data

30, 66, 40, 36, 54, 39, 47, 40, 45, 43.

If 1 2 3, , ,..., nx x x x are n observations, then the mean deviation about the median is given by:

1

1M.D. M

n

i

i

M xN

, where M = median of the data.

First, we find the median.

To find the median, we arrange the data in ascending order, to get,

30, 36, 39, 40, 40, 43, 45, 47, 54, 66.

The number of observations, N = 10.

Since the number of observations is even, to find the median, we use the formula

observation 1 observation2 2

M .2

th thN N

Thus, substituting 10 for N in the formula, we get

10 10observation 1 observation

2 2M

2

th th

5 observation 6 observation

.2

th th

From the ascending list of observations, we observe that 5th observation is 40 and the 6th observation is 43.

Substituting these values on the right hand side of 5 observation 6 observation

M2

th th , we get

40 43

M2

41.5.

The deviation of each observation is obtained by subtracting the median from it.

Therefore, the deviations are:

1 30 41.5x M

MB1115: Statistics


11.5,

2 36 41.5x M

5.5,

3 39 41.5x M

2.5,

4 40 41.5x M

1.5,

5 40 41.5x M

1.5,

6 43 41.5x M

1.5,

7 45 41.5x M

3.5,

8 47 41.5x M

5.5,

9 54 41.5x M

12.5,

10 66 41.5x M

24.5.

Now,

10

1

1M.D. M.i

i

M xN


found.

Thus, we find the mean deviation about median as,

1

M.D. 11.5 5.5 2.5 1.5 1.5 1.5 3.5 5.5 12.5 24.510

M

70

10

7.

MB1115: Statistics


Thus, the mean deviation about the median for the given data is 7.

MB1115: Statistics


5: Find the mean deviation about the mean for the data.

xi 3 6 9 12 15

fi 5 2 4 1 3


1

1M.D.

n

i

i

x x xN


Let us make a table of the given data and append other columns after calculations:

xi fi fi xi ix x

i if x x

3 5 15 5 25

6 2 12 2 4

9 4 36 1 4

12 1 12 4 4

15 3 45 7 21

15 120 58

To find the mean for the given data used in column 4, we have used the formula

1

1 n

i i

i

x f xN

, where

1

.n

i

i

N f

From the table, we can find the values as,

5 5

1 1

15 and 120.i i ii i

N f f x

Substituting these values in 1

1 n

i i

i

x f xN

, we get

1120

15

8.

x

MB1115: Statistics


This value of the mean is used in the table to find the deviations, ix x , from the mean.

Also, from the table, we find that 5

1

56.i ii

f x x

Substituting the values of 5

1

i i

i

f x x

and N in 1

1M.D.

n

i

i

x x xN

, we get

5

1

1M.D. i i

i

x f x xN

158

15

3.87.

Thus, the mean deviation about the mean for the given data is approximately3.87 .

MB1115: Statistics


6: Find the mean deviation about the mean for the data?


1

1M.D.

n

i

i

x x xN


Let us make a table of the given data and append other columns after calculations:

xi fi fi xi ix x i if x x

5 2 10 40 80

25 12 300 20 240

45 14 630 0 0

65 8 520 20 160

85 4 340 40 160

40 1800 640

To find the mean for the given data used in column 4, we have used the formula,

1

1 n

i i

i

x f xN

, where

1

.n

i

i

N f

From the table, we can find the values as,

5 5

1 1

40 and 1800.i i ii i

N f f x

Substituting these values in 1

1 n

i i

i

x f xN

, we get

5

1

1i i

i

x f xN

11800

40

xi 5 25 45 65 85

fi 2 12 14 8 4

MB1115: Statistics


45.

This value of the mean is used in the table to find the deviations, ix x , from the mean.

Also, from the table, we find that,

5

1

640.i ii

f x x

Substituting the values of 5

1

i i

i

f x x

and N in the formula for the mean deviation about the mean, we get

5

1

1M.D. i i

i

M f x xN

1640

40

16.

Thus, the mean deviation about the median for the given data is 16.

MB1115: Statistics


7: Find the mean and variance for the data 3, 4, 7, 9, 10, 1, 5, 9.

We know that the mean of a given raw data is given by,

1

n

i

i

x

xN

,

where, 1

n

i

i

x

denotes the sum of all the observations and N denotes the number of observations.

Observing the data, we can find that 8.N Thus, we calculate mean ( x ) as, 8

1

8

i

i

x

x

3 4 7 9 10 1 5 9

8

48

8

6.

Now, if 1 2 3, , ,... nx x x x be n observations and x is their mean, then the variance of these observations is denoted

by 2 and is given by

22

1

1.

n

i

i

x xN

Now, we make the following table by constructing columns for ix x and 2

ix x :

MB1115: Statistics


xi ix x 2

i

x x

3 3 9

4 2 4

7 1 1

9 3 9

10 4 16

1 5 25

5 1 1

9 3 9

82

1

74ii

x x

From the table, we can see that

28

1

74ii

x x

.

Substituting this value and the value of N in the formula of variance that is

282

1

1i

i

x xN

, we get

282

1

1i

i

x xN

174

8

9.25.

Hence, the mean and the variance for the data are 6 and 9.25 respectively.

MB1115: Statistics


8: Find the mean and variance for the first 10 multiples of 2.

We know that the mean of a raw data is given by,

1 ,

n

i

i

x

xN

where , 1

n

i

i

x

denotes the sum of all the observations and N denotes the number of observations.

The first 10 multiples of 2 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.

We see that the number of observation is n = 10.

Thus, we find the required mean as 10

1

10

i

i

x

x

2+ 4+6+8+10+12+14+16+18+20

10

110

10

11.

Now, if 1 2 3, , ,... nx x x x be n observations and x is their mean, then the variance of these observations is denoted

by 2 and is given by

22

1

1.

n

i

i

x xN

To find the variance, we construct the following table in which we find -ix x for each observation by using

11x :

MB1115: Statistics


xi ix x 2

i

x x

2 9 81

4 7 49

6 5 25

8 3 9

10 1 1

12 1 1

14 3 9

16 5 25

18 7 49

20 9 81

102

1

330ii

x x

From the table, we find that,

10

2

1

330ii

x x

.

Substituting this value and the value of N in the formula for variance,

22

1

1,

n

i

i

x xN

we get

2 1 33010

33.

Hence, the mean and variance for the first 10 multiples of 2 are 11 and 33 respectively.

MB1115: Statistics


9: Find the mean and variance for the data.

The given data is in the form of discrete frequency distribution. For discrete frequency distribution, the mean

and variance are given by,

1

n

i i

i

f x

xN

,

and

22

1

1 n

i i

i

f x xN

,

where x is the mean, 2 is the variance and

1

n

i

i

N f

.

Thus, we construct a table shown below:

xi f i fixi ix x 2

i

x x 2

i if x x

4 1 4 12.5 156.25 156.25

8 3 24 8.5 72.25 216.75

12 6 72 4.5 20.25 121.5

16 11 176 0.5 0.25 2.75

20 7 140 3.5 12.25 85.75

24 3 72 -7.5 56.25 168.75

28 2 56 11.5 132.25 264.5

7

1

33ii

f

7

1

544i ii

f x

7

2

1

1016.25i ii

f x x

From the table, we can find that

7

1

33ii

N f

and 7

1

544.i ii

f x

Substituting these values in the formula for mean ( x ), we get

xi 4 8 12 16 20 24 28

f i 1 3 6 11 7 3 2

MB1115: Statistics


7

1

i i

i

f x

xN

544

33

16.5

This value of x is used in the table to find ix x for each observation ix .

From the table, we also find that,

7

2

1

1016.25.i ii

f x x

Using this value and the value of N in the formula for variance, 22

1

1,

n

i i

i

f x xN

we get

2 1 1016.2533

30.80.

Hence, the mean and variance of the given data are approximately 16.5 and 30.80 respectively.

MB1115: Statistics


10: Find the mean and variance for the data.

The given data is in the form of discrete frequency distribution. For discrete frequency distribution, the mean

and variance are given by

1

n

i i

i

f x

xN

,

and

22

1

1 n

i i

i

f x xN

,

where x is the mean, 2 is the variance and

1

.n

i

i

N f

Thus, we construct a table shown below:

xi f i fixi ix x 2

i

x x 2

i if x x

90 2 180 8.3 68.89 137.78

91 1 91 7.3 53.29 53.29

95 2 190 3.3 10.89 21.78

96 1 96 2.3 5.29 5.29

100 5 500 1.7 2.89 14.45

102 2 204 3.7 13.69 27.38

107 2 214 8.7 75.69 151.38

7

1

15ii

f

7

1

1475i ii

f x

7

2

1

411.35i ii

f x x

From the table, we can find that, 7

1

15ii

N f

and 7

1

1475.i ii

f x

Substituting these values in the formula for mean ( x ), 7

1

1,i i

i

x f xN

we get

xi 90 91 95 96 100 102 107

f i 2 1 2 1 5 2 2

MB1115: Statistics


11475

15x

98.3.

This value of x is used in the table to find ix x for each observation ix .

From the table, we also find that,

7

2

1

411.35i ii

f x x

.

Using this value and the value of N in the formula for variance, we get

7

22

1

1i i

i

f x xN

1411.35

15

27.42.

Hence, the mean and variance of the given data are 98.3 and 27.42 .

MB1115: Statistics


11: An analysis of monthly salary paid to employees in two companies A and B, belonging to the same

work, gives the following results:

Company A Company B

No. of employees 580 642

Mean of monthly salary `6263 `6263

Variance of the distribution of salary 121 144

(i) Which company A or B pays larger amount as monthly salary?

(ii) Which company, A or B, shows greater variability in individual salary?

(i)

We know that,

Sum of the observationsMean = .

Totol number of observations

Therefore, we have

Sum of the observations = Mean Totol number of observations.

Applying the above statement in this problem, we have

Amount paid as monthly salary = Mean of monthly salary No. of employees.

For company A, we have

No. of employees = 586 and Mean of monthly salary = Rs. 6263.

Substituting these values in Amount paid as monthly salary = Mean of monthly salary No. of employees, we get

Amount paid as monthly salary = 6263 580.

For Firm B, we have

No. of wage earners = 642 and Mean of monthly wages = Rs. 6263.

Substituting these values in Amount paid as monthly salary = Mean of monthly salary No. of employees, we

get

MB1115: Statistics


Amount paid as monthly salary = 6263 642.

On comparing the amount paid as monthly wages for the two companies, we conclude that company B pays the

larger amount as monthly salary than company A.

(ii)

To compare the variability in individual salary for the two companies, we compare their standard deviations.

The company that has greater standard deviation will show more variability in individual salary.

We are given variance for both the companies. Therefore, to find standard deviation, we use the formula,

2 2, where is variance.

For company A, 2 is 121. Thus, we have

121

11.

And, for company B, 2 is 141. Thus, we have

144

12.

Now the coefficient of variations (C.V) are given by (C.V) for company A is

standard deviation of company 11100 100

mean 6263

0 175

A

. ,

(C.V) for company B is

standard deviation of company 12100 100

mean 6263

0 192

B

. .

Clearly C.V for company B is greater than the C.V for company A.

Thus, we can say that company B show more variability than company A.

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