Back to Study material
M3


Unit-3


Statistics

Question-1: Find the arithmetic mean for the following distribution:

Class

0-10

10-20

20-30

30-40

40-50

Frequency

7

8

20

10

5

Solution. Let assumed mean (a) = 25

Class

Midvalue

Frequency

40— 50

Total

 

 

Question-2: Find the median of the following dataset-

Sol.

Class interval

Frequency

Cumulative frequency

0 - 10

3

3

10 – 20

5

8

20 – 30

7

15

30 – 40

9

24

40 – 50

4

28

So that median class is 20-30.

Now putting the values in the formula-

So that the median is 28.57

Question-3: Find the mode from the following data:

Age

0-6

6-12

12-18

18-24

24-30

30-36

36-42

Frequency

6

11

25

35

18

12

6

Solution.

Age

Frequency

Cumulative frequency

0-6

6-12

12-18

24-30

30-36

36-42

6

11

25

35

12

6

6

17

42

77

95

107

113

Mode

Question-4: Find the quartile deviation of the following data-

Class interval

0-10

10-20

20-30

30-40

40-50

Frequency

3

5

7

9

4

Sol.

Here N/4 = 28/4 = 7 so that the 7’th observation falls in class 10 – 20.

And

3N/4 = 21, and 21’st observation falls in the interval 30 – 40 which is the third quartile.

The quartiles can be calculated as below-

And

Hence the quartile deviation is-

Question-5: Find the mean deviation from the mean of the following data-

Class interval

0-6

6-12

12-18

18-24

24-30

Frequency

8

10

12

9

5

Sol.

Class interval

Mid-value

Frequency

d = x - a

f.d

|x - 14|

f |x - 14|

0-6

3

8

-12

-96

11

88

6-12

9

10

-6

-60

5

50

12-18

15

12

0

0

1

12

18-24

21

9

6

54

7

63

24-30

27

5

12

60

13

65

Total

 

44

 

-42

 

278

Then mean deviation from mean-

Question-6: Calculate the standard deviation of the following frequency distribution-

Weight

60 – 62

63 – 65

66 – 68

69 – 71

72 – 74

item

5

18

42

27

8

Sol.

Weight

Item (f)

X

d = x – 67

f.d

60 – 62

5

61

-6

-30

180

63 – 65

18

64

-3

-54

162

66 – 68

42

67

0

0

0

69 – 71

27

70

3

81

243

72 – 74

8

73

6

48

288

 

Total

 

100

 

 

 

45

 

873

Question-7: Calculate coefficient variation for the following frequency distribution.

Wages in Rupees earned per day

0-10

10-20

20-30

30-40

40-50

50-60

No. of Labourers

5

9

15

12

10

3

Solution:

We already calculated

Now, 

A.M    

A.M    

Coefficient of Variation

Question-8: Calculate Karl Pearson’s coefficient of skewness of marks obtained by 150 students.

Sol. The mode is not well defined so that first we calculate mean and median-

Class

f

x

CF

fd

0-10

10

5

10

-3

-30

90

10-20

40

15

50

-2

-80

160

20-30

20

25

70

-1

-20

20

30-40

0

35

70

0

0

0

40-50

10

45

80

1

10

10

50-60

40

55

120

2

80

160

60-70

16

65

136

3

48

144

70-80

14

75

150

4

56

244

Now,


 

And

Standard deviation-

Then-

Question-9: By the method of least squats, find the straight line that best fits the following data:

1

2

3

4

5

14

27

40

55

68

Solution:  Let the equation of the straight line best fit be   …. (1)

1

14

14

1

2

27

54

4

3

40

120

9

4

55

220

16

5

68

340

25

 

 

 

 

Normal equations are

Putting the values of in (2) and (3), we have

On solving (4) and (5), we get

On Substituting the values of a and b in (1), we get

Question-10: Fit a second-degree parabola to the following data by the least-squares method.

1929

1930

1931

1932

1933

1934

1935

1936

1937

352

356

357

358

360

361

361

360

359

Solution: Taking

Taking 

The equation is transformed into

1929

-4

352

-5

20

16

-80

-64

256

1930

-3

360

-1

3

9

-9

-27

81

1931

-2

357

0

0

4

0

-8

16

1932

-1

358

1

-1

1

1

-1

1

1933

0

360

3

0

0

0

0

0

1934

1

361

4

4

1

4

1

1

1935

2

361

4

8

4

16

8

16

1936

3

360

3

9

9

27

27

81

1937

4

359

2

8

16

32

64

256

Total

 

Normal equations are

On solving these equations, we get

Question-11: Find the correlation coefficient between age and weight of the following data-

Age

30

44

45

43

34

44

Weight

56

55

60

64

62

63

Sol.

X

Y

(

)
)

30

56

-10

100

-4

16

40

44

55

4

16

-5

25

-20

45

60

5

25

0

0

0

43

64

3

9

4

16

12

34

62

-6

36

2

4

-12

44

63

4

16

3

9

12

 

Sum= 240

 

360

 

0

 

202

 

0

 

70

 

 

32

Karl Pearson’s coefficient of correlation-

Here the correlation coefficient is 0.27.which is the positive correlation (weak positive correlation), this indicates that as age increases, the weight also increases

Question-12: Compute the Spearman’s rank correlation coefficient of the dataset given below-

Person

A

B

C

D

E

F

G

H

I

J

Rank in test-1

9

10

6

5

7

2

4

8

1

3

Rank in test-2

1

2

3

4

5

6

7

8

9

10

Sol.

Person

Rank in test-1

Rank in test-2

d =

A

9

1

8

64

B

10

2

8

64

C

6

3

3

9

D

5

4

1

1

E

7

5

2

4

F

2

6

-4

16

G

4

7

-3

9

H

8

8

0

0

I

1

9

-8

64

J

3

10

-7

49

Sum

 

 

 

280

Question-13: Find the regression line of y on x for the given dataset.

X

4.3

4.5

5.9

5.6

6.1

5.2

3.8

2.1

Y

12.6

12.1

11.6

11.8

11.4

11.8

13.2

14.1

Sol.

Let y = a + bx is the line of regression of y on x, where ‘a’ and ‘b’ are given as-

We will make the following table-

x

y

Xy

4.3

12.6

54.18

18.49

4.5

12.1

54.45

20.25

5.9

11.6

68.44

34.81

5.6

11.8

66.08

31.36

6.1

11.4

69.54

37.21

5.2

11.8

61.36

27.04

3.8

13.2

50.16

14.44

2.1

14.1

29.61

4.41

Sum = 37.5

98.6

453.82

188.01

Using the above equations we get-

On solving these both equations, we get-

a = 15.49 and b = -0.675

So that the regression line is –

y = 15.49 – 0.675x 

Question-14: Discuss the Reliability of Regression Estimates:

A

45

38

59

64

72

B

60

48

82

93

45

Solution:

For A,

45

38

59

64

72

2025

1444

3481

4096

5184

For B,

60

48

82

93

45

2025

1444

3481

4096

5184

Now,

45

38

59

64

72

60

48

82

93

45

2700

1824

4838

5952

3240

The standard error of Regression of estimates of y on x is

…..(Standard error of Regression of estimates of y on x is )