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M3

Unit-3

Statistics

 

Question-1: Find the mean of the following dataset.

x

20

30

40

f

5

6

4

 

Sol.

We have the following table-

X

F

Fx

20

5

100

30

6

180

40

7

160

 

Sum = 15

Sum = 440

 

Then Mean will be-

 

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: Compute the variance and standard deviation.

Class

Frequency

0-10

3

10-20

5

20-30

7

30-40

9

40-50

4

 

Sol.

Class

Mid-value (x)

Frequency (f)

0-10

5

3

1470.924

10-20

15

5

737.250

20-30

25

7

32.1441

30-40

35

9

555.606

40-50

45

4

1275.504

Sum

 

4071.428

 

Then standard deviation,

 

Question-7: If the coefficient of skewness is 0.64. The standard deviation is 13 and mean is 59.2, then find the mode and median.

Sol.

We know that-

So that-

And we also know that-

 

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: Find the best values of a and b so that fit the data given in the table.

X

0

1

2

3

4

Y

1

2.9

4.8

6.7

8.6

 

 

Solution: 

0

1

0

0

1

2.9

2.9

1

2

4.8

9.6

4

3

6.7

20.1

9

4

8.6

13.4

16

 

 

 

 

 

Normal equations      …. (2)

  …. (3)

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

    …. (4)

   …. (5)

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

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

 

Question-10: Find the least-squares approximation of the second degree for the discrete data.             

-2

-1

0

1

2

15

1

1

3

19

Solution:

Let the equation of second-degree polynomial be

-2

15

-30

4

60

-8

16

-1

1

-1

1

1

-1

1

0

1

0

0

0

0

0

1

3

3

1

3

1

1

2

19

38

4

76

8

16

 

 

 

 

 

 

 

 

Normal equations are     

On putting the values of in equations (2), (3), (4), we have

On solving (5), (6), (7), we get

The required polynomial of the second degree is

Question-11: Fit a second-degree parabola to the following data:

x

0

1

2

3

4

y

1

1.8

1.3

2.5

6.3

 

Solution: Let and so that the parabola of fit becomes

  …. (i)

The normal equations are

Saving these as simultaneous equations we get

(i) becomes

Or 

Hence

 

Question-12: 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-13: Find the correlation coefficient between the values X and Y of the dataset given below by using the short-cut method-

X

10

20

30

40

50

Y

90

85

80

60

45

 

Sol.

X

Y

10

90

-20

400

20

400

-400

20

85

-10

100

15

225

-150

30

80

0

0

10

100

0

40

60

10

100

-10

100

-100

50

45

20

400

-25

625

-500

 

Sum = 150

 

360

 

0

 

1000

 

10

 

1450

 

-1150

 

Short-cut method to calculate correlation coefficient-

Question-14: Two variables X and Y are given in the dataset below, find the two lines of regression.

x

65

66

67

67

68

69

70

71

y

66

68

65

69

74

73

72

70

 

Sol.

The two lines of regression can be expressed as-

And

x

y

xy

65

66

4225

4356

4290

66

68

4356

4624

4488

67

65

4489

4225

4355

67

69

4489

4761

4623

68

74

4624

5476

5032

69

73

4761

5329

5037

70

72

4900

5184

5040

71

70

5041

4900

4970

Sum = 543

557

36885

38855

37835

 

Now-

And

The standard deviation of x-

Similarly-

Correlation coefficient-

 

Put these values in the regression line equation, we get

Regression line y on x-

Regression line x on y-