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NCERT solution class 11 chapter 15 Statistics exercise 15.1 mathematics

EXERCISE 15.1


Page No 360:

Question 1:

Find the mean deviation about the mean for the data

4, 7, 8, 9, 10, 12, 13, 17

Answer:

The given data is

4, 7, 8, 9, 10, 12, 13, 17

Mean of the data, 

The deviations of the respective observations from the mean are

–6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e., are

6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is

Question 2:

Find the mean deviation about the mean for the data

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Answer:

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data,

The deviations of the respective observations from the mean are

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

The absolute values of the deviations, i.e. , are

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

The required mean deviation about the mean is

Question 3:

Find the mean deviation about the median for the data.

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Answer:

The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

The deviations of the respective observations from the median, i.e.are

–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations,, are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is

Question 4:

Find the mean deviation about the median for the data

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Answer:

The given data is

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

The deviations of the respective observations from the median, i.e.are

–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations,, are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Thus, the required mean deviation about the median is

Question 5:

Find the mean deviation about the mean for the data.

xi

5

10

15

20

25

fi

7

4

6

3

5

Answer:

xi

fi

fi xi

5

7

35

9

63

10

4

40

4

16

15

6

90

1

6

20

3

60

6

18

25

5

125

11

55

25

350

158

Question 6:

Find the mean deviation about the mean for the data

xi

10

30

50

70

90

fi

4

24

28

16

8

Answer:

xi

fi

fi xi

10

4

40

40

160

30

24

720

20

480

50

28

1400

0

0

70

16

1120

20

320

90

8

720

40

320

80

4000

1280

Question 7:

Find the mean deviation about the median for the data.

xi

5

7

9

10

12

15

fi

8

6

2

2

2

6

Answer:

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xi

fi

c.f.

5

8

8

7

6

14

9

2

16

10

2

18

12

2

20

15

6

26

Here, N = 26, which is even.

Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

The absolute values of the deviations from median, i.e.are

|xi – M|

2

0

2

3

5

8

fi

8

6

2

2

2

6

fi |xi – M|

16

0

4

6

10

48

and

Question 8:

Find the mean deviation about the median for the data

xi

15

21

27

30

35

fi

3

5

6

7

8

Answer:

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xifi

c.f.

15

3

3

21

5

8

27

6

14

30

7

21

35

8

29

Here, N = 29, which is odd.

observation = 15th observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

∴ Median = 30

The absolute values of the deviations from median, i.e.are

  |xi – M|

15

9

3

0

5

 fi

3

5

6

7

8

 fi |xi – M|

45

45

18

0

40

∴ 

Page No 361:

Question 9:

Find the mean deviation about the mean for the data.

Income per day

Number of persons

0-100

4

100-200

8

200-300

9

300-400

10

400-500

7

500-600

5

600-700

4

700-800

3

Answer:

The following table is formed.

Income per day

Number of persons fi

Mid-point xi

fi xi

0 – 100

4

50

200

308

1232

100 – 200

8

150

1200

208

1664

200 – 300

9

250

2250

108

972

300 – 400

10

350

3500

8

80

400 – 500

7

450

3150

92

644

500 – 600

5

550

2750

192

960

600 – 700

4

650

2600

292

1168

700 – 800

3

750

2250

392

1176

50

17900

7896

Here,

Question 10:

Find the mean deviation about the mean for the data

Height in cms

Number of boys

95-105

9

105-115

13

115-125

26

125-135

30

135-145

12

145-155

10

Answer:

The following table is formed.

Height in cms

Number of boys fi

Mid-point xi

fi xi

95-105

9

100

900

25.3

227.7

105-115

13

110

1430

15.3

198.9

115-125

26

120

3120

5.3

137.8

125-135

30

130

3900

4.7

141

135-145

12

140

1680

14.7

176.4

145-155

10

150

1500

24.7

247

Here,

Question 11:

Find the mean deviation about median for the following data:

Marks

Number of girls

0-10

6

10-20

8

20-30

14

30-40

16

40-50

4

50-60

2

Answer:

The following table is formed.

Marks

Number of girls fi

Cumulative frequency (c.f.)

Mid-point xi

|xi â€“ Med.|

fi |xi â€“ Med.|

0-10

6

6

5

22.85

137.1

10-20

8

14

15

12.85

102.8

20-30

14

28

25

2.85

39.9

30-40

16

44

35

7.15

114.4

40-50

4

48

45

17.15

68.6

50-60

2

50

55

27.15

54.3

50

517.1

The class interval containing theor 25th item is 20 – 30.

Therefore, 20 – 30 is the median class.

It is known that,

Here, l = 20, C = 14, f = 14, h = 10, and N = 50

∴ Median =

Thus, mean deviation about the median is given by,

Question 12:

Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Age

Number

16-20

5

21-25

6

26-30

12

31-35

14

36-40

26

41-45

12

46-50

16

51-55

9

Answer:

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age

Number fi

Cumulative frequency (c.f.)

Mid-point xi

|xi – Med.|

fi |xi – Med.|

15.5-20.5

5

5

18

20

100

20.5-25.5

6

11

23

15

90

25.5-30.5

12

23

28

10

120

30.5-35.5

14

37

33

5

70

35.5-40.5

26

63

38

0

0

40.5-45.5

12

75

43

5

60

45.5-50.5

16

91

48

10

160

50.5-55.5

9

100

53

15

135

100

735

The class interval containing theor 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here, l = 35.5, C = 37, = 26, h = 5, and N = 100

 

Thus, mean deviation about the median is given by,


 

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