NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.1

NCERT Solutions for Class 11 Maths Chapter 15 Statistics, contains solutions for all exercise 15.1 questions. These solutions help students to clear their concepts and solving difficult problems at their own pace. Students can download the NCERT Solutions of Class 11 maths and practice offline to improve their skills.

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Access other exercise solutions of Class 11 Maths Chapter 15 Statistics

Exercise 15.2 Solutions : 10 Questions

Exercise 15.3 Solutions : 5 Questions

Miscellaneous Exercise Solutions: 7 Questions

Access Solutions for Class 11 Maths Chapter 15.1 Exercise

Find the mean deviation about the mean for the data in Exercises 1 and 2.

1. 4, 7, 8, 9, 10, 12, 13, 17

Solution:-

First we have to find (x̅) of the given data

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 1

So, the respective values of the deviations from mean,

i.e., xi – x̅ are, 10 – 4 = 6, 10 – 7 = 3, 10 – 8 = 2, 10 – 10 = 0,

10 – 12 = – 2, 10 – 13 = – 3, 10 – 17 = – 7

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

Now absolute values of the deviations,

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 2

MD = sum of deviations/ number of observations

= 24/8

= 3

So, the mean deviation for the given data is 3.

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution:-

First we have to find (x̅) of the given data

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 3

So, the respective values of the deviations from mean,

i.e., xi – x̅ are, 50 – 38 = -12, 50 -70 = -20, 50 – 48 = 2, 50 – 40 = 10, 50 – 42 = 8,

50 – 55 = – 5, 50 – 63 = – 13, 50 – 46 = 4, 50 – 54 = -4, 50 – 44 = 6

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

Now absolute values of the deviations,

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 4

MD = sum of deviations/ number of observations

= 84/10

= 8.4

So, the mean deviation for the given data is 8.4.

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution:-

First we have to arrange the given observations into ascending order,

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

The number of observations is 12

Then,

Median = ((12/2)th observation + ((12/2)+ 1)th observation)/2

(12/2)th observation = 6th = 13

(12/2)+ 1)th observation = 6 + 1

= 7th = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |xi – M| 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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 5

Mean Deviation,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 6

= (1/12) × 28

= 2.33

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

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution:-

First we have to arrange the given observations into ascending order,

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

The number of observations is 10

Then,

Median = ((10/2)th observation + ((10/2)+ 1)th observation)/2

(10/2)th observation = 5th = 46

(10/2)+ 1)th observation = 5 + 1

= 6th = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 7

Mean Deviation,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 8

= (1/10) × 70

= 7

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

Find the mean deviation about the mean for the data in Exercises 5 and 6.

5.

xi

5

10

15

20

25

fi

7

4

6

3

5

Solution:-

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

Xi

fi

fixi

|xi – x̅|

fi |xi – x̅|

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

The sum of calculated data,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 9

The absolute values of the deviations from the mean, i.e., |xi – x̅|, as shown in the table.

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 10

6.

xi

10

30

50

70

90

fi

4

24

28

16

8

Solution:-

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

Xi

fi

fixi

|xi – x̅|

fi |xi – x̅|

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 11

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 12

Find the mean deviation about the median for the data in Exercises 7 and 8.

7.

xi

5

7

9

10

12

15

fi

8

6

2

2

2

6

Solution:-

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

Xi

fi

c.f.

|xi – M|

fi |xi – M|

5

8

8

2

16

7

6

14

0

0

9

2

16

2

4

10

2

18

3

6

12

2

20

5

10

15

6

26

8

48

Now, N = 26, which is even.

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

Then,

Median = (13th observation + 14th observation)/2

= (7 + 7)/2

= 14/2

= 7

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 13

8.

xi

15

21

27

30

35

fi

3

5

6

7

8

Solution:-

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

Xi

fi

c.f.

|xi – M|

fi |xi – M|

15

3

3

13.5

40.5

21

5

8

7.5

37.5

27

6

14

1.5

9

30

7

21

1.5

10.5

35

8

29

6.5

52

Now, N = 30, which is even.

Median is the mean of the 15th and 16th observations. Both of these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Then,

Median = (15th observation + 16th observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 14

Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

Income per day in ₹

0 – 100

100 – 200

200 – 300

300 – 400

400 – 500

500 – 600

600 – 700

700 – 800

Number of persons

4

8

9

10

7

5

4

3

Solution:-

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

Income per day in ₹

Number of persons fi

Mid – points

xi

fixi

|xi – x̅|

fi|xi – x̅|

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

1160

700 – 800

3

750

2250

392

1176

50

17900

7896

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 15

10.

Height in cms

95 – 105

105 – 115

115 – 125

125 – 135

135 – 145

145 – 155

Number of boys

9

13

26

30

12

10

Solution:-

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

Height in cms

Number of boys fi

Mid – points

xi

fixi

|xi – x̅|

fi|xi – x̅|

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

100

12530

1128.8

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 16

11. Find the mean deviation about median for the following data:

Marks

0 -10

10 -20

20 – 30

30 – 40

40 – 50

50 – 60

Number of girls

6

8

14

16

4

2

Solution:-

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

Marks

Number of Girls fi

Cumulative frequency (c.f.)

Mid – points

xi

|xi – Med|

fi|xi – 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 Nth/2 or 25th item is 20-30

So, 20-30 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 20, c = 14, f = 14, h = 10 and n = 50

Median = 20 + (((25 – 14))/14) × 10

= 20 + 7.85

= 27.85

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 17

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

Age

(in years)

16 – 20

21 – 25

26 – 30

31 – 35

36 – 40

41 – 45

46 – 50

51 – 55

Number

5

6

12

14

26

12

16

9

[Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

Solution:-

The given data is converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding the 0.5 to the upper limit of each class intervals and append other columns after calculations.

Age

Number fi

Cumulative frequency (c.f.)

Mid – points

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 Nth/2 or 50th item is 35.5 – 40.5

So, 35.5 – 40.5 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100

Median = 35.5 + (((50 – 37))/26) × 5

= 35.5 + 2.5

= 38

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 18

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