NCERT Solutions Class 11 Mathematics Chapter 15 Statistics

NCERT Solutions Class 11 Mathematics Chapter 15 Statistics have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 11 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 11 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 11 Mathematics are an important part of exams for Class 11 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 11 Mathematics and also download more latest study material for all subjects. Chapter 15 Statistics is an important topic in Class 11, please refer to answers provided below to help you score better in exams

Chapter 15 Statistics Class 11 Mathematics NCERT Solutions

Class 11 Mathematics students should refer to the following NCERT questions with answers for Chapter 15 Statistics in Class 11. These NCERT Solutions with answers for Class 11 Mathematics will come in exams and help you to score good marks

Chapter 15 Statistics NCERT Solutions Class 11 Mathematics

Exercise 15.1

Question. 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, 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics

Question. 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,

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-1

Question. 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

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-2

The deviations of the respective observations 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
The absolute values of the deviations, |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
The required mean deviation about the median is

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-3

Question. 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

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-4

Question. Find the mean deviation about the mean for the data.

xi510152025
fi74635

Answer :

xi

fi

fixi

  

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

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-5

Question. 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

fxi

NCERT Solutions for Class 11 Maths Chapter 15 Statistics

NCERT Solutions for Class 11 Maths Chapter 15 Statistics

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-Solutions-Class-11-Mathematics-Chapter-15-Statistics-6

Question. Find the mean deviation about the median for the data.

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-13

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

f

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.

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-14

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

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-12

Question. 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. 

xi 

fi

c.f.

15

3

3

21

5

8

27

6

14

30

7

21

35

8

29

Here, N = 29, which is odd. 
∴ Median = [(29 + 1)/2 ]th 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. |xi – M|, are  

|xi – M|

15

9

3

0

5

fi

3

5

6

7

8

fi |xi – M|

45

45

18

0

40

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-11

Question. Find the mean deviation about the mean for the data.

Income per dayNumber of persons
0-1004
100-2008
200-3009
300-40010
400-5007
500-6005
600-7004
700-8003

Answer :
The following table is formed. 

Income per day

Number of person 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

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-10

Question. Find the mean deviation about the mean for the data

Height in cmsNumber of boys
95-1059
105-11513
115-12526
125-13530
135-14512
145-15510

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

 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-9

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

MarksNumber of girls
0-106
10-208
20-3014
30-4016
40-504
50-602

Answer :
The following table is formed. 

Marks

Number of boys  fi

Cumulative frequency (c. f.)

Mid – point 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

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-8

Question. 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 

37

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 the (N/2)th or  50th item is  35.5 - 40.5 . 
Therefore, 35.5 - 40.5 is the median class.
It is known that, 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-7

Exercise 15.2

Question. Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12.
Answer :

6, 7, 10, 12, 13, 4, 8, 12

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-15

The following table is obtained. 

xi

(xi – x̅)

(xi – x̅)2

6

–3

9

7

–2

4

10

–1

1

12

3

9

13

4

16

4

–5

25

8

–1

1

12

3

9

 

 

74

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-16

Question. Find the mean and variance for the first natural numbers 
Answer :

The mean of first n natural numbers is calculated as follows. 
Mean = (Sum of all observations)/(Number of observations) 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-17

Question. Find the mean and variance for the first 10 multiples of 3.
Answer :

The first 10 multiples of 3 are  
3, 6, 9, 12, 15, 18, 21, 24, 27, 30 
Here, number of observations, n = 10 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-18

The following table is obtained. 

xi

(xi – x̅)

(xi – x̅)2

3

–13.5

182.25

6

–10.5

110.25

9

–7.5

56.25

12

–4.5

20.25

15

–1.5

2.25

18

1.5

2.25

21

4.5

20.25

24

7.5

56.25

27

10.5

110.25

30

13.5

182.25

 

 

742.5

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-19

Question. Find the mean and variance for the data

xi

6

10

14

18

24

28

30

f i

2

4

7

12

8

4

3

Answer :
The data is obtained in tabular form as follows. 

xi

f i

fixi

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)2

6

2

12

–13

169

338

10

4

40

–9

81

324

14

7

98

–5

25

175

18

12

216

–1

1

12

24

8

192

5

25

200

28

4

112

9

81

324

30

3

90

11

121

363

 

40

760

  

1736

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-20

Question. Find the mean and variance for the data

xi

92

93

97

98

102

104

109

f i

3

2

3

2

6

3

3

Answer :
The data is obtained in tabular form as follows.

xi

i

fixi

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)2

92

3

276

–8

64

192

93

2

186

–7

49

98

97

3

291

–3

9

27

98

2

196

–2

4

8

102

6

612

2

4

24

104

3

312

4

16

48

109

3

327

9

81

243

 

22

2200

  

640

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-21

Question. Find the mean and standard deviation using short-cut method.

xi

60

61

62

63

64

65

66

67

68

fi

2

1

12

29

25

12

10

4

5

Answer :
The data is obtained in tabular form as follows.

xi

fi

Yi = (xi – A)/h

yi2

fiyi

fiyi2

60

2

–4

16

–8

32

61

1

–3

9

–3

9

62

12

–2

4

–24

48

63

29

–1

1

–29

29

64

25

0

0

0

0

65

12

1

1

12

12

66

10

2

4

20

40

67

4

3

9

12

36

68

5

4

16

20

80

 

100

220

 

0

286

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-22

Question. Find the mean and variance for the following frequency distribution.

Classes

0-30

30-60

60-90

90-120

120-150

150-180

180-210

Frequencies

2

3

5

10

3

5

2

Answer :

Class

Frequency fi

Mid-point xi

fixi

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)2

0-30

2

15

30

-92

8464

16928

30-60

3

45

135

-62

3844

11532

60-90

5

75

375

-32

1024

5120

90-120

10

105

1050

-2

4

40

120-150

3

135

405

28

784

2352

150-180

5

165

825

58

3364

16820

180-210

2

195

390

88

7744

15488

 

30

 3210 

2

68280

Question. Find the mean and variance for the following frequency distribution.

Classes

0-10

10-20

20-30

30-40

40-50

Frequencies

5

8

15

16

6

Answer :

Class

Frequency fi

Mid-point xi

fixi

(xi – x̅)

(xi – x̅)2

fi(xi – x̅)2

0-10

5

5

25

-22

484

2420

10-20

8

15

120

-12

144

1152

20-30

15

25

375

-2

4

60

30-40

16

35

560

8

64

1024

40-50

6

45

270

18

324

1944

 

50

 1350  

6600

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-25

Question. Find the mean, variance and standard deviation using short-cut method

Height in cms

No. of children

70-75

3

75-80

4

80-85

7

85-90

7

90-95

15

95-100

9

100-105

6

105-110

6

110-115

3

Answer :

Class Interval

Frequency fi

Mid-point xi

Yi = (xi – A)/h

yi2

fiyi

fiyi2

70-75

3

72.5

–4

16

–12

48

75-80

4

77.5

–3

9

–12

36

80-85

7

82.5

–2

4

–14

28

85-90

7

87.5

–1

1

–7

7

90-95

15

92.5

0

0

0

0

95-100

9

97.5

1

1

9

9

100-105

6

102.5

2

4

12

24

105-110

6

107.5

3

9

18

54

110-115

3

112.5

4

16

12

48

 

60

   

6

254

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-24

Question. The diameters of circles (in mm) drawn in a design are given below:

Diameters

No. of children

33-36

15

37-40

17

41-44

21

45-48

22

49-52

25

Answer :

Class Interval

Frequency fi

Mid-point xi

Yi = (xi – A)/h

fi2

fiyi

fiyi2

32.5-36.5

15

34.5

–2

4

–30

60

36.5-40.5

17

38.5

–1

1

–17

17

40.5-44.5

21

42.5

0

0

0

0

44.5-48.5

22

46.5

1

1

22

22

48.5-52.5

25

50.5

2

4

50

100

 

100

 

 

 

25

199

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-23

Exercise 15.3

Question. From the data given below state which group is more variable, A or B?

Marks

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Group A

9

17

32

33

40

10

9

Group B

10

20

30

25

43

15

7

Answer :
Firstly, the standard deviation of group A is calculated as follows. 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-51

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-52

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-53

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-54

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
Thus, group B has more variability in the marks.

Question. From the prices of shares X and Y below, find out which is more stable in value:

X

35

54

52

53

56

58

52

50

51

49

Y

108

107

105

105

106

107

104

103

104

101

Answer :
The prices of the shares X are 
35, 54, 52, 53, 56, 58, 52, 50, 51, 49 
Here, the number of observations, N = 10 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-55

Question. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results::

 

Firm A

Firm B

No. of wage earners

586

648

Mean of monthly wages

Rs 5253

Rs 5253

Variance of the distribution of wages

100

121

(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?
Answer :

(i) Monthly wages of firm A = Rs 5253
Number of wage earners in firm A = 586
∴Total amount paid = Rs 5253 × 586
Monthly wages of firm B = Rs 5253
Number of wage earners in firm B = 648
∴Total amount paid = Rs 5253 × 648
Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A (σ12) = 100
∴ Standard deviation of the distribution of wages in firm
A ((σ1) = √100 = 10
Variance of the distribution of wages in firm B (σ22) = 121
∴ Standard deviation of the distribution of wages in firm B (σ22) = √121 = 11
The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater
standard deviation will have more variability.
Thus, firm B has greater variability in the individual wages.

Question. The following is the record of goals scored by team A in a football session:

No. of goals scored

0

1

2

3

4

No. of matches

1

9

7

5

3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Answer :

The Mean and the standard deviation of goals scored by team A are calculated as follows. 

No. of goals scored

No. of matches

fi xi

xi2

fixi2

0

1

0

0

0

1

9

9

1

9

2

7

14

4

28

3

5

15

9

45

4

3

12

16

48

 

25

50

 

130

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-59

The standard deviation of team B is 1.25 goals.
The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.
Thus, team A is more consistent than team B.

Question. The sum and sum of squares corresponding to length (in cm) and weight (in gm) of 50 plant products are given below:

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-56

Which is more varying, the length or weight?
Answer :

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-57

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-58

Thus, C.V. of weights is greater than the c.v. of lengths. Therefore, weights vary more than the lengths.

Miscellaneous Solutions

Question. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Answer :
Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, xy.

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-26

From (1), we obtain
x2 + y2 + 2xy = 144 …(3)
From (2) and (3), we obtain
2xy = 64 … (4)
Subtracting (4) from (2), we obtain
x2 + y– 2xy = 80 – 64 = 16
⇒ x – y = ± 4 …(5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 4, when x – y = 4
x = 4 and y = 8, when x – y = –4
Thus, the remaining observations are 4 and 8.

Question. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.
Answer :

Let the observations be x1, x2, x3, x4, x5, and x6.
It is given that mean is 8 and standard deviation is 4.

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-27

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-28

Question. Given that x is the mean and σ2 is the variance of observations x1x2 … xn. Prove that the mean and variance of the observations ax1ax2ax3 …axare ax and a2 σ2, respectively (a ≠ 0).
Answer :

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-29

Question. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted.
(ii) If it is replaced by 12.
Answer :

(i) Number of observations (n) = 20
Incorrect mean = 10
Incorrect standard deviation = 2

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-30

That is, incorrect sum of observations = 200 
Correct sum of observations  = 200 - 8 = 192 
Correct mean = (Correct sum )/19 = 192/19 = 10.1 

(ii) When 8 is replaced by 12, 
Incorrect sum of observations = 200
Correct sum of observations = 200 - 8 + 12 = 204 
∴ Correct mean = (Correct sum)/20 = 204/20 = 10.2 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-32

Question. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

SubjectMathematicsPhysicsChemistry
Mean423240.9
Standard deviation121520

Which of the three subjects shows the highest variability in marks and which shows the lowest?
Answer :

Standard deviation of Mathematics = 12
Standard deviation of Physics = 15
Standard deviation of Chemistry = 20
The coefficient of variation (C.V.) is given by (Standard deviation )/Mean  × 100 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-33

The subject with greater C.V. is more variable than others.
Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

Question. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
Answer :

Number of observations (n) = 100 
Incorrect mean x = 20 
Incorrect standard deviation (σ) = 3 

""NCERT-Solutions-Class-11-Mathematics-Chapter-15-Statistics-34

NCERT Solutions Class 11 Mathematics Chapter 15 Statistics

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Class 11 NCERT Solution Mathematics Chapter 15 Statistics

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