NCERT Exemplar Solutions for Class 11 Maths Chapter 15 Statistics

NCERT Solutions Exemplar for Class 11 Maths Chapter 15 Statistics are provided here, which is considered to be one of the most important study materials for students studying in Class 11. The solutions provided at BYJU’S are formulated in such a way that every step is explained clearly and in detail. The Solutions for Class 11 Maths NCERT Exemplar are prepared by the subject experts to help students in their board exam preparation. These solutions can be helpful not only for the exam preparation but also in solving homework and assignments. NCERT Exemplar Solutions provided here are detailed explanations, which helps to reassure students and makes the exam preparation effective for them.

Students can access NCERT Exemplar Solutions for Class 11 Maths Chapter 15 Statistics from the below provided links. This Chapter is mainly focused on the measures of dispersion. Some of the essential topics of this chapter are listed below.

  • Measures of dispersion
  • Definition and meaning of range
  • Mean Deviation
  • Mean deviation for ungrouped data
  • Mean deviation for discrete frequency distribution
  • Mean deviation for continuous frequency distribution (Grouped data)
  • Variance
  • Standard Deviation
  • Standard deviation for a discrete frequency distribution
  • Standard deviation of a continuous frequency distribution (grouped data)
  • Coefficient of variation

Download the PDF of NCERT Exemplar Solutions For Class 11 Maths Chapter 15 Statistics

 

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Access answers to Maths NCERT Exemplar Solutions For Class 11 Chapter 15 – Statistics

Exercise Page No: 278

Short Answer type:

1. Find the mean deviation about the mean of the distribution:

Size

20

21

22

23

24

Frequency

6

4

5

1

4

Solution:

Given data distribution

Now we have to find the mean deviation about the mean of the distribution

Construct a table of the given data

Size (xi)

Frequency (fi)

fixi

20

6

20 × 6 = 120

2

4

21 × 4 = 84

22

5

22 × 5 = 110

23

1

23 × 1 = 23

24

4

24 × 4 = 96

Total

20

433

We know that mean, 
NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 1

To find mean deviation we have to construct another table

Size (xi)

Frequency (fi)

fixi

di = |xi – mean|

fidi

20

6

120

1.65

9.90

21

4

84

0.65

2.60

22

5

110

0.35

1.75

23

1

23

1.35

1.35

24

4

96

2.35

9.40

Total

20

433

6.35

25.00

Hence Mean Deviation becomes,

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 2

Therefore, the mean deviation about the mean of the distribution is 1.25

2. Find the mean deviation about the median of the following distribution:

Marks obtained

10

11

12

14

15

No. of Students

2

3

8

3

4

Solution:

Given data distribution

Now we have to find the mean deviation about the median

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

Marks Obtained (xi)

Number of Students (fi)

Cumulative frequency (c. f)

10

2

2

11

3

2 + 3 = 5

12

8

5 + 8 = 13

14

3

13 + 3 = 16

15

4

16 + 4 = 20

Total

20

Now, here N=20, which is even.

Here median,

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 3

So the above table with more columns is as shown below,

Marks Obtained

Number of Students

Cumulative frequency

di = |xi – M|

fidi

10

2

2

2

4

11

3

2 + 3 = 5

1

3

12

8

5 + 8 = 13

0

0

14

3

13 + 3 = 16

2

6

15

4

16 + 4 = 20

3

12

Total

20

6.35

25

Hence Mean Deviation becomes,

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 4

Therefore, the mean deviation about the median of the distribution is 1.25

3. Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 5

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 6

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 7

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 8

4. Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 9

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 10

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 11

5. Find the standard deviation of the first n natural numbers.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 12

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 13

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 14

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 15

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 16

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 17

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 18

7. The mean and standard deviation of a set of n1 observations are 
NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 19 and s1, respectively while the mean and standard deviation of another set of n2 observations are NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 20 and s2, respectively. Show that the standard deviation of the combined set of (n1 + n2) observations is given by
NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 21

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 22

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 23

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 24

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 25

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 26

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 27

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 28

8. Two sets each of 20 observations have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 29

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 30

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 31

9. The frequency distribution:

x

A

2A

3A

4A

5A

6A

F

2

1

1

1

1

1

Where A is a positive integer, has a variance of 160. Determine the value of A.

Solution:

Given frequency distribution table, where variance =160

Now we have to find the value of A, where A is a positive number

Now we have to construct a table of the given data

Size (xi)

Frequency (fi)

fixi

fixi2

A

2

2A

2A2

2A

1

2A

4A2

3A

1

3A

9A2

4A

1

4A

16A2

5A

1

5A

25A2

6A

1

6A

36A2

Total

7

22A

92A2

And we know variance is

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 32

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 33

10. For the frequency distribution:

x

2

3

4

5

6

7

f

4

9

16

14

11

6

Find the standard distribution.

Solution:

Given frequency distribution table

Now we have to find the standard deviation

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

Size (xi)

Frequency (fi)

fixi

fixi2

2

4

8

16

3

9

27

81

4

16

64

256

5

14

70

350

6

11

6

396

7

6

42

294

Total

60

277

1393

And we know standard deviation is

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 34

11. There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

Marks

0

1

2

3

4

5

Frequency

x – 2

x

x2

(x + 1)2

2x

x + 1

Where x is a positive integer. Determine the mean and standard deviation of the marks.

Solution:

Given there are 60 students in a class. The frequency distribution of the marks obtained by the students in a test is also given.

Now we have to find the mean and standard deviation of the marks.

It is given there are 60 students in the class, so

∑fi=60

⇒ (x – 2) + x + x2 + (x + 1)2 + 2x + x + 1 = 60

⇒ 5x – 1 + x2 + x2 + 2x + 1 = 60

⇒ 2x2 +7x = 60

⇒ 2x2 +7x – 60 = 0

Splitting the middle term, we get

⇒ 2x2 + 15x – 8x – 60 = 0

⇒ x (2x + 15) – 4(2x + 15) = 0

⇒ (2x + 15) (x – 4) = 0

⇒ 2x + 15 = 0 or x – 4 = 0

⇒ 2x = -15 or x = 4

Given x is a positive number, so x can take 4 as the only value.

And let assumed mean, a=3.

Now put x = 4 and a = 3 in the frequency distribution table and add other columns after calculations, we get

Marks (xi)

Frequency (fi)

di = xi – a

fidi

fidi2

0

x – 2 = 4 – 2 = 2

-3

-6

18

1

x = 4

-2

-8

16

2

x2 = 42 = 16

-1

-16

16

3

(x + 1)2 = 25

0

0

0

4

2x = 8

1

4

8

5

x + 1 = 5

2

10

20

Total

60

-12

78

And we know standard deviation is

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 35

12. The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.

Solution:

Given the mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours

Now we have to find the overall standard deviation

As per given criteria, in first set of samples,

Number of sample bulbs, n1=60

Standard deviation, s1=8hrs

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 36

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 37

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 38

Or, σ=8.9

Hence the standard deviation of the set obtained by combining the given two sets is 8.9

13. Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all items and the sum of the squares of the items.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 39

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 40

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 41

14. If for a distribution ∑ (x −5)=3, ∑ (x −5)2 = 43 and the total number of item is 18, find the mean and standard deviation.

Solution:

Given for a distribution ∑ (x −5) = 3, ∑ (x −5)2 = 43 and the total number of item is 18

Now we have to find the mean and standard deviation.

As per given criteria,

Number of items, n=18

And given ∑(x – 5) = 3,

And also given, ∑(x −5)2 = 43

But we know mean can be written as,

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 42

15. Find the mean and variance of the frequency distribution given below:

x

1 ≤ x < 3

3 ≤ x < 5

5 ≤ x < 7

7 ≤ x < 10

f

6

4

5

1

Solution:

Given the frequency distribution

Now we have to find the mean and variance

Converting the ranges of x to groups, the given table can be rewritten as shown below,

x (Class)

fi

xi

fixi

fixi2

1 – 3

6

2

12

24

3 – 5

4

4

16

64

5 – 7

5

6

30

180

7 – 10

1

8.5

8.5

72.25

Total

16

66.5

340.25

And we know variance can be written as

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 43

16. Calculate the mean deviation about the mean for the following frequency distribution:

Class interval

0 – 4

4 – 8

8 -12

12 – 16

16 – 20

Frequency

4

6

8

5

2

Solution:

Given the frequency distribution

Now we have to find the mean deviation about the mean

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

Class interval

Mid – Value (xi)

Frequency (fi)

fixi

0 – 4

2

4

8

4 – 8

6

6

36

8 – 12

8

8

80

12 – 16

5

5

70

16 – 20

2

2

36

total

25

230

Here mean, 
NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 44

Now we have to find mean deviation

Class interval

Mid – Value (xi)

Frequency (fi)

fixi

di=|xi – mean|

fidi

0 – 4

2

4

8

7.2

28.8

4 – 8

6

6

36

3.2

19.2

8 – 12

8

8

80

0.8

6.4

12 – 16

5

5

70

1.8

24.4

16 – 20

2

2

36

8.8

17.6

total

25

230

96

Hence Mean Deviation becomes,

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 45

Therefore, the mean deviation about the mean of the distribution is 3.84

17. Calculate the mean deviation from the median of the following data:

Class interval

0 – 6

6 – 12

12 – 18

18 – 24

24 – 30

Frequency

4

6

8

5

2

Solution:

Given the frequency distribution

Now we have to find the mean deviation from the median

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

Class interval

Mid – Value (xi)

Frequency (fi)

fixi

0 – 6

3

4

4

6 – 12

9

5

9

12 – 18

15

3

12

18 – 24

21

6

18

24 – 30

27

2

20

total

20

Now, here N=20, which is even.

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 46

⇒M = 12 + 2 = 14

Class interval

Mid – Value (xi)

Frequency (fi)

fixi

di=|xi – mean|

fidi

0 – 6

3

4

4

11

44

6 – 12

9

5

9

5

25

12 – 18

15

3

12

1

3

18 – 24

21

6

18

7

42

24 – 30

27

2

20

13

26

total

20

140

Hence Mean Deviation becomes,

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 47

Therefore, the mean deviation about the median of the distribution is 7

18. Determine the mean and standard deviation for the following distribution:
NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 48

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 49

So the above table with more columns is as shown below,

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 50

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 51

19. The weights of coffee in 70 jars are shown in the following table:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 52
Determine variance and standard deviation of the above distribution.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 53

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 54

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 55

20. Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 56

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 57

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 58

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 59

21. Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 60
Who is more intelligent and who is more consistent?

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 61

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 62

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 63

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 64

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 65

22. Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 66

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 67

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 68

σ=10.24

Hence the corrected standard deviation is 10.24.

23. While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 69

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 70

NCERT Exemplar Solutions for Class 11 Maths Chapter 15 - Image 71

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