# NCERT Exemplar Solutions for Class 11 Maths Chapter 15 Statistics

NCERT Solutions Exemplar for Class 11 Maths Chapter 15 Statistics, provided here, is one of the most important study materials for students studying in Class 11. The solutions provided at BYJUâ€™S are formulated in such a way that every step is explained clearly and in detail. The Class 11 Maths NCERT Exemplar Solutions are prepared by the subject experts to help students in their annual exam preparation. These solutions can be helpful not only for exam preparation but also in completing homework and assignments. NCERT Exemplar Solutions provided here consist of detailed explanations, which will make the exam preparation effective for students.

Students can access NCERT Exemplar Solutions for Class 11 Maths Chapter 15 Statistics from the below-provided link. 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

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### Access Answers to Maths NCERT Exemplar Solutions for Class 11 Chapter 15 – Statistics

Exercise Page No: 278

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 Ã— 6 = 120 2 4 21 Ã— 4 = 84 22 5 22 Ã— 5 = 110 23 1 23 Ã— 1 = 23 24 4 24 Ã— 4 = 96 Total 20 433

We know that mean,

To find the mean deviation, we have to construct another table

 Size (xi) Frequency (fi) fixi di = |xi â€“ 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,

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 the 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,

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

 Marks Obtained Number of Students Cumulative Frequency di = |xi â€“ 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,

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:

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

Solution:

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

Solution:

Solution:

7. The mean and standard deviation of a set of n1Â observations areÂ
Â and s1, respectively, while the mean and standard deviation of another set of n2Â observations areÂ Â and s2, respectively. Show that the standard deviation of the combined set of (n1Â + n2) observations is given by

Solution:

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:

9. The frequency distribution:

 x A 2A 3A 4A 5A 6A F 2 1 1 1 1 1

Where A is a positive integer and 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

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 the 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

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

âˆ‘fi=60

â‡’Â (x – 2) + x + x2 + (x + 1)2 + 2x + x + 1 = 60

â‡’Â 5x – 1 + x2 + x2 + 2x + 1 = 60

â‡’Â 2x2Â +7x = 60

â‡’Â 2x2Â +7x â€“ 60 = 0

Splitting the middle term, we get

â‡’Â 2x2Â + 15x â€“ 8x â€“ 60 = 0

â‡’Â x (2x + 15) â€“ 4(2x + 15) = 0

â‡’Â (2x + 15) (x â€“ 4) = 0

â‡’Â 2x + 15 = 0 or x â€“ 4 = 0

â‡’Â 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 columnsÂ after calculations, we get

 Marks (xi) Frequency (fi) di = xi – a fidi fidi2 0 x â€“ 2 = 4 â€“ 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

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

Or, Ïƒ=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:

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

Solution:

Given for a distribution âˆ‘ (x âˆ’5) = 3, âˆ‘ (x âˆ’5)2Â = 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 âˆ‘(x – 5) = 3,

And also given, âˆ‘(x âˆ’5)2Â = 43

But we know mean can be written as,

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

 x 1 â‰¤ x < 3 3 â‰¤ x < 5 5 â‰¤ x < 7 7 â‰¤ 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 â€“ 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

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 the calculations

 Class Interval Midâ€“value (xi) Frequency (fi) fixi 0 – 4 2 4 8 4 â€“ 8 6 6 36 8 â€“ 12 8 8 80 12 â€“ 16 5 5 70 16 – 20 2 2 36 total 25 230

Here mean,

Now we have to find mean deviation

 Class Interval Mid â€“ Value (xi) Frequency (fi) fixi di=|xi â€“ mean| fidi 0 – 4 2 4 8 7.2 28.8 4 â€“ 8 6 6 36 3.2 19.2 8 â€“ 12 8 8 80 0.8 6.4 12 â€“ 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,

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 the calculations

 Class interval Mid â€“ Value (xi) Frequency (fi) fixi 0 â€“ 6 3 4 4 6 â€“ 12 9 5 9 12 â€“ 18 15 3 12 18 â€“ 24 21 6 18 24 – 30 27 2 20 total 20

Now, here N=20, which is even.

â‡’M = 12 + 2 = 14

 Class interval Midâ€“value (xi) Frequency (fi) fixi di=|xi â€“ mean| fidi 0 â€“ 6 3 4 4 11 44 6 â€“ 12 9 5 9 5 25 12 â€“ 18 15 3 12 1 3 18 â€“ 24 21 6 18 7 42 24 – 30 27 2 20 13 26 total 20 140

Hence Mean Deviation becomes,

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

18. Determine the mean and standard deviation for the following distribution:

Solution:

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

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

Determine variance and standard deviation of the above distribution.

Solution:

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:

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

Who is more intelligent and who is more consistent?

Solution:

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:

Ïƒ=10.24

Hence the corrected standard deviation is 10.24.

23. While calculating the mean and variance of 10 readings, a student wrongly used 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: