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 BYJU’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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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,
To find the 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,
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 – 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
 and s1, respectively, while the mean and standard deviation of another set of n2 observations are 
 and s2, respectively. Show that the standard deviation of the combined set of (n1 + 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
∑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
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, 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:
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,
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
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 | 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,
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,
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 – 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.
⇒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,
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:
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 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:
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