 # Important Questions for Class 11 Maths Chapter 15 - Statistics

Important questions for class 11 Maths Chapter 15 – Statistics are provided here. These questions are taken from the previous year question paper. The questions are framed as per the syllabus of the CBSE board. These problems help you to score good marks in the final examination. The questions provided here are almost repeated in every year. It includes all types of questions such as very short answers, short answers, long answers, HOTS (Higher Order Thinking Skill-based questions), and VBA (Value-Based Questions). Get the important questions for class 11 Maths for all the chapters at BYJU’S.

Class 11 Maths Chapter 15 – Statistics includes important concepts, such as

• Mean deviation
• Mean deviation for the ungrouped data
• Mean deviation for the discrete frequency distribution
• Mean deviation for the continuous frequency distribution
• Variance
• Standard Deviation
• Shortcut methods for finding variance and standard deviation
• Measures of dispersion and range
• Frequency Distribution Analysis

Also, check:

## Class 11 Chapter 15 – Statistics Important Questions with Solutions

Class 11 Maths Chapter 16 probability important questions are given here with solutions. Go through and practice the below-given problems, as these problems are preferred from the previous year question papers.

Question 1:

The variance of the given data 2, 4, 5, 6, 8, 17 is 23.33. Then find the variance for the data 4, 8, 10, 12, 16, 34.

(a) 23.23 (b) 25.33 (c)46.66 (d)48.66

Solution:

A correct answer is an option (c)

Explanation:

For the given data: 2, 4, 5, 6, 8, 17, the variance is 23.33.

To find the variance for the data: 4, 8, 10, 12, 16, 34

If you notice the data which you have to find the variance, it is the multiple of the given data.

So, multiply the variance of the given data by 2,

It means that, 23.33 x 2 = 46.66

Thus, the variance of the data: 4, 8, 10, 12, 16, 34 is 46.66

Question 2:

Find the variance and the standard deviation for the following data: 57, 64, 43, 67, 49, 59, 44, 47, 61, 59

Solution:

Given data: 57, 64, 43, 67, 49, 59, 44, 47, 61, 59

To find mean (μ):

Mean (μ)= ( 57+ 64+ 43+ 67+ 49+ 59+ 44+ 47+ 61+ 59)/10

= 550/10

Mean = 55

To find Variance (σ2):

Variance(σ2) = (xi – μ)2/n

=(22+92+122+122+62+42+62+42+112+82)/10

= 662/10

=66.2

Therefore, variance(σ2) = 66.2

To find standard deviation (σ):

To find the standard deviation, take the square root of variance, we get

Standard Deviation(σ) = √(σ2)

= √66.2 = 8.13

Therefore, the standard deviation is 8.13

Question 3:

The coefficients of variations for the two distributions are 60 and 70 and its standard deviations are 21 and 16 respectively. Determine its arithmetic mean.

Solution:

Given that,

Coefficient of Variations (C.V of 1st distribution) = 60, σ1 = 21

Coefficient of Variations (C.V of 2nd distribution) = 70, σ2 = 16

Let μ1 and μ2 are the means of the 1st and the 2nd distribution.

We know that the formula to find the arithmetic mean is given as:

Coefficient of Variations(C.V) = (Standard Deviation/arithmetic Mean) x 100

Thus, Arithmetic Mean = (Standard Deviation/C.V)x100

Therefore, the arithmetic mean for the 1st deviation is given by:

μ1= [σ1 / (c.v of 1st distribution)]x100

μ1= (21/60)x100

μ1= 0.35×100

μ1= 35

Similarly for μ2:

μ2= [σ2 / (c.v of 2nd distribution)]x100

μ2= (16/70)x100

μ2= 0.2285×100

μ2= 22.85

Therefore, the arithmetic mean for the 1st and the 2nd distributions are 35 and 22.85 respectively.

Question 4:

Find the mean deviation about mean for the following data:

 Size (x) 1 3 5 7 9 11 13 15 Frequency (f) 3 3 4 14 7 4 3 4

Solution:

Let mean = μ

The formula to find mean is

μ = fi xi / N

N = 3+3+4+14+7+4+3+4 = 42

μ = (3+ 9+ 20+ 98+ 63+ 44+ 39+ 60)/42

μ = 336/42

μ = 8

Now, to find the mean deviation about mean:

The formula is:

M.D(μ) = fi |xi– μ| / N

M.D(μ) =[3(7)+3(5)+ 4(3)+ 14(1)+7(1)+ 4(3)+ 3(5)+ 4(7)]/42

= (21+ 15+ 12+ 14+ 7+12+ 15+ 28)/42

= 62/21

= 2.95

Therefore, the mean deviation about mean for the given data is 2.95

Question 5:

Determine the mean deviation about the median for the following data:

 Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 6 7 15 16 4 2

Solution:

From the given data:

 Class Frequency (fi) Cumulative frequency (c.f) Midpoints (xi) |xi – Median| fi|xi – Median| 0-10 10-20 20-30 30-40 40-50 50-60 6 7 15 16 4 2 6 13 28 44 48 50 5 15 25 35 45 55 23 13 3 7 17 27 138 91 45 112 68 54 50 508

From this, it is noticed that the class interval containing 25th item is 20-30. Therefore, the median is 20-30.

We know that the formula for the median is given as:

Median= l+ {[((N/2)-C)/f] x h}

Here l= 20, f=15, C=13, N=50, and h= 10

Now substitute these values in the formula, we get:

Median= 20+ {[(25-13)/15] x 10}

= 20+8

=28

Therefore, median is 28.

Hence, the mean deviation about the median is given by:

M.D(M)= $$\frac{1}{N}\sum_{i=1}^{6}f_{i}|x_{i}-M|$$

M.D(M)=(1/50)508

M.D(M) = 10.16

Hence, the meam deviation about the median is 10.16.

### Practice Problems for Class 11 Maths Chapter 15 Statistics

Solve the below-given problems:

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

2. Determine the mean deviation about the mean for the given frequency distribution:

 Class interval 0-4 4-8 8-12 12-16 16-20 Frequency 4 6 8 5 2

3. The mean and the standard deviation of the 100 observations are found to be 40 and 10 respectively. If at the time of calculation, two observations were taken wrongly as 30 and 70 in the place of 3 and 27 respectively. Determine the accurate standard deviation.

4. If for the distribution (x-5)= 3, (x-5)2 = 43, and the total number of items is 18, determine the mean and standard deviation.

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

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

6. The mean and the standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who mistook one observation as 50 instead of 40. What are the correct mean and standard deviation?

7. The mean of 2, 7, 4, 6, 8 and p is 7. Find the mean deviation about the median of these observations.

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