RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.2 (Updated for 2021-22) RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2: In this exercise, we’ll look at the mean deviation of a discrete frequency distribution and how to compute it using an algorithm. Students who are experiencing difficulty solving arithmetic questions can use the solutions prepared by experts, who have presented the solutions in the most straightforward manner possible for any pupil to understand. Students who are unable to resolve their doubts during class can refer the RD Sharma Class 11 Maths Solutions. Students can easily obtain the RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2 pdf by clicking on the links below.

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RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2

Access answers to RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2- Important Question with Answers

1. Calculate the mean deviation from the median of the following frequency distribution:

 Heights in inches 58 59 60 61 62 63 64 65 66 No. of students 15 20 32 35 35 22 20 10 8

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

We know, Median is the Middle term,

So, Median = 61

Let xi =Heights in inches

And, fi = Number of students

 xi fi Cumulative Frequency |di| = |xi – M| = |xi – 61| fi |di| 58 15 15 3 45 59 20 35 2 40 60 32 67 1 32 61 35 102 0 0 62 35 137 1 35 63 22 159 2 44 64 20 179 3 60 65 10 189 4 40 66 8 197 5 40 N = 197 Total = 336

N=197

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 1/197 × 336

= 1.70

∴ The mean deviation is 1.70.

2. The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution:

 Number of calls 0 1 2 3 4 5 6 7 Frequency 14 21 25 43 51 40 39 12

Compute the mean deviation about the median.

Solution:

To find the mean deviation from the median, firstly let us calculate the median.

We know, Median is the even term, 3+5/2 = 4

So, Median = 8

Let xi =Number of calls

And, fi = Frequency

 xi fi Cumulative Frequency |di| = |xi – M| = |xi – 61| fi |di| 0 14 14 4 56 1 21 35 3 63 2 25 60 2 50 3 43 103 1 43 4 51 154 0 0 5 40 194 1 40 6 39 233 2 78 7 12 245 3 36 Total = 366 Total = 245

N = 245

$$MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|$$

= 1/245 × 336

= 1.49

∴ The mean deviation is 1.49.

Statistics Ex 32.2 Q3 Statistics Ex 32.2 Q4(i) Statistics Ex 32.2 Q4(ii)  Statistics Ex 32.2 Q4(iii)  Statistics Ex 32.2 Q4(iv) Statistics Ex 32.2 Q5   We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2. If you have any queries related to CBSE Class 11, feel free to ask us in the comment section below.

FAQs on RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2

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There are a total of 5 questions in RD Sharma Class 11 Solutions Chapter 32 Exercise 32.2.

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