# RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.3 (Updated for 2021-22)

RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3: The mean deviation of a grouped or continuous frequency distribution is the focus of this exercise. Experts at Kopykitab have solved the exercise-by-exercise problems using shortcut approaches to help students do well on their board exams. These RD Sharma Class 11 Maths Solutions assist students in resolving any questions they may have concerning the concepts covered in this topic. Students can download the RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3 in pdf format based on their needs and begin practising on their own schedule.

## Download RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.3 Free PDF

RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3

### Access answers to RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3- Important Question with Answers

1. Compute the mean deviation from the median of the following distribution:

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 10 20 5 10

Solution:

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

Median is the middle term of the Xi,

Here, the middle term is 25

So, Median = 25

 Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di| 0-10 5 5 5 20 100 10-20 15 10 15 10 100 20-30 25 20 35 0 0 30-40 35 5 91 10 50 40-50 45 10 101 20 200 Total = 50 Total = 450

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

= 1/50 × 450

= 9

∴ The mean deviation is 9

2. Find the mean deviation from the mean for the following data:

i

 Classes 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 Frequencies 4 8 9 10 7 5 4 3

Solution:

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

By using the formula,

= 17900/50

= 358

 Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di| 0-100 50 4 200 308 1232 100-200 150 8 1200 208 1664 200-300 250 9 2250 108 972 300-400 350 10 3500 8 80 400-500 450 7 3150 92 644 500-600 550 5 2750 192 960 600-700 650 4 2600 292 1168 700-800 750 3 2250 392 1176 Total = 50 Total = 17900 Total = 7896

N = 50

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

= 1/50 × 7896

= 157.92

∴ The mean deviation is 157.92

ii

 Classes 95-105 105-115 115-125 125-135 135-145 145-155 Frequencies 9 13 16 26 30 12

Solution:

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

By using the formula,

= 13630/106

= 128.58

 Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di| 95-105 100 9 900 28.58 257.22 105-115 110 13 1430 18.58 241.54 115-125 120 16 1920 8.58 137.28 125-135 130 26 3380 1.42 36.92 135-145 140 30 4200 11.42 342.6 145-155 150 12 1800 21.42 257.04 N = 106 Total = 13630 Total = 1272.6

N = 106

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

= 1/106 × 1272.6

= 12.005

∴ The mean deviation is 12.005

Statistics Ex 32.3 Q2(iii)

Statistics Ex 32.3 Q3

Statistics Ex 32.3 Q4

Statistics Ex 32.3 Q5

Statistics Ex 32.3 Q6

Statistics Ex 32.3 Q7

Statistics Ex 32.3 Q8

We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3. 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 32.3

### How many questions are there in RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.3?

There are a total of 6 questions in RD Sharma Class 11 Solutions Chapter 32 Exercise 32.3.

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