# Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.3

• Last Updated : 11 Feb, 2021

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

Solution:

Calculating the median:

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Median is the middle term of the observation in ascending order, Xi,

Here, the middle term is 25.

Therefore, Median = 25

Mean Deviation, MD= 1/50 × 450

= 9

Therefore, mean deviation is 9.

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

(i)

(ii)

Solution:

(i) Mean = 17900/50

= 358

Also, the number of observations, N=50 = 1/50 × 7896

= 157.92

Therefore, mean deviation is 157.92.

(ii) Mean = 13630/106

= 128.58

Also, the number of observations, N=106 = 1/106 × 1272.6

= 12.005

### Question 3.Compute mean deviation from mean of the following distribution:

Solution: = 5390/110

= 49 = 1/110 × 1644

= 14.94

Therefore, mean deviation is 14.94.

### Calculate the mean deviation from the median age.

Solution:

Number of observations, N = 96

So, N/2 = 96/2 = 48

The cumulative frequency just greater than 48 is 59, and therefore, the corresponding value of x is 38.25

Hence, Median = 38.25

Number of observations, N = 96. = 1/96 × 980.5

= 10.21

∴ The mean deviation is 10.21

### Question 5.Find the mean deviation from the mean and from a median of the following distribution:

Solution:

Number of observations, N = 50

So, N/2 = 50/2 = 25

The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28.

Therefore, Median = 28

Now, = 1350/50

= 27

Mean deviation from the median of observation = 478/50 = 9.56

And, Mean deviation from mean of observation = 472/50 = 9.44

∴ The mean deviation from the median is 9.56 and from the mean is 9.44.

### Question 6. Calculate mean deviation about median age for the age distribution of 100 persons given below:

Solution:

Converting the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

We have, N = 100

So, N/2 = 100/2 = 50

The cumulative frequency just greater than N/2 is 63 and the corresponding class is 35.5-40.5.

l=35.5, f=26, h= 5, F =37

Therefore, Median = l + (N/2 – F)/f * h  = 35.5 + 50-37/26 * 5 =38 ### Question 7.  Calculate the mean deviation from the median of the following data:

Solution:

Mean = 230/25 =9.2

Mean Deviation=96/25 = 3.84

### Question 8. Calculate the mean deviation from the median of the following data:

Solution:

Mean = 282/20 =14.1

Mean Deviation= 139.8/20 = 6.99

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