Question: From the following data, find the standard deviation and coefficient of variation.


Class Interval0-100-200-300-400-50
Frequency818303840

Solution:

Using actual mean,

[We have, interval between two classes is 10.
To get the frequency, subtract the give frequency of the respective class from one step higher class.]

Arranging the data in a table,


Xfmfmd|x-21.5|fd²
0-10854016.5272.252178
0-20 (10-20)10151506.542.25422.5
0-30 (20-30)12253003.512.25147
0-40 (30-40)83528013.5182.251458
0-50 (40-50)2459023.5552.251104.5
N=40

\sumfm
 = 860
\sumfd²
=5310
Now,

Mean = \dfrac{\sum fm}{N}

= \dfrac{860}{40}

= 21.5
 

Standard Deviation (\sigma) = \sqrt{ \dfrac{\sum fd²}{N} }

= \sqrt{ \dfrac{5310}{40} }

= \sqrt{132.75}

= 11.52

And,

Coefficient of Variation = \dfrac{\sigma}{mean} × 100%

= \dfrac{11.52}{21.5} × 100%

= 0.5358 × 100%

= 53.58%

Hence, the required standard deviation using actual mean of the given data is 11.52 and the coefficient of variation is 53.58%.

Related Notes and Solutions:

Here is the website link to the notes of Statistics of Class 10.

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