Question: Find the standard deviation:


Class Interval0-100-200-300-400-50
Frequency25101315

Solution:

[Comment: assumed mean is easier way to perform standard deviation, so we use Assumed Mean method]

Taking Assumed Mean (A) = 25

Arranging the given data in a table;


XfMid-value
(m)
d(x-25)fdfd²
0-1025-20400-40800
0-20 (10-20)315-10100-30300
0-30
(20-30)
5250000
0-40 (30-40)
3351010030300
0-50 (40-50)2452040040800
N = 15$\sum$fd
= 0
$\sum$fd²
=2200

Now,

Standard Deviation ($\sigma$) = $\sqrt{ \dfrac{\sum fd²}{N} - \left ( \dfrac{\sum fd}{N} \right )}^2$

$= \sqrt{ \dfrac{2200}{15} - \left ( \dfrac{-0}{15} \right )}^2$

$= \sqrt{ 146.66}$

$= 12.11$

Hence, the required standard deviation of the give data is 12.11.


How to take Assumed Mean?

In this method, the mean is taken as the mid-value of the mid-class from the given data. Here, mid term is mid-class is 20-30 and the mid-value is 25.


Related Notes and Solutions:

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

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