Question: Find the standard deviation of the following data:


Class Interval0-1010-2020-3030-4040-50
Frequency5815166

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;


MarksFrequency
(f)
Mid-value
(m)
d(x-25)fdfd²
0-1055-20 400-1002000
10-20815-10 100-80800
20-3015250 000
30-40163510 1001601600
40-5064520 4001202400
N = 50 $\sum$fd
= 100
$\sum$fd²
=6800

Now,

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

$= \sqrt{ \dfrac{6800}{50} - \left ( \dfrac{100}{50} \right )}^2$

$= \sqrt{ 136 - (2)²}$

$= \sqrt{136 - 4}$

$= \sqrt{132}$

$= 11.48$

Hence, the required standard deviation of the give data is 11.48.


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.

#SciPiPupil