Question: Find the standard deviation of the following data:
Class Interval | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
Frequency | 5 | 4 | 6 | 7 | 3 |
Solution:
[Comment: assumed mean is easier way to perform standard deviation, so we use Assumed Mean method]
Taking Assumed Mean (A) = 50
Arranging the given data in a table;
Marks | Frequency (f) | Mid-value (m) | d(x-50) | d² | fd | fd² |
25-35 | 5 | 30 | -20 | 400 | -100 | 2000 |
35-45 | 4 | 40 | -10 | 100 | -40 | 400 |
45-55 | 6 | 50 | 0 | 0 | 0 | 0 |
55-65 | 7 | 60 | 10 | 100 | 70 | 700 |
65-75 | 3 | 70 | 20 | 400 | 60 | 1200 |
N = 25 | $\sum$fd = -10 | $\sum$fd² =4300 |
Now,
Standard Deviation ($\sigma$) = $\sqrt{ \dfrac{\sum fd²}{N} - \left ( \dfrac{\sum fd}{N} \right )}^2$
$= \sqrt{ \dfrac{4300}{25} - \left ( \dfrac{-10}{25} \right )}^2$
$= \sqrt{ 172 - (-0.4)²}$
$= \sqrt{172- 0.16}$
$= \sqrt{171.84}$
$= 13.108$
$= 13.11$
Hence, the required standard deviation of the give data is 13.08.
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 45-55 and the mid-value is 50.
Related Notes and Solutions:
Here is the website link to the notes of Statistics of Class 10.
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