Question: Find the standard deviation of the following data:
Wage (in RS.) | 60-62 | 63-65 | 66-68 | 69-71 | 72-74 |
No. Of Workers | 5 | 18 | 42 | 27 | 8 |
Solution:
Using assumed mean method [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 66-68 and the mid-value is 67]
Taking Assumed Mean (A) = 67
Arranging the given data in a table;
Marks | Frequency (f) | Mid-value (m) | d(x-50) | d² | fd | fd² |
60-62 | 5 | 61 | -6 | 36 | -30 | 180 |
63-65 | 18 | 64 | -3 | 9 | -54 | 162 |
66-68 | 42 | 67 | 0 | 0 | 0 | 0 |
69-71 | 27 | 70 | 3 | 9 | 81 | 243 |
72-74 | 8 | 73 | 6 | 36 | 45 | 288 |
N =100 | $\sum$fd = -45 | $\sum$fd² =883 |
Now,
Standard Deviation ($\sigma$) = $\sqrt{ \dfrac{\sum fd²}{N} - \left ( \dfrac{\sum fd}{N} \right )}^2$
$= \sqrt{ \dfrac{873}{100} - \left ( \dfrac{-45}{100} \right )}^2$
$= \sqrt{ 8.73 - (-0.45)²}$
$= \sqrt{8.73- 0.2025}$
$= \sqrt{8.5275}$
$= 2.92$
$= 13.11$
Hence, the required standard deviation of the give data using assumed mean is 2.92.
Related Notes and Solutions:
Here is the website link to the notes of Statistics of Class 10.
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