Question
Find out the standard deviation from the following data after finding mean:
| Class lnterval: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency: | 8 | 13 | 16 | 8 | 5 |
✓
Answer
| Class Interval | Frequency | M.V. (X) | $dx$ | $dx'$ | $dx'^2$ | $fdx'$ | $fdx'^2$ |
| 0-10 | 8 | 5 | -20 | -2 | 4 | -16 | 32 |
| 10-20 | 13 | 15 | -10 | -1 | 1 | -13 | 13 |
| 20-30 | 16 | 25 | 0 | 0 | 0 | 0 | 0 |
| 30-40 | 8 | 35 | 10 | 1 | 1 | 8 | 8 |
| 40-50 | 5 | 45 | 20 | 2 | 4 | 10 | 20 |
| $\Sigma\text{f}=50$ | $\Sigma\text{fdx}'=-11$ | $\Sigma\text{fdx}'^2=73$ |
$=25-\frac{(-11)}{50}\times10$
$=25-\frac{(-11)}{5}$
$=25-2.2=22.8$
S.D. $=\sqrt{\frac{\Sigma\text{fdx}'^2}{\Sigma\text{f}}-\Big(\frac{\Sigma\text{fdx}'}{\Sigma\text{f}}\Big)^2}\times\text{C}$
$=\sqrt{\frac{73}{50}-\Big(\frac{11}{50}\Big)^2}\times10$
$=\sqrt{1.46-(0.22)^2}\times10$
$=\sqrt{1.46-.05}\times10=\sqrt{1.41}\times10$
$=1.19\times10=11.9$
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