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Let $a_1, a_2, \ldots . a_{10}$ be $10$ observations such that $\sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to :
$5$
$\sqrt{5}$
$10$
$\sqrt{115}$
Solution
$ \sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50 $
$ \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50$ $………(i)$
$ \sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=1100 $ $………..(ii)$
$ \text { If } \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50$ .
$ \left(\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}\right)^2=2500 $
$\Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2+2 \sum_{\mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=2500$
$ \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=2500-2(1100) $
$ \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=300, \text { Standard deviation ' } \sigma \text { ' } $
$ \frac{\sum^{\frac{a_i^2}{2}}}{10}-\left(\frac{\sum \mathrm{a}_{\mathrm{i}}}{10}\right)^2=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2}$
$ =\sqrt{30-25}=\sqrt{5}$
Similar Questions
The diameters of circles (in mm) drawn in a design are given below:
| Diameters | $33-36$ | $37-40$ | $41-44$ | $45-48$ | $49-52$ |
| No. of circles | $15$ | $17$ | $21$ | $22$ | $25$ |
Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]
