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ધારો કે $a_1, a_2, \ldots a_{10}$ એવા $10$ અવલોકનો છે કે જેથી $\sum_{k=1}^{10} a_k=50$ અને $\sum_{k < j} a_k \cdot a_j=1100$, તો $a_1, a_2, \ldots, a_{10}$ નું પ્રમાણિત વિચલન ....................છે.
$5$
$\sqrt{5}$
$10$
$\sqrt{115}$
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
નીચે આપેલ માહિતી માટે પ્રમાણિત વિચલન શોધો :
| ${x_i}$ | $3$ | $8$ | $13$ | $18$ | $25$ |
| ${f_i}$ | $7$ | $10$ | $15$ | $10$ | $6$ |
