If Σlimits_{i = 1}^{18} {({xᵢ} - 8) = 9} and Σlimits_{i = 1}^{18} {({xᵢ} - 8)^2 =

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13.Statistics

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If $\sum\limits_{i = 1}^{18} {({x_i} - 8) = 9} $ and $\sum\limits_{i = 1}^{18} {({x_i} - 8)^2 = 45} $ then the standard deviation of $x_1, x_2, ...... x_{18}$ is :-

A

$4/9$

B

$9/4$

C

$3/2$

D

None of these

Solution

Varriance of observation $\left(\mathrm{x}_{1}-8\right) \forall \mathrm{i}=1,2,3, \ldots .18$

$=\frac{45}{18}-\left(\frac{9}{18}\right)^{2}=\frac{5}{2}-\frac{1}{4}=\frac{9}{4}$

then $S.D.$ of $x_{1} \forall i=1,2,3, \ldots . .18$

$=\sqrt{\frac{9}{4}}=\frac{3}{2}$

Standard 11

Mathematics

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