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Question

$\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}$

Let first four observation be $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$

Here, $\

Vedclass · Accepted Answer

$\frac{5}{4}$

Vedclass · Answer

$\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}$

Let first four observation be $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$

Here, $\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$.

Also, $\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4}{4}=\frac{7}{2}$

$\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4=14$

Now from eqn $-1$

$\mathrm{x}_5$=$10$

Now, $\sigma^2=\frac{194}{25}$

$ \frac{\mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2+\mathrm{x}_5^2}{5}-\frac{576}{25}=\frac{194}{25} $

$ \Rightarrow \mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2=54$

Now, variance of first $4$ observations

Var $=\frac{\sum_{i=1}^4 x_i^2}{4}-\left(\frac{\sum_{i=1}^4 x_i}{4}\right)^2$

$ =\frac{54}{4}-\frac{49}{4}=\frac{5}{4}$

$x_i$	$2$	$4$	$6$	$8$	$10$	$12$	$14$	$16$
$f_i$	$4$	$4$	$\alpha$	$15$	$8$	$\beta$	$4$	$5$

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