The mean of five observations is 5 and their | vedclass

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(B) Let the five observations be $1, 3, 8, x,$ and $y$.Given mean $\mu = 5$,so $\frac{1 + 3 + 8 + x + y}{5} = 5$.$12 + x + y = 25 \Rightarrow x + y =

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$4 : 9$

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(B) Let the five observations be $1, 3, 8, x,$ and $y$.Given mean $\mu = 5$,so $\frac{1 + 3 + 8 + x + y}{5} = 5$.$12 + x + y = 25 \Rightarrow x + y = 13$ (Equation $1$).Given variance $\sigma^2 = 9.20$,using the formula $\sigma^2 = \frac{\sum x_i^2}{N} - \mu^2$:$9.2 = \frac{1^2 + 3^2 + 8^2 + x^2 + y^2}{5} - 5^2$.$9.2 = \frac{1 + 9 + 64 + x^2 + y^2}{5} - 25$.$34.2 = \frac{74 + x^2 + y^2}{5} \Rightarrow 171 = 74 + x^2 + y^2$.$x^2 + y^2 = 97$ (Equation $2$).We know $(x + y)^2 = x^2 + y^2 + 2xy$,so $13^2 = 97 + 2xy$.$169 - 97 = 2xy$ $\Rightarrow 72 = 2xy$ $\Rightarrow xy = 36$.Solving $x + y = 13$ and $xy = 36$,the quadratic equation $t^2 - 13t + 36 = 0$ gives $(t - 4)(t - 9) = 0$.Thus,the two observations are $4$ and $9$.The ratio is $\frac{4}{9}$ or $\frac{9}{4}$.

$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

The mean of five observations is $5$ and their variance is $9.20$. If three of the given five observations are $1, 3$ and $8$,then the ratio of the other two observations is

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