The mean and variance of 5 observations are 4 | vedclass

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(C) Given $n = 5$,mean $\bar{x} = 4$,and variance $\sigma^2 = 5.2$. Let the other two observations be $x_1$ and $x_2$.The mean is given by $\frac{x_1

Vedclass · Accepted Answer

$4, 7$

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(C) Given $n = 5$,mean $\bar{x} = 4$,and variance $\sigma^2 = 5.2$. Let the other two observations be $x_1$ and $x_2$.The mean is given by $\frac{x_1 + x_2 + 1 + 2 + 6}{5} = 4$.$x_1 + x_2 + 9 = 20 \implies x_1 + x_2 = 11$.The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.$5.2 = \frac{x_1^2 + x_2^2 + 1^2 + 2^2 + 6^2}{5} - 4^2$.$5.2 = \frac{x_1^2 + x_2^2 + 1 + 4 + 36}{5} - 16$.$21.2 = \frac{x_1^2 + x_2^2 + 41}{5}$.$106 = x_1^2 + x_2^2 + 41 \implies x_1^2 + x_2^2 = 65$.We have $x_1 + x_2 = 11$ and $x_1^2 + x_2^2 = 65$.Using $(x_1 + x_2)^2 = x_1^2 + x_2^2 + 2x_1x_2$,we get $121 = 65 + 2x_1x_2 \implies 2x_1x_2 = 56 \implies x_1x_2 = 28$.The quadratic equation $t^2 - (x_1+x_2)t + x_1x_2 = 0$ becomes $t^2 - 11t + 28 = 0$.$(t - 4)(t - 7) = 0$.Thus,the observations are $4$ and $7$.

The mean and variance of $5$ observations are $4$ and $5.2$ respectively. If three of these observations are $1, 2,$ and $6,$ find the other two observations.

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