The mean of five observations is 4 and their | vedclass

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(C) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given $x_1=1, x_2=3, x_3=4$. Let the other two be $a$ and $b$.Mean $\bar{x} = \frac{1+3+4+

Vedclass · Accepted Answer

$35$

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(C) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given $x_1=1, x_2=3, x_3=4$. Let the other two be $a$ and $b$.Mean $\bar{x} = \frac{1+3+4+a+b}{5} = 4 \implies 8+a+b = 20 \implies a+b = 12$.Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 4$.$\frac{1^2+3^2+4^2+a^2+b^2}{5} - 4^2 = 4$.$\frac{1+9+16+a^2+b^2}{5} - 16 = 4 \implies \frac{26+a^2+b^2}{5} = 20$.$26+a^2+b^2 = 100 \implies a^2+b^2 = 74$.We know $(a+b)^2 = a^2+b^2+2ab$.$12^2 = 74+2ab \implies 144 = 74+2ab$.$2ab = 70 \implies ab = 35$.

The mean of five observations is $4$ and their variance is also $4$. If three of the five observations are $1, 3, 4$,then the product of the other two is:

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