The mean and variance of 8 observations are 1 | vedclass

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(A) Let the two remaining observations be $a$ and $b$.Given the mean $\bar{x} = 10$ for $8$ observations:$\frac{5+7+10+12+14+15+a+b}{8} = 10$$63 + a +

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$7$

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(A) Let the two remaining observations be $a$ and $b$.Given the mean $\bar{x} = 10$ for $8$ observations:$\frac{5+7+10+12+14+15+a+b}{8} = 10$$63 + a + b = 80 \Rightarrow a + b = 17 \quad (1)$Given the variance $\sigma^2 = 13.5$:$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$13.5 = \frac{5^2+7^2+10^2+12^2+14^2+15^2+a^2+b^2}{8} - 10^2$$113.5 = \frac{25+49+100+144+196+225+a^2+b^2}{8}$$908 = 739 + a^2 + b^2 \Rightarrow a^2 + b^2 = 169 \quad (2)$From $(a+b)^2 = a^2 + b^2 + 2ab$,we have $17^2 = 169 + 2ab$ $\Rightarrow 289 = 169 + 2ab$ $\Rightarrow 2ab = 120$ $\Rightarrow ab = 60$.Now,$(a-b)^2 = (a+b)^2 - 4ab = 17^2 - 4(60) = 289 - 240 = 49$.Thus,$|a-b| = \sqrt{49} = 7$.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
$f$	$17$	$f_1$	$32$	$f_2$	$19$

The mean and variance of $8$ observations are $10$ and $13.5,$ respectively. If $6$ of these observations are $5, 7, 10, 12, 14, 15,$ then the absolute difference of the remaining two observations is

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