The mean and variance of 7 observations are 8 | vedclass

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Question

$\frac{x_1+x_2+\ldots .+x_7}{7}=8$

$\frac{x_1+x_2+x_3 \ldots .+x_6+14}{7}=8$

$\Rightarrow x_1+x_2+\ldots .+x_6=42$

$	herefore \frac{x_1+x_2

Vedclass · Accepted Answer

$37$

Vedclass · Answer

$\frac{x_1+x_2+\ldots .+x_7}{7}=8$

$\frac{x_1+x_2+x_3 \ldots .+x_6+14}{7}=8$

$\Rightarrow x_1+x_2+\ldots .+x_6=42$

$	herefore \frac{x_1+x_2 \ldots .+x_6}{6}=\frac{42}{6}=7=a$

$\frac{\sum x_i^2}{7}-8^2=16$

$\Rightarrow x^2=560$

$\Rightarrow x_1^2+x_2^2+\ldots+x_6^2=364$

$b=\frac{x_1^2+x_2^2+\ldots . .+x_6^2}{6}-7^2$

$=\frac{364}{6}-49$

$b=\frac{70}{6}$

$a+3 b-5=7+3 	imes \frac{70}{6}-5$

$=37$

${x_i}$	$3$	$8$	$13$	$18$	$25$
${f_i}$	$7$	$10$	$15$	$10$	$6$

The mean and variance of $7$ observations are $8$ and $16$ respectively. If one observation $14$ is omitted a and $b$ are respectively mean and variance of remaining $6$ observation, then $a+3 b-5$ is equal to $..........$.

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