The mean and variance of 7 observations are 8 | vedclass

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(D) Let the $7$ observations be $x_1, x_2, \ldots, x_7$. Given $\bar{x} = 8$ and $\sigma^2 = 16$.$\frac{\sum_{i=1}^{7} x_i}{7} = 8 \Rightarrow \sum_{i

Vedclass · Accepted Answer

$37$

Vedclass · Answer

(D) Let the $7$ observations be $x_1, x_2, \ldots, x_7$. Given $\bar{x} = 8$ and $\sigma^2 = 16$.$\frac{\sum_{i=1}^{7} x_i}{7} = 8 \Rightarrow \sum_{i=1}^{7} x_i = 56$.If one observation $14$ is omitted,the sum of the remaining $6$ observations is $56 - 14 = 42$.Thus,the new mean $a = \frac{42}{6} = 7$.Given $\frac{\sum_{i=1}^{7} x_i^2}{7} - (8)^2 = 16 \Rightarrow \frac{\sum x_i^2}{7} = 16 + 64 = 80$.So,$\sum_{i=1}^{7} x_i^2 = 80 	imes 7 = 560$.The sum of squares of the remaining $6$ observations is $560 - (14)^2 = 560 - 196 = 364$.The new variance $b = \frac{\sum_{i=1}^{6} x_i^2}{6} - a^2 = \frac{364}{6} - (7)^2 = \frac{364}{6} - 49 = \frac{364 - 294}{6} = \frac{70}{6} = \frac{35}{3}$.Now,$a + 3b - 5 = 7 + 3 	imes (\frac{35}{3}) - 5 = 7 + 35 - 5 = 37$.

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

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