The mean and variance of 7 observations are 8 and 16 respectively.

Question

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 observations, then $a+3b-5$ is equal to \_\_\_\_\_\_\_.

Answer: 37

Accepted Answer

Sum of 7 obs $=56$. Removing 14: sum of remaining 6 $=42$, so $a = \dfrac{42}{6}=7$.

$\sum x_i^2 = 7(16+64)=560$. Removing $14^2=196$: sum of squares of remaining 6 $=364$.

$$b = \frac{364}{6}-7^2 = \frac{364}{6}-49 = \frac{70}{6}$$

$$a+3b-5 = 7+3\times\frac{70}{6}-5 = 7+35-5 = 37$$

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