Question

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

Answer 1

Answer: 37

$\frac{x_1+x_2+\dots +x_7}{7}=8$
$\frac{x_1+x_2+x_3\dots +x_6+14}{7}=8$
$\Rightarrow x_1+x_2+\dots .+x_6=42$
$\therefore \frac{x_1+x_2\dots .+x_6}{6}=\frac{42}{6}=7=a$
$\frac{\Sigma x_i^2}{7}-8^2=16$
$\Sigma x{i}^2=560$
$\Rightarrow x_1^2+x_2^2+\dots +x_6^2=364$
$b=\frac{x_1^2+x_2^2+\dots .+x_6^2}{6}-7^2$
$=\frac{364}{6}-49$
$b=\frac{70}{6}$
$a+3b-5=7+3\times \frac{70}{6}-5$
$=37$