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 $a+3 b-5$ is equal to ______

Answer 1

$ \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 $
$ \therefore \frac{ x _1+ x _2 \ldots .+ x _6}{6}=\frac{42}{6}=7= a $
$\frac{\Sigma x _{ i }^2}{7}-8^2=16 $
$ \Sigma xi ^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 \times \frac{70}{6}-5 $
$ =37$