Question

If $X$ is normally distributed with mean $6$ and standard deviation $5$, find:
(i) $P[0\le X\le 8]$
(ii) $P(|X-6|<10)$
Here $P(0<Z<1.2)=0.3849$
$P(0<Z<0.4)=0.1554$
$P(0<Z<2)=0.4772$

Answer 1

Given $\mu =6,\sigma =5$
(i) $P(0\le X\le 8)$
We know that $Z=\frac{X-\mu }{\sigma }$
When $X=0,Z=\frac{0-6}{5}=\frac{-6}{5}=-12$
When $X=8,Z=\frac{8-6}{5}=\frac{2}{5}=0.4$
$\therefore P(0\le X\le 8)=P(-1.2<Z<0.4)=P(0<Z<1.2)+P(0<Z<4)$
(due to symmetry)
$=0.3849+0.1554=0.5403$
(ii) $P(|X-6|<10)=P(-10<(X-6)<10)$
$\Rightarrow P(-4<X<16)$
When $X=-4,Z=\frac{-4-6}{5}=\frac{-10}{5}=-2$
When $X=16,Z=\frac{16-6}{5}=\frac{10}{5}=2$
$(-4<X<16)=P(-2<Z<2)=2P(0<Z<2)$
$=2\left(0.4772\right)=0.9544$