For a normal distribution 30% items are below 45 and 8% are above 64. Find the mean and variance of the normal distribution.

Solution:

Let $m$ and $\sigma$ be the mean and standard deviation of the distribution. $z=\cfrac{X-m}{\sigma}$ --------------(1) **Case 1:** For X=45 at $z_{1}$ $z=z_{1}=\cfrac{45-m}{\sigma}$ ----------------(2) $\therefore P(X < 45)= 30 \%$ $P(z < z_{1})=0.30$ ![enter image description here][1] $=0.5 -$ Area between $z=0 \ to z=-z_{1}$ is 0.30 $\therefore$ Area between …