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 $z=0 \ to z=-z_{1}$ is 0.20 $\therefore$ From the 'z' table, $z_{1}=-0.5244$ Equation (2) becomes $\therefore z_{1}=\cfrac{45-m}{\sigma}$ $-0.5244=\cfrac{45-m}{\sigma}$ $m-0.5244 \sigma=45$ ----------------(3) **Case 2:** For X=64 at $z_{2}$ $z=z_{2}=\cfrac{64-m}{\sigma}$ ----------------(4) $\therefore P(X > 64)= 8 \%$ $P(z > z_{2})=0.08$ $=0.5 -$ Area between $z=0 \ to z=-z_{2}$ is 0.08 ![enter image description here][2] $\therefore$ Area between $z=0 \ to z=-z_{2}$ is 0.42 $\therefore$ From the 'z' table, $z_{2}=1.4051$ Equation (4) becomes $\therefore z_{2}=\cfrac{64-m}{\sigma}$ $1.4051=\cfrac{64-m}{\sigma}$ $m+1.4051 \sigma=64$ ----------------(5) Solving (3) and (5) simultaneously, we get, $\sigma=50.16 \approx 50$ $\sigma=9.847 \approx 10$