Answer to Question #129216 in Statistics and Probability for Aadhi

Question

In a normal distribution 15% of items are below 35 and 10% are above 65. Find the mean.

Answer 1

Let $X$ be a normal random variable with mean $m$ and variance $\sigma^2$ such that

$P(X<35)=15\%=0.15$ and $P(X>65)=10\%=0.1$, then

$P(X<65)=1-0.1=0.9.$

Since $X$ is a normal random variable, then

$Z=\frac{X-m}{\sigma}$ is the standard normal random variable.

$P(X<35)=P(\frac{X-m}{\sigma}<\frac{35-m}{\sigma})=\\ P(Z<\frac{35-m}{\sigma})=0.15,$

$P(X<65)=P(\frac{X-m}{\sigma}<\frac{65-m}{\sigma})=\\ P(Z<\frac{65-m}{\sigma})=0.9.$

Using quantile function $Q$ for the standard normal random variable,

$\frac{35-m}{\sigma}=Q(0.15)=-1.04$,

$\frac{65-m}{\sigma}=Q(0.9)=1.28$,

hence

$35-m=-1.04\sigma$,

$65-m=1.28\sigma$,

$65-m-(35-m)=1.28\sigma+1.04\sigma,$

$2.32\sigma=30, \sigma=\frac{30}{2.32}\approx12.931,$

$65-m=1.28\cdot12.931\approx 16.552, \\ m=65-16.552=48.448.$

Answer: $48.448.$