Normal distribution in which 90% of samples are between 2.99 and 3.01; what is the standard deviation?

Question

Steel rods are manufactured to be 3 inches in diameter but they are acceptable if they are inside the limit 2.99 inches and 3.01 inches. It is observed that 5% are rejected as oversized and 5% are rejected undersized. Assuming that the diameters are normally distributed, find the standard deviation of the distribution.

Answer 1

Since $X$ is normally (hence symmetric with respect of its mean) distributed, the equation $P(X>3.01)=P(X<2.99)$ allows the conclusion that $\mathbb E(X)=\frac{1}{2}(3.01+2.99)=3$.

Then $X=3+\sigma U$ where $U\sim Norm(0,1)$, so that $P(U>\frac{0.01}{\sigma})=P(X>3.01)=0.05$.

This equation can be written as $1-\Phi\left(\frac{0.01}{\sigma}\right)=0.05$ where $\Phi$ denotes the CDF of the standard normal distribution. It enables you to find $\sigma$ by means of a table.

Answer 2

First you have to transform the random variable: $Z=\frac{X-\mu}{\sigma}$

What you are looking for is $P(2.99 \leq X \leq 3.01)=2\Phi(\frac{3.01-3}{\sigma})-1=0.9$

$\Phi(.)$ is the Standard normal distribution. What you have to do next is to solve this equation for $\sigma$. The value for $\Phi(.)^{-1}$ (inverse function) can be looked up in a table.