Question

A normal random variable x has an unknown mean and standard deviation. The probability that x exceeds 4 is .9772, and the probability that x exceeds 5 is .9332. Find the mean and standard deviation.

Answer 1

You're given that \(X\) follows a normal distribution, and
\[\begin{cases}P(X>4)=0.9772\\P(X>5)=0.9332\end{cases}\]
Change \(X\) to \(Z\), the standard normal distribution, using the transformation \(Z=\dfrac{X-\mu}{\sigma}\).
\[\begin{cases}P\left(\dfrac{X-\mu}{\sigma}>\dfrac{4-\mu}{\sigma}\right)=0.9772\\P\left(\dfrac{X-\mu}{\sigma}>\dfrac{5-\mu}{\sigma}\right)=0.9332\end{cases}~~\Rightarrow~~\begin{cases}P\left(Z>\dfrac{4-\mu}{\sigma}\right)=0.9772\\P\left(Z>\dfrac{5-\mu}{\sigma}\right)=0.9332\end{cases}\]
Find the corresponding \(z\)-values in a table. The first probability gives you \(z_1=-2.0\), and the second gives \(z_2=-1.5\).

Now solve for \(\mu\) and \(\sigma\):
\[\begin{cases}\dfrac{4-\mu}{\sigma}=-2\\\dfrac{5-\mu}{\sigma}=-1.5\end{cases}\]