Question

A normally distributed variable, x, has a standard are deviation of 3.9. Also 71.37% of the data are larger than 249.8. Find the mean of x.

Answer 1

Denote the normally distributed variable by \(Y\) and the standard normal distribution by another random variable \(Z\). You want to find \(x\) such that
\[P(Y>249.8)=P\left(\frac{Y-x}{3.9}>\frac{249.8-x}{3.9}\right)=P\left(Z>\frac{249.8-x}{3.9}\right)=0.7137\]
Refer to a \(z\)-table (like the one here: ` http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf `).

This table gives you left-tail probabilities, but we're dealing with a right-tail value. In this table, we have \(0.7137\) associated with a left-tail \(z\) score between \(0.56\) and \(0.57\). Let's call it the midpoint \(0.565\). To make this a right-tail value, we simply take advantage of the fact that the distribution is symmetric, so we would use \(z=-0.565\).

In other words, \(P(Z>-0.565)=0.7137\), which means you can find the mean \(x\) by solving
\[\frac{249.8-x}{3.9}=-0.565\]
Here's a plot showing that this is a close approximation for the actual \(z\) score: http://www.wolframalpha.com/input/?i=P%28Z%3E-0.565%29