Question:

$\displaystyle X$ has a normal distribution, and $\displaystyle P( X > 73.05 = 0.0289 )$. Give that the variance of the distribution is $\displaystyle 18$, find the mean.

Attempt:

$\displaystyle = 1 - 0.0289$

$\displaystyle = 0.9711$

$\displaystyle = -1.897$

$\displaystyle \sigma^2 = 18$ , $\displaystyle \sigma = 3\sqrt2$

$\displaystyle P\left( \frac{X - \mu}{\sigma} > 73.05 \right) = -1.897$

Don't know how to find the mean...