# Probability - Normal Distribution without giving mean or variance

The annual profit of a life insurance company is normally distributed.

The probability that the annual profit does not exceed 2000 is 0.7642. The probability that the annual profit does not exceed 3000 is 0.9066. Calculate the probability that the annual profit does not exceed 1000.

I've tried to approach this question using the formula $$\displaystyle Z = \frac{X-\mu}{\sigma}$$ but I cant seem to find the mean or variance.

I would greatly appreciate any help that can point me in the right direction for this question.

• What are the $Z$s corresponding to probabilities of $0.7642$ and $0.9066$? You know the $X$s are $2000$ and $3000$ so that gives you two equations in two unknowns you can solve – Henry Sep 29 '19 at 9:41

$$P( X \le 2000) = 0.7642 \iff P( Z \le \frac{2000-\mu}{\sigma}) = 0.7642$$
You should be able to compute what should$$\frac{2000-\mu}{\sigma}$$
be, and obtain a linear equation in $$\mu$$ and $$\sigma$$.
Do similar thing to the other information and you can solve the linear system of equations for $$\mu$$ and $$\sigma$$.