Determine the values of $\mu$ and $\sigma^2$. [closed]

Question

I have the following exercise which I'm not sure how to solve:

Suppose that the random variable $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$,i.e. $X\sim N(\mu,\sigma^2)$. You are told that:

$$E(X^2)=10 \text{ and } P(X>4)=0.1587$$

Determine the values of $\mu$ and $\sigma^2$.

I'm am not sure how to use these facts to derive $\mu$ and $\sigma^2$. Do I have to use software or tables? Any hint is welcome

Answer 1

You know that

$$P(X\leq 4)=0.8413$$

Knowing also that $E(X^2)-\mu^2=\sigma^2$ you get

$$P\left(Z \leq\frac{4-\mu}{\sqrt{10-\mu^2} } \right)=0.8413$$

Using the tables you get

$$\frac{4-\mu}{\sqrt{10-\mu^2}}=1$$

Solve it w.r.t. $\mu$ and you get your mean: $\mu=0.75$

and immediately the variance $\sigma^2=10-0.75^2=9.4375$

Answer 2

Normal distribution $N(\mu,\sigma^2)$ is $$f(x)=\frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} $$ $P(X>4)$ is calculated via $$\int_4^\infty f(x) \ dx $$ and $E[X^2]$ is calculated via $$\int_{-\infty}^\infty x^2f(x) \ dx $$

So you can obtain your answer by simply calculating these.