$X$ is a normally distributed (N($\mu$,$\sigma^2$)) random variable with $\mathbb{P}(X\leq0)=\frac{1}{3}$ and $\mathbb{P}(X\leq1)=\frac{2}{3}$. Determine values of $\mu$ and $\sigma^2$.
I've tried solving $$\int_{-\infty}^0\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}dx = \frac{1}{3}$$ and $$\int_{-\infty}^1\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}dx = \frac{2}{3}$$ simultaneously for $\mu$ and $\sigma$ but that didn't seem to work. Any idea on how to solve this problem? Do I need to use the central limit theorem?
Update
So I can say that if $Z = \frac{X-\mu}{\sigma}$ then $$\mathbb{P}(Z\leq \frac{-\mu}{\sigma}) = \frac{1}{3} = P(Z>1-\frac{\mu}{\sigma})$$ For the first part of the equation we have a negative value $-\frac{\mu}{\sigma}$ So we rewrite this as $$\mathbb{P}(Z\leq \frac{-\mu}{\sigma}) = \mathbb{P}(Z> \frac{\mu}{\sigma}) = 1- \mathbb{P}(Z\leq \frac{\mu}{\sigma})$$
And therefore in the normal table: 1 - $\Phi(\frac{\mu}{\sigma}) = \frac{1}{3}$ and 1 - $\Phi(1-\frac{\mu}{\sigma})= \frac{1}{3}$. However, if I look up the values in the table I find that these equations seem to contradict each other. Where do I make a mistake?
Hint: you have: $$\mathsf P(X\leqslant 0)~=~\tfrac 1 {~3~}~=~\mathsf P(X> 1)$$
Or, if $Z:=\frac{X-\mu}{\sigma}$, so that $Z\sim\mathcal N(0,1)$
$$\mathsf P(Z\leqslant \frac{-\mu}{\sigma})~=~\tfrac 1 {~3~}~=~\mathsf P(Z> \frac{1-\mu}{\sigma})$$