Suppose X has normal $(μ,σ^{2})$ distribution, and $P(X \leq 0)=1/3, P(X \leq 1)=2/3$
What are the values of μ and σ?
$\frac{1}{3}=P(X \leq 0)=P(\frac{X-μ}{σ} \leq \frac{0-μ}{σ})=φ(\frac{-μ}{σ})$
$φ(\frac{-μ}{σ})=\frac{1}{3} → φ(\frac{μ}{σ})=\frac{2}{3} → \frac{μ}{σ}=φ^{-1}(\frac{2}{3})=0.43$
I don't understand how this $φ(\frac{-μ}{σ})=\frac{1}{3} → φ(\frac{μ}{σ})=\frac{2}{3}$ happened.
You will have to make use of a special property of the CDF of standard normal distributions, namely $\Phi(x)+\Phi(-x)=1\forall x\in\Bbb R$. It is easy to see why this is true:$$\begin{align*}\Phi(x)+\Phi(-x)&=\int_{-\infty}^x f(x)dx+\int_{-\infty}^{-x}f(x)dx\\&=-\int_{-\infty}^x f(-(-x))d(-x)+\int_{-\infty}^{-x}f(x)dx\\&=-\int_\infty^{-x} f(-y)dy+\int_{-\infty}^{-x}f(x)dx\end{align*}$$ where $y=-x$ and $f(x)$ is the PDF of standard normal distribution. $f(x)$ is even so $f(-y)=f(y)$, giving$$\int^\infty_{-x} f(y)dy+\int_{-\infty}^{-x}f(x)dx=1$$You can extend this concept to general normal distributions:$$F(\mu-x)+F(\mu+x)=1$$