$X\sim N(\mu,\sigma^{2})$. It is also known that $P(X ≤ 160) = \frac{1}{2}$ and $P(X ≤ 140) = \frac{1}{4}.$ What are the values of $µ$ and $σ$?

A random variable $X$ is know to be distributed according to a normal distribution. It is also known that $P(X ≤ 160) = \frac{1}{2}$ and $P(X ≤ 140) = \frac{1}{4}.$

What are the values of $µ$ and $σ$?

My solution :

$P(X\leq160)=P(\frac{X-\mu}{\sigma}\leq\frac{160-\mu}{\sigma})\overset{Z=\frac{X-\mu}{\sigma}}{=}$ $P(Z\leq\frac{160-\mu}{\sigma})=\phi(\frac{160-\mu}{\sigma})=\frac{1}{2}$

$P(X\leq140)=P(\frac{X-\mu}{\sigma}\leq\frac{140-\mu}{\sigma})\overset{Z=\frac{X-\mu}{\sigma}}{=}$ $P(Z\leq\frac{140-\mu}{\sigma})=\phi(\frac{140-\mu}{\sigma})=\frac{1}{4}$

I think I need to find the values with this table:

I don't know how to do it and be grateful for some help , Thank you !

From $P(X\leq 160)=1/2$, we deduce that the normally distributed random variable $X$ has mean $\mu=160$. Then, from $P(X\leq 140)=1/4$, we get

$$P\Big(\frac{X-160}{\sigma}\leq \frac{-20}{\sigma}\Big)=\frac{1}{4}.$$ By the symmetry of standard normal curve, this means that $$P\Big(\frac{X-160}{\sigma}\leq \frac{20}{\sigma}\Big)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}=0.75$$ According to your table, this means that $20/\sigma\approx 0.675$, and solving for $\sigma$, we get $\sigma\approx 29.63$. So, $X$ has mean $\mu=160$ and standard deviation $\sigma\approx29.63$.