A student leaves home at 8 a.m. every morning in order to arrive at the University at 9 a.m.
He finds that over a long period he is late once in forty times. ($\frac{1}{40}$)
He then tries leaving home at 7.55 a.m. and finds that over a similar period he is late once in one hundred times. ($\frac{1}{100}$)
Assuming that the time of his journey has a normal distribution, before what time should he leave home in order not to be late more than once in two hundred times?
Let random variable $X$ be the length of the trip, $\mu$ be the mean of $X$, and $\sigma$ the standard deviation, all in minutes.
We have $\Pr(X\gt 60)=\frac{1}{40}$. It follows (normal table) that $60$ is $1.96$ standard deviation units below the mean, so $$\mu-60 \approx -1.96\sigma.$$
Similarly, $$\mu-65\approx -2.33\sigma.$$
Treat $\approx$ as if it were $=$, and solve the system of two linear equations for $\mu$ and $\sigma$. Once we know these, the full distribution is known, and I expect you can take care of the rest.