Question

A lawyer travels daily from his home to his oce in town. On average, the one-way trip takes 20 minutes, with a standard deviation of 3 minutes. What is the probability that the lawyer's one way trip time is between 18 and 27 minutes. Assume the distribution of trip times to be normally distributed.

Answer 1

Let $X$ be a random variable representing the lawyer's one way trip then, $\mu=20 minutes$ and $\sigma=3minutes$. We need to find the probability that the lawyer's one way trip time is between 18 and 27 minutes given as,

$p(18\lt X\lt27)$ and we standardize in order to use the standard normal tables to find the required probability.

Now,

$p(18\lt X\lt27)=p((18-\mu)/\sigma \lt Z\lt(27-\mu)/\sigma)=p((18-20)/3\lt Z\lt (27-20)/3)=p(0.67\lt Z\lt 2.33)$

We can obtain this probability from standard normal tables as,

$p(0.67\lt Z\lt 2.33)=\phi(2.33)-\phi(0.67)=0.9901-0.7486=0.2415$

Therefore,  the probability that the lawyer's one way trip time is between 18 and 27 minutes is 0.2415.