Suppose that the travel time from your home to your office is normally distributed with mean $40$minutes and standard deviation $7$minutes. If you want to be $95$percent certain that you will not be late for an office appointment at $1$p.m., what is the latest time that you should leave home?

Assume your commute time from home to work is regularly distributed, with a mean of $40$minutes and a standard deviation of $7$minutes. What is the latest time you should leave home if you want to be $95\%$sure you won't be late for a $1$p.m. office appointment

That's something we're given $X~\mathcal{N}\left(40,7^2\right)$. We must pick a time $t$that will allow us to complete our task.

$P(X\le t)=0.95$

Because in such instance, if we leave home on time or early, we will be on time in our office with a probability of $0.95$The preceding expression is identical to

$$\begin{gathered} 0.95=P\left(\frac{X-40}{7}\le \frac{t-40}{7}\right) \\ =\Phi \left(\frac{t-40}{7}\right) \end{gathered}$$

We can see from the table of standard normal cumulative distribution that

$$\begin{gathered} \frac{t-40}{7}=1.64 \\ t=51.515 \end{gathered}$$

That suggests that to arrive at work on time with a chance of $0.95$, we must leave the house at least $12:08$a.m.