81. The$90th$ percentile sample average wait time (in minutes) for a sample of 100 riders is:

a. $315.0$

b.$40.3$

c.$38.5$

d.$65.2$

Consider $X$be the continuous random variable which shows the waiting time is uniformly distributed. It should be expressed as:

$X~U(0,75)$

Where,

$a=0$

$b=75$

Let's compute the average waiting time as follow:

${\mu }_x=\frac{b-a}{2}$

$=\frac{75-0}{2}$

$=37.5$Minutes

Standard deviation of the given distribution is:

${\sigma }_x=\sqrt{\frac{{(b-a)}^2}{12}}$

$=\sqrt{\frac{{(75-0)}^2}{12}}$

$=21.650$

The sample size is greater than 30.

Hence, according to Central Limit Theorem

$$\begin{gathered} \overline{X}~N\left(37.5,\frac{21.650}{\sqrt{100}}\right) \\ \text{where,}n=100 \end{gathered}$$

Let's use Ti-83 calculator to compute the ${90}^{th}$percentile for sample average waiting time.

For this, Click on $2^{nd}$.

Then DISTR and scroll down to the invNorm option and enter the provided values of mean $(37.5),$standard deviation $\left(\frac{21.65}{\sqrt{100}}\right)$and the percentile,

After this, click on ENTER button of calculator to have the desired result.

The screenshot is given as below:

Therefore, ${90}^{th}$percentile sample average waiting time is approximately $40.27$hours.

Thus, the correct option is 'b'.