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Q. 81

Expert-verifiedFound in: Page 434

Book edition
OER 2018

Author(s)
Barbara Illowsky, Susan Dean

Pages
902 pages

ISBN
9781938168208

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.$65.2$

The ${90}^{th}$percentile sample average wait time for a sample of 100 riders is option "b" $40.3$

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

$\overline{X}~N\left(37.5,\frac{21.650}{\sqrt{100}}\right)\text{}\phantom{\rule{0ex}{0ex}}\text{where,}n=100$

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'.

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