Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q. 81

Expert-verified

Introductory Statistics

Found in: Page 434

Book edition OER 2018

Author(s) Barbara Illowsky, Susan Dean

Pages 902 pages

ISBN 9781938168208

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later

Short Answer

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

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,

$a = 0$

$b = 75$

Step 2:Final answer

Let's compute the average waiting time as follow:

$μ_{x} = \frac{b - a}{2}$

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

$= 37.5$ Minutes

Standard deviation of the given distribution is:

$σ_{x} = \sqrt{\frac{{(b - a)}^{2}}{12}}$

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

$= 21.650$

The sample size is greater than 30.

Hence, according to Central Limit Theorem

$\bar{X} ~ N (37.5, \frac{21.650}{\sqrt{100}}) where, n = 100$

Step 3: Calculate the 90th percentile

Let's use Ti-83 calculator to compute the $90^{t h}$ percentile for sample average waiting time.

For this, Click on $2^{n d}$ .

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

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

The screenshot is given as below:

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

Thus, the correct option is 'b'.

Most popular questions for Math Textbooks

What is the mean, standard deviation, and sample size?

An unknown distribution has a mean of $80$ and a standard deviation of $12$ . A sample size of $95$ is drawn randomly from the population.

Find the probability that the sum of the $95$ values is greater than $7650$ .

Based on data from the National Health Survey, women between the ages of $18$ and $24$ have an average systolic blood pressures (in mm Hg) of $114.8$ with a standard deviation of $13.1 .$ Systolic blood pressure for women between the ages of $18$ to $24$ follow a normal distribution.
a. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than $120$ .b. If $40$ women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than $120$ .c. If the sample were four women between the ages of $18$ to $24$ and we did not know the original distribution, could the central limit theorem be used?

Find the percentage of sums between 1.5 standard deviations below the mean and one standard deviation above the mean.

Find the probability that the sums will fall between the $z -$ scores $- 2$ and $1$ .

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free