/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 81 81. The90th percentile sample av... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

81. The90th 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

Short Answer

Expert verified

The 90thpercentile sample average wait time for a sample of 100 riders is option "b"40.3

Step by step solution

01

Given information

Consider Xbe 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

02

Step 2:Final answer

Let's compute the average waiting time as follow:

μx=b-a2

=75-02

=37.5Minutes

Standard deviation of the given distribution is:

σx=(b-a)212

=(75-0)212

=21.650

The sample size is greater than 30.

Hence, according to Central Limit Theorem

X¯~N37.5,21.650100where,n=100

03

Calculate the 90th percentile

Let's use Ti-83 calculator to compute the 90thpercentile for sample average waiting time.

For this, Click on 2nd.

Then DISTR and scroll down to the invNorm option and enter the provided values of mean (37.5),standard deviation 21.65100and the percentile,

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

The screenshot is given as below:

Therefore, 90thpercentile sample average waiting time is approximately 40.27hours.

Thus, the correct option is 'b'.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36and a standard deviation of about ten. Suppose that 16individuals are randomly chosen. Let role="math" localid="1648361500255" X¯=average percent of fat calories.

a. X¯~_____ (______, ______)

b. For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined.

c. Find the first quartile for the average percent of fat calories.

70. Which of the following is NOT TRUE about the distribution for averages?

a. The mean, median, and mode are equal.

b. The area under the curve is one.

c. The curve never touches the x-axis.

d. The curve is skewed to the right.

Your company has a contract to perform preventive maintenance on thousands of air-conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will serve a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?

Find the probability that the sum of the 40 values is less than 7,000.

The mean number of minutes for app engagement by a table use is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.

a. What is the probability that the sum of the sample is between seven hours and ten hours? What does this mean in context of the problem?

b. Find the 84thand 16thpercentiles for the sum of the sample. Interpret these values in context.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.