/*! 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.95 Your company has a contract to p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified

Accordingly, compute probability average time of servicing is 1.1hours per technician to finish the work is approximately 0.8. Thus, it means that 80% chance that the service time to do the work will be less than1.1 hours.

Step by step solution

01

Given Information

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?

02

Explanation

According to the given details, the average time of servicing is 1hour, the standard deviation is 1 hour and the sample size is 70units. To compute the probability average time of servicing is 1.1hours per technician to complete the work, use the Ti-83calculator, for this, click on 2nd, then DISTR and then scroll down to the normal CDF option and enter the furnished details. After this, click on ENTER button of the calculator to have the desired outcome. The screenshot is given as below:

03

Explanation

Accordingly, compute probability average time of servicing is 1.1hours per technician to finish the work is approximately 0.8. Thus, it means that 80% chance that the service time to do the work will be less than1.1 hours.

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

Use the information in Example \(7.9\), but change the sample size to \(144\).

a. Find \(P(20<\bar{x}<30)\).

b. Find \(P(\sum x\) is at least \(3,000)\).

c. Find the \(75th\) percentile for the sample mean excess time of \(144\) customers.

A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.

Find the 90th percentile for the total weight of the 100 weights.

Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of \(0.88. Suppose that we randomly pick 25daytime statistics students.

a. In words,Χ=____________

b.Χ~_____(_____,_____)

c.role="math" localid="1651578876947" Inwords,X=____________

d. X~______(______,______)

e. Find the probability that an individual had between \)0.80and\(1.00. Graph the situation, and shade in the area to be determined.

f. Find the probability that the average of the 25 students was between \)0.80and$1.00. Graph the situation, and shade in the area to be determined.

g. Explain why there is a difference in part e and part f.

Yoonie is a personnel manager in a large corporation. Each month she must review 16of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let x-be the random variable representing the meantime to complete the 16reviews. Assume that the 16 reviews represent a random set of reviews.

Find the probability that the mean of a month’s reviews will take Yoonie from 3.5to 4.25hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

a.

b. P(________________) = _______

76. The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds.

a. In words, X=

b. X~

c. In wordsX-=

d. X-~

e. Before doing any calculations, which do you think will be higher? Explain why.

i. The probability that an individual attention span is less than ten minutes.

ii. The probability that the average attention span for the 60 children in less than ten minutes?

f. Calculate the probabilities in part e.

g. Explain why the distribution for X- is not exponential.

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.