/*! 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} Problem 89 Stress on the job is a major con... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Stress on the job is a major concern of a large number of people who go into managerial positions. It is estimated that \(80 \%\) of the managers of all companies suffer from job-related stress. a. What is the probability that in a sample of 200 managers of companies, exactly 150 suffer from job-related stress? b. Find the probability that in a sample of 200 managers of companies, at least 170 suffer from job-related stress. c. What is the probability that in a sample of 200 managers of companies, 165 or fewer suffer from job-related stress? d. Find the probability that in a sample of 200 managers of companies, 164 to 172 suffer from job-related stress.

Short Answer

Expert verified
The exact probabilities will depend on the results of the computations in the four steps. Keep in mind that calculating those sums might require a statistical calculator or a computer program.

Step by step solution

01

Calculate the probability of exactly 150 managers suffering from job-related stress.

Following the formula, we find that \( P(X=150) = C(200,150) \cdot 0.8^{150} \cdot (1-0.8)^{200-150} \)
02

Calculate the probability of at least 170 managers suffering from job-related stress.

Here we need to sum the probabilities from 170 to 200: \( P(X\geq 170) = \sum_{k=170}^{200} C(200,k) \cdot 0.8^k \cdot (1-0.8)^{200-k} \)
03

Calculate the probability of 165 or fewer managers suffering from job-related stress.

Here we need to sum the probabilities from 0 to 165: \( P(X\leq 165) = \sum_{k=0}^{165} C(200,k) \cdot 0.8^k \cdot (1-0.8)^{200-k} \)
04

Calculate the probability of 164 to 172 managers suffering from job-related stress.

Here we need to sum the probabilities from 164 to 172: \( P(164 \leq X\leq 172) = \sum_{k=164}^{172} C(200,k) \cdot 0.8^k \cdot (1-0.8)^{200-k} \)

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

Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is \(2.4\) minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a \(50 \%\) discount on the charges. The company wants to limit this discount to at most \(5 \%\) of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.

The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every customer who spends more than a certain amount per visit at this supermarket. The expectation of the management is that after this promotional policy is advertised, the expenditures for all customers at this supermarket will be normally distributed with a mean of \(\$ 95\) and a standard deviation of \(\$ 20 .\) If the management decides to give free gifts to all those customers who spend more than \(\$ 130\) at this supermarket during a visit, what percentage of the customers are expected to get free gifts?

Mong Corporation makes auto batteries. The company claims that \(80 \%\) of its LL70 batteries are good for 70 months or longer. a. What is the probability that in a sample of 100 such batteries, exactly 85 will be good for 70 months or longer? b. Find the probability that in a sample of 100 such batteries, at most 74 will be good for 70 months or longer. c. What is the probability that in a sample of 100 such batteries, 75 to 87 will be good for 70 months or longer? d. Find the probability that in a sample of 100 such batteries, 72 to 77 will be good for 70 months or longer.

According to the records of an electric company serving the Boston area, the mean electricity consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electricity consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. a. Find the probability that the monthly electricity consumption during winter by a randomly selected household from this area is less than 1950 kilowatt- hours. b. What percentage of the households in this area have a monthly electricity consumption of 900 to 1300 kilowatt-hours?

According to a November 8,2010 report on www.teleread.com, \(7 \%\) of U.S. adults with online services currently read e-books. Assume that this percentage is true for the current population of U.S. adults with online services. Find the probability that in a random sample of 600 U.S. adults with online services, the number who read e-books is a. exactly 45 b. at most 53 c. 30 to 50

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.