/*! 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 A subject is taught to do a task... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 89

A subject is taught to do a task in two different ways. Studies have shown that when subjected to mental strain and asked to perform the task, the subject most often reverts to the method first learned, regardless of whether it was easier or more difficult. If the probability that a subject returns to the first method learned is .8 and six subjects are tested, what is the probability that at least five of the subjects revert to their first learned method when asked to perform their task under stress?

Short Answer

Expert verified
Based on our calculation using the binomial distribution, the probability that at least five out of six subjects revert to their first-learned method when asked to perform their task under stress is approximately 0.507904.

Step by step solution

01

Identify the parameters of the binomial distribution

The total number of subjects (n) is 6, and the probability of reverting to the first method (p) is 0.8. We are looking for the probability of at least 5 successes, which means we need to calculate the probabilities of getting exactly 5 and exactly 6 successes.
02

Calculate the probability of X = 5 successes

Using the binomial formula, we will calculate the probability of getting exactly 5 successes (X = 5): \[P(X=5) = C(n, k) * p^k * (1-p)^{(n-k)}\] Where: - n = 6 (total trials) - k = 5 (number of successes) - p = 0.8 (probability of success) - C(n, k) = combination(n, k) = n! / (k! * (n-k)!) Plugging the values into the formula: \[P(X=5) = C(6, 5) * 0.8^5 * (1-0.8)^{(6-5)}\] Calculating the combination: \[C(6, 5) = \frac{6!}{5!(6-5)!} = 6\] Now, calculating the probability: \[P(X=5) = 6 * 0.8^5 * (1-0.8)^{(6-5)} = 6 * 0.8^5 * 0.2^1 = 0.24576\]
03

Calculate the probability of X = 6 successes

Using the binomial formula, we will calculate the probability of getting exactly 6 successes (X = 6): \[P(X=6) = C(6, 6) * 0.8^6 * (1-0.8)^{(6-6)}\] Calculating the combination: \[C(6, 6) = \frac{6!}{6!(6-6)!} = 1\] Now, calculating the probability: \[P(X=6) = 1 * 0.8^6 * (1-0.8)^{(6-6)} = 1 * 0.8^6 * 0.2^0 = 0.262144\]
04

Calculate the probability of at least 5 successes

Now that we have the probabilities of exactly 5 and exactly 6 successes, we can add them together to find the probability of at least 5 successes: \[P(X \geq 5) = P(X=5) + P(X=6) = 0.24576 + 0.262144 = 0.507904\] The probability that at least five out of six subjects revert to their first-learned method when asked to perform their task under stress is approximately 0.507904.

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 91影视!

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

If a person is given the choice of an integer from 0 to 9 , is it more likely that he or she will choose an integer near the middle of the sequence than one at either end? a. If the integers are equally likely to be chosen, find the probability distribution for \(x\), the number chosen. b. What is the probability that a person will choose a \(4,5,\) or \(6 ?\) c. What is the probability that a person will not choose a \(4,5,\) or \(6 ?\)

According to the Humane Society of the United States, there are approximately 65 million owned dogs in the United States, and approximately \(40 \%\) of all U.S. households own at least one dog. \({ }^{4}\) Suppose that the \(40 \%\) figure is correct and that 15 households are randomly selected for a pet ownership survey. a. What is the probability that exactly eight of the households have at least one dog? b. What is the probability that at most four of the households have at least one dog? c. What is the probability that more than 10 households have at least one dog?

A psychiatrist believes that \(80 \%\) of all people who visit doctors have problems of a psychosomatic nature. She decides to select 25 patients at random to test her theory. a. Assuming that the psychiatrist's theory is true, what is the expected value of \(x\), the number of the 25 patients who have psychosomatic problems? b. What is the variance of \(x\), assuming that the theory is true? c. Find \(P(x \leq 14)\). (Use tables and assume that the theory is true.) d. Based on the probability in part \(\mathrm{c}\), if only 14 of the 25 sampled had psychosomatic problems, what conclusions would you make about the psychiatrist's theory? Explain.

Security Systems A home security system is designed to have a \(99 \%\) reliability rate. Suppose that nine homes equipped with this system experience an attempted burglary. Find the probabilities of these events: a. At least one of the alarms is triggered. b. More than seven of the alarms are triggered. c. Eight or fewer alarms are triggered.

Find the mean and standard deviation for a binomial distribution with these values: a. \(n=1000, p=.3\) b. \(n=400, p=.01\) c. \(n=500, p=.5\) d. \(n=1600, p=.8\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation
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.