/*! 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 552 Suppose that \(75 \%\) of the st... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 552

Suppose that \(75 \%\) of the students taking statistics pass the course. In a class of 40 students, what is the expected number who will pass. Find the variance and standard deviation.

Short Answer

Expert verified
In the given statistics class, we expect 30 students to pass the course, with a variance of 7.5 and a standard deviation of approximately 2.74 students.

Step by step solution

01

Calculate the expected number of students who will pass

Using the formula for the expected value of a binomial distribution: Expected value (mean) = n * p = 40 students * 0.75 probability of passing = 30 students The expected number of students who will pass the course is 30.
02

Calculate the variance

Next, we use the formula for the variance of a binomial distribution: Variance = n * p * (1-p) = 40 students * 0.75 probability of passing * 0.25 probability of failing = 7.5 The variance is 7.5.
03

Calculate the standard deviation

Finally, we use the formula for the standard deviation, which is the square root of the variance: Standard deviation = sqrt(variance) = sqrt(7.5) 鈮� 2.74 The standard deviation is approximately 2.74 students. In conclusion, we expect 30 students to pass the course, with a variance of 7.5, and a standard deviation of approximately 2.74 students.

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

Find the probability that a person flipping a balanced coin requires four tosses to get a head.

\(\mathrm{Z}\) is a standard normal random variable. \(\mathrm{U}\) is chi-square with \(\mathrm{k}\) degrees of freedom. Assume \(\mathrm{Z}\) and \(\mathrm{U}\) are independent. Using the change of variable technique, find the distribution of \(\mathrm{X}=[\mathrm{Z} / \sqrt{(\mathrm{u} / \mathrm{k})}]\)

In testing a hypothesis concerned with the value of a population mean, first the level of significance to be used in the test is specified and then the regions of acceptance and rejection for evaluating the obtained sample mean are determined. If the 1 percent level of significance is used, indicate the percentages of sample means in each of the areas of the normal curve, assuming that the population hypothesis is correct, and the test is two-tailed.

Two continuous random variables \(\mathrm{X}\) and \(\mathrm{Y}\) may also be jointly distributed. Suppose \((X, Y)\) has a distribution which is uniform over a unit circle centered at \((0,0)\). Find the joint density of \((\mathrm{X}, \mathrm{Y})\) and the marginal densities of \(\mathrm{X}\) and \(\mathrm{Y}\). Are \(\mathrm{X}\) and \(\mathrm{Y}\) independent?

The State of New Hampshire conducts an annual lottery to raise funds for the school districts in the state. Assume a million tickets are sold. One ticket is the winning ticket and the winner receives \(\$ 10,000\). If each ticket costs \(\$ .25\), find the expected value of a randomly purchased ticket and the revenue that the lottery generates for the school districts in the state.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

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.