Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Suppose that the length of research papers is uniformly distributed from ten to \(25\) pages. We survey a class in which \(55\) research papers were turned in to a professor. The \(55\) research papers are considered a random collection of all papers. We are interested in the average length of the research papers.

a. In words, \(X= \)

b. \(X= ( , )\)

c. \(\mu_{x}= \)

Short Answer

Expert verified

Part a. \(X=\) length of the research papers.

Part b. \(X=U(10,25)\)

Part c. \(\mu_{x}=17.5\)

Step by step solution

01

Part a. Step 1. Given information

The length of the research papers is uniformly distributed from \(10\) to \(25\) pages. We survey a class in which \(55\) research papers were turned in to a professor. The \(55\) research papers are considered a random collection of all papers. We are interested in average length of research papers.

02

Part a. Step 2. Calculation

The random variable \(X\) is defined by

\(X=\) length of the research papers

Hence, the random variable \(X\) is length of the research papers.

03

Part b. Step 1. Calculation

The distribution of \(X\) is uniform.

The length of the research papers is uniformly distributed from \(10\) to \(2\) pages.

\(a = 10\)

\(b = 25\)

Therefore, \(X=U(a,b)=X=U(10,25)\)

\(X=U(10,25)\)

04

Part c. Step 1. Calculation

\(X=U(10,25)\)

Mean, \(\mu_{x}=\frac{a+b}{2}\)

\(=\frac{10+25}{2}\)

\(=\frac{35}{2}\)

\(=17.5\)

Hence, \(\mu_{x}=17.5\)

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 Vaia!

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

According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040is 10.53hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36taxpayers.

a. In words, ฮง=_____________

b. In words,X=_____________

c. Xยฏ~_____(_____,_____)

d. Would you be surprised if the 36taxpayers finished their Form 1040s in an average of more than 12hours? Explain why or why not in complete sentences.

e. Would you be surprised if one taxpayer finished his or her Form 1040in more than 12hours? In a complete sentence, explain why.

The distribution of income in some Third World countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge shaped distribution. Let the average salary be \(2,000per year with a standard deviation of \)8,000. We randomly survey 1,000residents of that country.

a. In words,ฮง=_____________

b. In words,X=_____________

c.Xยฏ~_____(_____,_____)

d. How is it possible for the standard deviation to be greater than the average?

e. Why is it more likely that the average of the 1,000residents will be from \(2,000to \)2,100than from \(2,100to\)2,200?

A typical adult has an average IQ score of 105with a standard deviation of 20. If 20randomly selected adults are given an IQ test, what are the probability that the sample mean scores will be between 85 and 125 points?

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

Draw the graph from Exercise 7.39

For the sums of distribution to approach a normal distribution, what must be true?

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.

Sign-up for free