Logout Open In App
Open In App
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

Chapter 8: Problem 14

The time that a randomly selected individual waits for an elevator in an office building has a uniform distribution over the interval from 0 to \(1 \mathrm{~min}\). It can be shown that for this distribution \(\mu=0.5\) and \(\sigma=0.289\). a. Let \(\bar{x}\) be the sample average waiting time for a random sample of 16 individuals. What are the mean and standard deviation of the sampling distribution of \(\bar{x}\) ? b. Answer Part (a) for a random sample of 50 individuals. In this case, sketch a picture of a good approximation to the actual \(\bar{x}\) distribution.

Short Answer

Expert verified
a) For a sample of 16 individuals, the mean is 0.5 min and the standard deviation is 0.07225 min. b) For a sample of 50 individuals, the mean is 0.5 min and the standard deviation is 0.04086 min. The sampling distribution approximates a normal distribution centered at 0.5 min with a standard deviation of 0.04086 min.

Step by step solution

01

Find the Mean of the Sampling Distribution

The mean of the sampling distribution of the sample mean is always equal to the mean of the population. Thus, for both samples of 16 and 50 individuals, the mean of the sampling distribution of \(\bar{x}\) will be the same as the population mean, \(\mu = 0.5\) min.
02

Find the Standard Deviation of the Sampling Distribution for a Sample of 16 Individuals

The standard deviation of the sampling distribution of \(\bar{x}\) is equal to the population standard deviation \(\sigma\) divided by the square root of the sample size \(n\). Therefore, for a sample of 16 individuals, the standard deviation of the sampling distribution of \(\bar{x}\) is \(\sigma / \sqrt{n} = 0.289 / \sqrt{16} = 0.07225\) min.
03

Find the Standard Deviation of the Sampling Distribution for a Sample of 50 Individuals

Again use the formula \(\sigma / \sqrt{n}\), but this time with a sample size of 50. The standard deviation of the sampling distribution of \(\bar{x}\) for a sample of 50 individuals is \(\sigma / \sqrt{n} = 0.289 / \sqrt{50} = 0.04086\) min.
04

Sketch the Sampling Distribution for a Sample of 50 Individuals

The sampling distribution for a sample of 50 individuals approximates a normal distribution because the sample size is large. The distribution is centered around the mean of 0.5 min with a standard deviation of 0.04086 min. The graph should be bell-shaped.

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.

Start your free trial

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

A manufacturing process is designed to produce bolts with a 0.5-in. diameter. Once each day, a random sample of 36 bolts is selected and the diameters recorded. If the resulting sample mean is less than \(0.49\) in. or greater than \(0.51\) in., the process is shut down for adjustment. The standard deviation for diameter is \(0.02\) in. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an \(\bar{x}\) in the shutdown range when the true process mean really is \(0.5\) in.)

A random sample is selected from a population with mean \(\mu=100\) and standard deviation \(\sigma=10 .\) Determine the mean and standard deviation of the \(\bar{x}\) sampling distribution for each of the following sample sizes: a. \(n=9\) b. \(n=15\) c. \(n=36\) d. \(n=50\) c. \(n=100\) f. \(n=400\)

Consider a population consisting of the following five values, which represent the number of video rentals during the academic year for each of five housemates: \(\begin{array}{lll} 8 & 14&16 & 10 & 11\end{array}\) a. Compute the mean of this population. b. Select a random sample of size 2 by writing the numbers on slips of paper, mixing them, and then selecting \(2 .\) Compute the mean of your sample. c. Repeatedly select samples of size 2 , and compute the \(\bar{x}\) value for each sample until you have the results of 25 samples. d. Construct a density histogram using the \(25 \bar{x}\) values. Are most of the \(\bar{x}\) values near the population mean? Do the \(\bar{x}\) values differ a lot from sample to sample, or do they tend to be similar?

The article "Thrillers" (Newsweek, April 22,1985 ) stated, "Surveys tell us that more than half of America's college graduates are avid readers of mystery novels." Let \(\pi\) denote the actual proportion of college graduates who are avid readers of mystery novels. Consider a sample proportion \(p\) that is based on a random sample of 225 college graduates. a. If \(\pi=.5\), what are the mean value and standard deviation of \(p ?\) Answer this question for \(\pi=.6\). Does \(p\) have approximately a normal distribution in both cases? Explain. b. Calculate \(P(p \geq .6)\) for both \(\pi=.5\) and \(\pi=.6\). c. Without doing any calculations, how do you think the probabilities in Part (b) would change if \(n\) were 400 rather than \(225 ?\)

Suppose that a particular candidate for public office is in fact favored by \(48 \%\) of all registered voters in the district. A polling organization will take a random sample of 500 voters and will use \(p\), the sample proportion, to estimate \(\pi\). What is the approximate probability that \(p\) will be greater than .5, causing the polling organization to incorrectly predict the result of the upcoming election?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

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.

Sign-up for free