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Chapter 8: Problem 16

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there is a weight limit of \(2500 \mathrm{lb}\). Assume that the average weight of students, faculty, and staff on campus is \(150 \mathrm{lb}\), that the standard deviation is \(27 \mathrm{lb}\), and that the distribution of weights of individuals on campus is approximately normal. If a random sample of 16 persons from the campus is to be taken: a. What is the expected value of the sample mean of their weights? b. What is the standard deviation of the sampling distribution of the sample mean weight? c. What average weights for a sample of 16 people will result in the total weight exceeding the weight limit of \(2500 \mathrm{lb} ?\) d. What is the chance that a random sample of 16 persons on the elevator will exceed the weight limit?

Short Answer

Expert verified
Hence the expected value of the sample mean is 150 lb with a standard deviation of 6.75 lb. The average weight for a sample of 16 should be less than or equal to 156.25 lb to avoid exceeding the weight limit of 2500 lb. The chance that 16 randomly chosen persons will exceed the weight limit is approximately 0.176 or 17.6%.

Step by step solution

01

Expected Value

The expected value of the sample mean is the same as the mean of the population. In this instance, it is given as 150 lb.
02

Standard Deviation

The standard deviation of the sample mean weights, also known as the standard error, is the standard deviation of the population divided by the square root of the sample size. The formula is: \(\sigma/\sqrt{n}\), where \(\sigma\) is the standard deviation of the population and \(n\) is the sample size. Substituting the given values, \(\sigma = 27, n = 16\), we get \((27/\sqrt{16}) = 6.75\) lb.
03

Average Weights

To find the average weight that would make a sample of 16 exceed 2500 lb, we divide the total weight by the sample size (16) which is \(2500 / 16 = 156.25\) lb.
04

Chance of Exceeding the weight limit

Firstly, find the Z score by subtracting the mean from the weight limit and then dividing by the standard error like so: \((156.25 - 150) / 6.75 = 0.93\) The Z score table shows that the proportion corresponding to this Z score is 0.8238. However, we are interested in the chance that the weight will exceed 2500 lb so we must take this away from 1, i.e., \(1 - 0.8238 = 0.1762\).

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Most popular questions from this chapter

Let \(x\) denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of \(x\) are \(\mu=2 \mathrm{~min}\) and \(\sigma=0.8\) min, respectively. a. If \(\bar{x}\) is the sample average time for a random sample of \(n=9\) students, where is the \(\bar{x}\) distribution centered, and how much does it spread out about the center (as described by its standard deviation)? b. Repeat Part (a) for a sample of size of \(n=20\) and again for a sample of size \(n=100\). How do the centers and spreads of the three \(\bar{x}\) distributions compare to one another? Which sample size would be most likely to result in an \(\bar{x}\) value close to \(\mu\), and why?

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Explain the difference between \(\sigma\) and \(\sigma_{\bar{x}}\) and between \(\mu\) and \(\mu_{\bar{x}}\)

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