91Ó°ÊÓ

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?

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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value

In the context of sampling, the expected value, often referred to as the mean, is a key concept in helping us understand what to anticipate on average from our data. When dealing with a sample, the expected value of the sample mean is actually equal to the mean of the entire population. This is a fundamental principle.

• For example, if you have a population of campus individuals with an average weight of 150 pounds, then the expected value or mean of any sample taken from this population will also be 150 pounds.
• This consistency shows us that despite variation among individuals in the population, our samples tend towards the population mean.

Understanding expected value is essential because it creates a benchmark for comparing actual outcomes from samples.

Standard Error

The standard error is a measure that indicates the amount of variability in the sampling distribution of the sample mean. It's an important concept in statistics because it helps us understand how much the sample mean might differ from the population mean.

• The formula to calculate the standard error ( ext{SE}) of the sample mean is: \[ SE = \frac{\sigma}{\sqrt{n}} \]where \(\sigma\) represents the standard deviation of the population and \(n\) is the size of the sample.
• For our example, with a population standard deviation of 27 pounds and a sample size of 16, the standard error ends up being 6.75 pounds. This means the sample mean will typically vary from the population mean by 6.75 pounds.

Knowing the standard error is crucial because it provides insight into the reliability of the sample mean as an estimate of the population mean.

Normal Distribution

Normal distribution is a foundational concept in statistics, characterized by its familiar bell-shaped curve. It describes how the values of a variable are distributed. Many natural phenomena, including human characteristics, follow this pattern, which is symmetrically distributed around the mean.

• In the given exercise, it's mentioned that individual weights are approximately normally distributed. This assumption allows us to use standard statistical methods to make predictions about these weights.
• Normal distribution implies that most data points are concentrated around the mean, with fewer and fewer points farther away, known as the tails.

The normal distribution assumption simplifies statistical calculations, making it easier to apply the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normal if the sample size is large enough.

Z-score

A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It's expressed in terms of standard deviations from the mean.

• To calculate a Z-score, you subtract the mean from your value of interest and then divide by the standard deviation of the distribution. \[ Z = \frac{(X - \mu)}{SE} \]where \(X\) is the value of interest, \(\mu\) is the mean, and \(SE\) is the standard error.
• In our elevator example, the Z-score calculation helps determine how likely it is that a sample mean weight would exceed the elevator's 2500-pound limit.

Z-scores are vital in statistics as they allow us to understand the standard deviation distance of a data point or sample mean from the mean, providing a basis for hypothesis testing and probability determination.