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A sign in the elevator of a college library indicates a limit of 16 persons. In addition, there is a weight limit of 2500 pounds. Assume that the average weight of students, faculty, and staff at this college is 150 pounds, that the standard deviation is 27 pounds, and that the distribution of weights of individuals on campus is approximately normal. A random sample of 16 persons from the campus will be selected. a. What is the mean of the sampling distribution of \(\bar{x} ?\) b. What is the standard deviation of the sampling distribution of \(\bar{x} ?\) c. What average weights for a sample of 16 people will result in the total weight exceeding the weight limit of 2500 pounds? d. What is the probability that a random sample of 16 people will exceed the weight limit?

Short Answer

Expert verified
a. The mean of the sampling distribution of \(\bar{x}\) is 150 pounds. b. The standard deviation of the sampling distribution of \(\bar{x}\) is 6.75 pounds. c. If the average weight of a sample of 16 people exceeds 156.25 pounds, their total weight will exceed 2500 pounds. d. The probability that a random sample of 16 people will exceed the weight limit of 2500 pounds is approximately 17.7%.

Step by step solution

01

a. Mean of the sampling distribution of \(\bar{x}\)

The mean of the sampling distribution of the sample mean is equal to the mean of the population. In this case, the mean of the weight distribution of individuals on campus, which is 150 pounds. Therefore, the mean of the sampling distribution of \(\bar{x}\) is 150 pounds.
02

b. Standard deviation of the sampling distribution of \(\bar{x}\)

To calculate the standard deviation of the sampling distribution of the sample mean, we will use the following formula: \[ \frac{\text{population standard deviation}}{\sqrt{n}} \] Given that the population standard deviation is 27 pounds and the sample size is 16, the standard deviation of the sampling distribution of \(\bar{x}\) is: \[ \frac{27}{\sqrt{16}} = \frac{27}{4} = 6.75\; \text{pounds} \]
03

c. Average weights for a sample of 16 people exceeding the weight limit

To find the average weight for a sample that will result in a total weight exceeding 2500 pounds, we will divide the weight limit by the sample size: \[ \frac{2500}{16} = 156.25\; \text{pounds} \] Thus, if the average weight of a sample of 16 people exceeds 156.25 pounds, their total weight will exceed 2500 pounds.
04

d. Probability that a random sample of 16 people will exceed the weight limit

To calculate the probability that a random sample of 16 people will exceed the weight limit of 2500 pounds, we need to find the z-score corresponding to 156.25 pounds, using the mean and standard deviation of the sampling distribution of \(\bar{x}\), and then consult the standard normal distribution table: Z-score formula: \[ z = \frac{x - \mu}{\sigma} \] Z-score for 156.25 pounds: \[ z = \frac{156.25 - 150}{6.75} \approx 0.926 \] Using the standard normal distribution table, the probability \(P(Z > 0.926)\) is approximately 0.177, or 17.7%. Therefore, the probability that a random sample of 16 people will exceed the weight limit of 2500 pounds is approximately 17.7%.

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Key Concepts

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

Mean of the Sampling Distribution
Understanding the mean of the sampling distribution is fundamental in statistics. It represents the average of the sample means that would be obtained if we were to take all possible samples of a certain size from a population. In the context of the elevator weight limit problem, the mean of the sampling distribution symbolizes the expected average weight of a group of 16 people.

As seen in the exercise, the mean of the sampling distribution of \(\bar{x}\) is equal to the population mean, which here is 150 pounds. This indicates that if we repeatedly took samples of 16 people from the campus population, the mean weight of these samples would average out to 150 pounds over the long run. Although individual sample means may vary, the central point of their distribution will always be the population mean.
Standard Deviation of the Sampling Distribution
When it comes to the standard deviation of the sampling distribution, also known as the standard error, it measures how much variability exists within the means of various samples taken from the same population. A lower standard error suggests that the sample means are tightly clustered around the population mean.

The formula to determine the standard deviation of the sampling distribution is the population standard deviation divided by the square root of the sample size (\(n\)). In our exercise, since the population standard deviation is 27 pounds and our sample size is 16, we find a standard deviation of the sampling distribution to be 6.75 pounds. This value helps us understand the expected variation in average weights we would encounter from different samples of 16 people.
Normal Distribution
A normal distribution, which is sometimes referred to as the bell curve due to its shape, describes a distribution where most of the data points cluster around the mean, with fewer and fewer appearing as you move away from the center.

In practice, many variables are normally distributed, which allows statisticians to make inferences about populations based on sample data. The problem at hand assumes that the weights are normally distributed across the campus population. This normality ensures that the sampling distribution of the sample mean is also normally distributed if the sample size is sufficiently large, due to the central limit theorem. The mean and standard deviation of the sampling distribution help us to depict this curve for sample means and predict probabilities associated with the weights.
Probability Calculation
Probability calculation is integral to statistics as it enables the quantification of uncertainty. It involves determining the chance of a particular outcome occurring within a set of possible outcomes.

In the context of our elevator example, the probability calculation was used to find the likelihood that the sample of 16 people would exceed the weight limit of 2500 pounds. By utilizing the Z-score, we were able to normalize our question about the weight limit into a question about standard deviations from the mean. After calculating the Z-score to be approximately 0.926, we then interpret this value using the standard normal distribution table. The resulting probability of approximately 17.7% tells us there’s about a 1 in 5 chance that a random sample of 16 individuals from the campus will have a combined weight exceeding the elevator's limit.

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Most popular questions from this chapter

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(n=16, t=1.6\) b. Upper-tailed test, \(n=14, t=3.2\) c. Lower-tailed test, \(n=20, t=-5.1\) d. Two-tailed test, \(n=16, t=6.3\)

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