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At the downtown office of First National Bank there are five tellers. Last week the tellers made the following number of errors each: \(2,3,5,3,\) and \(5 .\) a. How many different samples of 2 tellers are possible? b. List all possible samples of size 2 and compute the mean of each. c. Compute the mean of the sample means and compare it to the population mean.

Short Answer

Expert verified
a. 10 samples; b. Means: 2.5, 3.5, 2.5, 3.5, 4, 3, 4, 4, 5, 4; c. Both means are 3.6.

Step by step solution

01

Calculate Possible Combinations for Samples

To determine how many different samples of 2 tellers are possible, use the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of tellers (5 in this case) and \( k \) is the size of the sample (2). Thus, \( C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \). There are 10 different samples of 2 tellers possible.
02

List All Possible Samples of Size 2

The possible pairs (samples) of 2 tellers from the set \( \{2,3,5,3,5\} \) are: (2,3), (2,5), (2,3), (2,5), (3,5), (3,3), (3,5), (5,3), (5,5), (3,5).
03

Compute the Mean of Each Sample

Calculate the mean for each pair: - (2,3): \( \frac{2+3}{2} = 2.5 \)- (2,5): \( \frac{2+5}{2} = 3.5 \)- (2,3): \( \frac{2+3}{2} = 2.5 \)- (2,5): \( \frac{2+5}{2} = 3.5 \)- (3,5): \( \frac{3+5}{2} = 4 \)- (3,3): \( \frac{3+3}{2} = 3 \)- (3,5): \( \frac{3+5}{2} = 4 \)- (5,3): \( \frac{5+3}{2} = 4 \)- (5,5): \( \frac{5+5}{2} = 5 \)- (3,5): \( \frac{3+5}{2} = 4 \).The means of these samples are: 2.5, 3.5, 2.5, 3.5, 4, 3, 4, 4, 5, 4.
04

Compute the Mean of the Sample Means

The mean of the sample means is calculated by averaging the list of means obtained: \( \frac{2.5 + 3.5 + 2.5 + 3.5 + 4 + 3 + 4 + 4 + 5 + 4}{10} = \frac{36}{10} = 3.6 \).
05

Calculate the Population Mean

The population mean is calculated by averaging the errors made by all five tellers: \( \frac{2+3+5+3+5}{5} = \frac{18}{5} = 3.6 \).
06

Compare the Mean of the Sample Means to the Population Mean

The mean of the sample means (3.6) is equal to the population mean (3.6), demonstrating how the sample means can represent the population mean accurately.

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Key Concepts

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

Combination Formula
When you need to find how many different groups you can form from a larger set, you use the combination formula. This formula is useful because it provides a way to count options without actually listing them all. The formula to calculate the number of combinations is:
  • \( C(n, k) = \frac{n!}{k!(n-k)!} \)
Here, \( n \) represents the total number of items in the set, and \( k \) represents the number of items to choose. The exclamation mark (!) signifies a factorial, which is the product of all positive integers up to that number.
For example, in the exercise, you want to know how many ways you can choose 2 tellers from 5. By using \( C(5, 2) \), you calculate that there are 10 different possible combinations, which represent the different "samples" of 2 tellers.
Population Mean
The population mean is a measure of central tendency that tells you the average value of an entire set of items. It's the sum of all values in the population divided by the number of values.
In the exercise, the population is the number of errors made by 5 tellers: 2, 3, 5, 3, and 5. To find the population mean, add these values together and divide by 5.
  • Population Mean \( = \frac{2 + 3 + 5 + 3 + 5}{5} \)
  • Which simplifies to \( \frac{18}{5} = 3.6 \).
The population mean of 3.6 represents the average number of errors made by the tellers. By comparing this mean to other calculations, you understand how well sample statistics estimate the population values.
Statistical Samples
A statistical sample is a subset of items from a larger set, known as a population. By studying samples, we can make inferences or generalizations about the entire population. In this exercise, we're considering all possible pairs of tellers as samples.
Listing all possible samples of size 2 from the tellers results in groups such as (2,3), (2,5), and so on. Calculating the mean for each of these pairs gives us insight into the characteristics of the population. Sampling in this way allows us to choose a manageable number of observations to analyze, instead of attempting to study every detail of the population.
By using each sample, you compute the sample mean, which is the average of the values in a specific sample. These sample means can be used to approximate or estimate the population mean.
Mean Comparison
Mean comparison involves analyzing the relationship between the mean of sample means and the population mean. It provides a way to assess the accuracy and reliability of sample statistics as estimators of population parameters.
In this exercise, after computing the means for all possible pairs, you find the mean of these sample means:
  • Mean of Sample Means \( = \frac{36}{10} = 3.6 \).
This result is then compared to the population mean of 3.6. The fact that they are equal demonstrates the power of statistical sampling: even though it's often impractical to measure an entire population, well-constructed samples can offer precise estimates.

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Most popular questions from this chapter

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is \(\$ 110,000 .\) This distribution is positively skewed. The standard deviation of the population is not known. a. A random sample of 50 households revealed a mean of \(\$ 112,000\) and a standard deviation of \(\$ 40,000 .\) What is the standard error of the mean? b. Suppose that you selected 50 samples of households. What is the expected shape of the distribution of the sample mean? c. What is the likelihood of selecting a sample with a mean of at least \(\$ 112,000 ?\) d. What is the likelihood of selecting a sample with a mean of more than \(\$ 100,000 ?\) e. Find the likelihood of selecting a sample with a mean of more than \(\$ 100,000\) but less than \(\$ 112,000\)

The mean SAT score for Division I student-athletes is 947 with a standard deviation of \(205 .\) If you select a random sample of 60 of these students, what is the probability the mean is below \(900 ?\)

Recent studies indicate that the typical 50 -year-old woman spends \(\$ 350\) per year for personal-care products. The distribution of the amounts spent is positively skewed. We select a random sample of 40 women. The mean amount spent for those sampled is \(\$ 335\), and the standard deviation of the sample is \(\$ 45 .\) What is the likelihood of finding a sample mean this large or larger from the specified population?

The mean rent for a one-bedroom apartment in Southern California is \(\$ 2,200\) per month. The distribution of the monthly costs does not follow the normal distribution. In fact, it is positively skewed. What is the probability of selecting a sample of 50 one-bedroom apartments and finding the mean to be at least \(\$ 1,950\) per month? The standard deviation of the sample is \(\$ 250 .\)

The Crossett Trucking Company claims that the mean weight of their delivery trucks when they are fully loaded is 6,000 pounds and the standard deviation is 150 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weiqhed. Within what limits will 95 percent of the sample means occur?

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