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According to home-water-works.org, the average shower in the United States lasts \(8.2\) minutes. Assume the standard deviation of shower times is 2 minutes and the distribution of shower times is right-skewed. Which of the following questions can be answered using the Central Limit Theorem for sample means as needed? If the question can be answered, do so. If the question cannot be answered, explain why the Central Limit Theorem cannot be applied. a. Find the probability that a randomly selected shower lasts more than 9 minutes. b. If five showers are randomly selected, find the probability that the mean length of the sample is more than 9 minutes. c. If 50 showers are randomly selected, find the probability that the mean length of the sample is more than 9 minutes.

Short Answer

Expert verified
a. We cannot apply the Central Limit Theorem. b. We should not apply the Central Limit Theorem due to small sample size and skewness of data. c. Yes, we can apply the Central Limit Theorem due to the large sample size and can find the probability after calculating the Z-score.

Step by step solution

01

Answering part a

As mentioned, the Central Limit Theorem is not applicable for a single instance (or a very small sample). Therefore, we can't answer this question using the Central Limit Theorem.
02

Answering part b

Here, we have a sample size of 5. The Central Limit Theorem is usually applied for larger sample sizes (around 30 or more), so we cannot use it reliably here. However, for the sake of argument, if we applied the theorem, we'd find a Z-score by subtracting the mean from 9, then dividing by the standard deviation divided by the square root of the sample size. The Z-score = (9-8.2) / (2/√5). Once we have the Z-score, we could find the probability using a standard normal table. But considering the small sample size and skewness of data, this isn't appropriate here.
03

Answering part c

In this case, the sample size is 50, which is large enough for us to apply the Central Limit Theorem despite the right skewness. We find the Z-score as above, substituting 50 for the sample size. Then we use a standard normal table or Z-table to find the probability that the mean length of a sample of 50 showers is more than 9 minutes.

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Key Concepts

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

Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes for an experiment. It's a way to describe how likely it is that any given outcome will occur. For example, the probability distribution of shower times could tell us what percentage of showers are expected to last between, say, 6 to 8 minutes, or more than 9 minutes. When data is right-skewed, as in our exercise, it means that most of the data points are clustered toward the left side with fewer large values trailing off to the right. This affects our calculations because standard probability distributions assume a normal (bell-shaped) distribution which is not the case with skewed data. However, the Central Limit Theorem comes to our aid when working with sample means.When considering improvements in understanding this concept, it’s important to recognize that not all probability distributions are the same. The Central Limit Theorem specifically deals with the distribution of sample means, which tends to become normal as the sample size grows, even if the original data is skewed.
Sample Means
Sample means play a crucial role in the application of the Central Limit Theorem. A sample mean is simply the average of a set of observations. When you take numerous samples from a population and calculate their means, you get a distribution of sample means. This distribution tends to resemble a normal distribution as the number of samples increases, which is at the heart of the Central Limit Theorem. It's essential to understand that the theorem does not apply to just one instance or a small sample size, as was shown in part a and b of our exercise. Larger samples, as mentioned, should be 30 or more to yield a commendable approximation to a normal distribution of means, even if the original population distribution is not normal.
Standard Deviation
The standard deviation is a statistic that measures the dispersion of a dataset relative to its mean. It calculates how much variation or 'spread' exists from the average (mean), or expected value. A low standard deviation indicates that the data points tend to be close to the mean; conversely, a high standard deviation indicates that the data points are spread out over a wider range of values. In the context of our exercise, the standard deviation of shower times is given as 2 minutes. To utilize this in the Central Limit Theorem for sample means, we would adjust this value depending on the number of observations in the sample by dividing it by the square root of the sample size, which gives us the standard error of the mean—a critical value for finding the Z-score.
Z-score
The Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations away from the mean. To calculate a Z-score in a Central Limit Theorem context, we subtract the population mean from the sample mean and then divide this difference by the standard error. The Z-score is a pivotal part of understanding probability regarding sample means. As illustrated in the solution to part c of our exercise, with a large enough sample size, we can calculate the Z-score and thereby find the probability associated with a sample mean being greater than a certain value, even with a non-normal distribution such as a right-skewed one. For optimization, noting the transformation from a possibly skewed distribution to a more normal distribution via the Central Limit Theorem helps students grasp why larger sample sizes matter when calculating probabilities using the Z-score.

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Most popular questions from this chapter

In exercise \(9.31\), two intervals were given for the same data, one for \(95 \%\) confidence and one for \(90 \%\) confidence. a. How would a \(99 \%\) interval compare? Would it be narrower than both. wider than both, or between the two in width. Explain. b. If we wanted to use a \(99 \%\) confidence level and get a narrower width. how could we change our data collection?

State whether each of the following changes would make a confidence interval wider or narrower. (Assume that nothing else changes.) a. Changing from a \(95 \%\) level of confidence to a \(90 \%\) level of confidence b. Changing from a sample size of 30 to a sample size of 20 c. Changing from a standard deviation of 3 inches to a standard deviation of \(2.5\) inches

A random sample of 10 independent healthy people showed the following body temperatures (in degrees Fahrenheit): $$98.5,98.299 .0,96.3,98.3,98.7,97.2,99.1,98.7,97.2$$ Test the hypothesis that the population mean is not \(98.6^{\circ} \mathrm{F}\), using a significance level of \(0.05 .\) See page 500 for guidance.

In the United States, the population mean height for 10 -year-old girls is \(54.5\) inches. Suppose a random sample of 1510 -year-old girls from Brazil is taken and that these girls had a sample mean height of \(53.2\) inches with a standard deviation of \(2.5\) inches. Assume that heights are Normally distributed. (Source: cdc.gov) a. Determine whether the population mean for height for 10 -year-old girls from Brazil is significantly different from the U.S. population mean. Use a significance level of \(0.05\). b. Now suppose the sample consists of 40 girls instead of \(15 .\) Repeat the test. c. Explain why the \(t\) -values and \(\mathrm{p}\) -value for parts \(\mathrm{a}\) and \(\mathrm{b}\) are different.

In the 2015 AFC Championship game, there was a charge the New England Patriots deflated their footballs for an advantage. The balls should be inflated to between \(12.5\) and \(13.5\) pounds per square inch. The measurements were \(11.50,10.85,11.15,10.70,11.10,11.60,11.85,11.10,10.95\), \(10.50\), and \(10.90\) psi (pounds per square inch). (Source: http:// online.wsj.com/public/resources/documents/Deflategate.pdf) a. Test the hypothesis that the population mean is less than \(12.5\) psi using a significance level of \(0.05 .\) State clearly whether the Patriots' balls are deflated or not. Assume the conditions for a hypothesis test are satisfied. b. Use the data to construct a \(95 \%\) confidence interval for the mean psi for the Patriots' footballs. How does this confidence interval support your conclusion in part a?

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