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Assume women's heights are approximately Normally distributed with a mean of 65 inches and a standard deviation of \(2.5\) inches. 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 woman is less than 63 inches tall. b. If five women are randomly selected, find the probability that the mean height of the sample is less than 63 inches. c. If 30 women are randomly selected, find the probability that the mean height of the sample is less than 63 inches.

Short Answer

Expert verified
The answers to the problems are: a. The probability that a randomly selected woman is less than 63 inches tall is the corresponding probability for Z equals -0.8 from the Z-table. b. The probability for a sample of 5 women is equal to the corresponding probability for Z equals -1.79. c. Finally, for a sample size of 30, the probability is the corresponding value for Z is -4.35 from the Z-table.

Step by step solution

01

- Solving Part a

The first part asks for the probability that a randomly selected woman is less than 63 inches tall. On the normal distribution curve, we find the normalized deviation for a given measurement (the Z-score), which is calculated by subtracting the mean from the individual value and dividing by the standard deviation. In this case, \(Z = (63-65)/2.5 = -0.8\). Once the Z-score is calculated, we can find the respective probability from the Z-table as any value that corresponds to Z equals -0.8.
02

- Solving Part b

The second part asks for the probability that the mean height of five randomly selected women is less than 63 inches. Here, the Central Limit Theorem doesn't apply as the sample size is small (less than 30). Nonetheless, we can calculate the probability using the standard error (SE) instead of the standard deviation. SE is calculated by dividing the standard deviation by the square root of the sample size. So, the new standard deviation becomes \(2.5/\sqrt{5} = 1.12\). Then, we find the Z-score by \(Z = (63-65)/1.12 = -1.79\). The respective probability can be found on the Z-table.
03

- Solving Part c

The final part c asks about the sample mean height of 30 randomly selected women. The Central Limit Theorem comes into play here since our sample size is large (over 30). Again, we calculate the SE as \(2.5/\sqrt{30} = 0.46\), and find the Z-score by \(Z = (63-65)/0.46 = -4.35\). The respective probability is found by referring to the Z-table.

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Key Concepts

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

Normal distribution
A normal distribution is a continuous probability distribution characterized by its bell-shaped curve, which is symmetric around the mean. Most values cluster around a central region, with probabilities tapering off as you move further from the mean. Key features of a normal distribution include:
  • Mean: The average value of the dataset, situated at the center of the distribution.
  • Standard deviation: A measure of the data's spread, indicating how much the values deviate from the mean.
A classic example is women's heights, which can be modeled by a normal distribution with specific mean and standard deviation values, such as 65 inches for the mean and 2.5 inches for the standard deviation as given in the problem.
This allows for the calculation of probabilities for given heights using the Z-scores.
Z-score
The Z-score is a critical component when working with normal distributions. It measures how many standard deviations a particular value is from the mean. This score offers a standard way to compare different data points or groups.
Calculating a Z-score is straightforward using the formula:
  • Subtract the mean from the value of interest.
  • Divide the result by the standard deviation.
For instance, in the example provided, calculating the Z-score for a height of 63 inches is done by \[Z = \frac{63 - 65}{2.5} = -0.8 \]This score indicates that 63 inches is 0.8 standard deviations below the mean height of 65 inches.
The Z-score helps in determining probabilities by referring to standard Z-tables, which give the probability that a statistic is less than a given Z-score.
Standard error
The standard error (SE) is an essential concept when dealing with sample means. It gives a measure of the variability of a sampling distribution or how much the sample mean is expected to fluctuate from the true population mean. Standard error is calculated using:
  • Standard deviation divided by the square root of the sample size.
An important consideration in statistics, especially under the Central Limit Theorem, is that as the sample size increases, the sample mean will approximate the population mean more closely, reducing the standard error.
For example, the original problem involves calculating the SE when considering a sample of 30 women:\[SE = \frac{2.5}{\sqrt{30}} \approx 0.46\]This smaller standard error compared to smaller sample sizes (e.g., 5 women) signifies greater reliability that the sample mean approximates the population mean.
Probability
Probability is the measure of how likely an event is to occur, ranging from 0 (impossible) to 1 (certainly). In the context of a normal distribution, probabilities help determine the likelihood of a random variable falling within a certain range.
  • The total area under the normal distribution curve equals 1.
  • The area to the left or right of a Z-score represents cumulative probability.
Using Z-scores and Z-tables, we can find probabilities for specified conditions. For example, the probability of randomly picking a woman who is shorter than 63 inches, with a Z-score of -0.8, can be found in the Z-table. This process aids in answering questions like those in the original exercise, for both individual cases and means of samples, helping us understand statistical behaviors and outcomes under normal distribution.

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Most popular questions from this chapter

Several times during the year, the U.S. Census Bureau takes random samples from the population. One such survey is the American Community Survey. The most recent such survey, based on a large (several thousand) sample of randomly selected households, estimates the mean retirement income in the United States to be \(\$ 21,201\) per year. Suppose we were to make a histogram of all of the retirement incomes from this sample. Would the histogram be a display of the population distribution, the distribution of a sample, or the sampling distribution of means?

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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State whether each situation has independent or paired (dependent) samples. a. A researcher wants to know whether pulse rates of people go down after brief meditation. She collects the pulse rates of a random sample of people before meditation and then collects their pulse rates after meditation. b. A researcher wants to know whether professors with tenure have fewer posted office hours than professors without tenure do. She observes the number of office hours posted on the doors of tenured and untenured professors.

According to a 2018 Money magazine article, the average income in Kansas is $$\$ 53,906$$. Suppose the standard deviation is $$\$ 3000$$ and the distribution of income is rightskewed. Repeated random samples of 400 Kansas residents are taken, and the sample mean of incomes is calculated for each sample. a. The population distribution is right-skewed. Will the distribution of sample means be Normal? Why or why not? b. Find and interpret a \(z\) -score that corresponds with a sample mean of $$\$ 53,606 .$$ c. Would it be unusual to find a sample mean of $$\$ 54,500 ?$$ Why or why not?

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