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Showers According to home-water-works.org, the average shower in the United States lasts \(8.2\) minutes. Assume this is correct, and assume the standard deviation of 2 minutes. a. Do you expect the shape of the distribution of shower lengths to be Normal, right-skewed, or left-skewed? Explain. b. Suppose that we survey a random sample of 100 people to find the length of their last shower. We calculate the mean length from the sample and record the value. We repeat this 500 times. What will be the shape of the distribution of these sample means? c. Refer to part b. What will be the mean and the standard deviation of the distribution of these sample means?

Short Answer

Expert verified
a. The shape is expected to be normal. b. The shape of the distribution of these sample means will be approximately normal due to the central limit theorem. c. The mean of these sample means will be 8.2 minutes, and the standard deviation will be 0.2 minutes.

Step by step solution

01

Determine the shape of the distribution

Given that the average shower length is 8.2 minutes with a standard deviation of 2 minutes, we expect the shape of the distribution of shower lengths to be normal. This assumption simplifies calculations. In reality, it might be slightly right-skewed as very long showers are more likely than very short ones.
02

Evaluate the shape of the distribution of sample means

The central limit theorem states that the distribution of sample means will approach normal distribution as the sample size increases, regardless of the shape of the original population distribution. In this case, we have a sample size of 100 which is sufficiently large for the theorem to hold. Therefore, regardless of the shape from Step 1, the distribution of these 500 sample means will be approximately normal.
03

Calculate the mean and standard deviation of the sample means

The mean of the sample means will be equal to the population mean, which is 8.2 minutes. The standard deviation of the sample means is calculated using the formula \(\sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}\) where \(\sigma_{\overline{x}}\) is the standard deviation of the sample mean, \(\sigma\) is the standard deviation of the population, and 'n' is the sample size. Here, using the given standard deviation of the population, \(2\) and the sample size, \(100\), we find \(\sigma_{\overline{x}} = \frac{2}{\sqrt{100}} = 0.2\). So, the standard deviation of sample means is 0.2 minutes.

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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 one of the most frequently occurring and important forms of data distribution. It is characterized by a symmetric, bell-shaped curve where most of the observations cluster around the central peak, and the probabilities for values taper off equally on both sides. This means that the likelihood of observing a value decreases the further away it is from the average. In many real-world situations, especially those involving biological, psychological, and economic phenomena, variables tend to follow a normal distribution.

For example, if shower lengths in the United States follow a normal distribution, then most people are likely to have shower lengths close to the average of 8.2 minutes. However, some people will have significantly shorter or longer showers, contributing to the tails of the distribution.
Sample Mean
The Sample Mean is a statistical term that refers to the average value calculated from a sample of observations taken from a larger population. To find the sample mean, you simply sum all the observations in your sample, and divide by the number of observations. This measure is central to many statistical methods and allows us to estimate population parameters.

In the context of our shower length exercise, if we survey a sample of 100 people, the sample mean is the average shower length reported by these 100 individuals. As we repeat the process 500 times, we record multiple sample means which help us understand the variability and reliability of our estimates of the population mean. Over repeated samples, these sample means tend to justify the application of the Central Limit Theorem.
Standard Deviation of Sample Means
The Standard Deviation of Sample Means is an important concept when assessing the variation in sample means derived from repeated sampling. It is also known as the Standard Error. This standard deviation is crucial because it informs us about the reliability of the sample mean as an estimate of the population mean.

Mathematically, it is given by the formula: \(\sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size.

In our exercise, with a population standard deviation of 2 minutes and a sample size of 100, the standard deviation of sample means turns out to be 0.2 minutes. This smaller standard deviation signifies a more precise estimate of the population mean from the sample means. As the sample size increases, the standard deviation of the sample mean decreases, thus providing a narrower range for the mean estimate.
Right-Skewed Distribution
A Right-Skewed Distribution, also known as positively skewed, is a distribution where the tail on the right side is longer or fatter than the left side. In simpler terms, most of the data falls to the left of the average with a few higher values stretching out on the right.

This type of distribution can occur when there is a natural limit on data measurements at the low end and more variability on the high end. For example, in the shower length context, very short showers are unlikely because they might not adequately serve their purpose, leading to most shower lengths clustering around the mean, with some exceptionally long showers skewing the data to the right.

Getting a proper understanding of skewness helps in applying appropriate statistical methods and in realistic interpretation of data, especially in estimating other quantities from the distribution.

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Most popular questions from this chapter

Ages A study of all the students at a small college showed a mean age of \(20.7\) and a standard deviation of \(2.5\) years. Are these numbers statistics or parameters? Explain. b. Label both numbers with their appropriate symbol (such as \(\bar{x}, \mu, s\), or \(\sigma)\).

Exam Scores The distribution of the scores on a certain exam is \(N(100,10)\) which means that the exam scores are Normally distributed with a mean of 100 and a standard deviation of \(10 .\) a. Sketch or use technology to create the curve and label on the \(x\) -axis the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score is between 90 and 110 . Shade the region under the Normal curve whose area corresponds to this probability.

Medical School Acceptance Rates The acceptance rate for a random sample of 15 medical schools in the United States was taken. The mean acceptance rate for this sample was \(5.77\) with a standard error of \(0.56 .\) Assume the distribution of acceptance rates is Normal. (Source: Accepted.com) a. Decide whether cach of the following statements is worded correctly for the confidence interval. Fill in the blanks for the correctly worded one(s). Explain the error for the ones that are incorrectly worded. i. We are \(95 \%\) confident that the sample mean is between and ii. We are \(95 \%\) confident that the population mean is between \(\quad\) and iii. There is a \(95 \%\) probability that the population mean is between and b. Based on your confidence interval, would you believe the average acceptance rate for medical schools is \(6.5\) ? Explain.

Babies Weights (Example 2) Some sources report that the weights of full-term newborn babies have a mean of 7 pounds and a standard deviation of \(0.6\) pound and are Normally distributed. a. What is the probability that one newborn baby will have a weight within \(0.6\) pound of the mean - that is, between \(6.4\) and \(7.6\) pounds, or within one standard deviation of the mean? b. What is the probability the average of four babies' weights will be within \(0.6\) pound of the mean - that is, between \(6.4\) and \(7.6\) pounds? c. Explain the difference between a and b.

Pulse Difference The following numbers are the differences in pulse rate (beats per minute) before and after running for 12 randomly selected people. $$ 24,12,14,12,16,10,0,4,13,42,4, \text { and } 16 $$ Positive numbers mean the pulse rate went up. Test the hypothesis that the mean difference in pulse rate was more than 0 , using a significance level of \(0.05 .\) Assume the population distribution is Normal.

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