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Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5 . a. What are the mean and standard deviation of the \(\bar{x}\) sampling distribution? Describe the shape of the \(\bar{x}\) sampling distribution. b. What is the approximate probability that \(\bar{x}\) will be within \(0.5\) of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

Short Answer

Expert verified
The mean of the \(\bar{x}\) sample distribution is 40, and the standard deviation is 0.625. The distribution is approximately normal. The probability that \(\bar{x}\) is within 0.5 of \(\mu\) is approximately 0.5764, and the probability that \(\bar{x}\) differs by more than 0.7 from \(\mu\) is approximately 0.2628.

Step by step solution

01

Find the Mean of the Sample Mean

The mean of the sample mean (\(\bar{x}\)) is equal to the population mean (\(\mu\)). So, the mean of the \(\bar{x}\) sampling distribution is 40.
02

Find the Standard Deviation of the Sample Mean

The standard deviation of the sample mean (\(\bar{x}\)) can be calculated by dividing the population standard deviation (\(\sigma\)) by the square root of the sample size (n). So, we have \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{64}} = \frac{5}{8} = 0.625\).
03

Describe the Shape

According to the Central Limit Theorem, if the sample size is large (generally \(n\geq30\)), the sampling distribution of the mean will be approximately normally distributed, regardless of the shape of the population distribution. Since our sample size is 64, an amount greater than 30, the shape of the \(\bar{x}\) sampling distribution should be approximately normal.
04

Find the Probability that \(\bar{x}\) is within 0.5 of the population mean

We will calculate the z-score for the values 40.5 and 39.5 as this range is within \(\pm 0.5\) of the mean 40. The z-score is calculated by subtracting the population mean from the sample mean, then dividing by the standard deviation of the sample mean (\(\sigma_{\bar{x}}\)). Hence, the z-scores are \(z_1 = \frac{39.5 - 40}{0.625} = -0.8\) and \(z_2 = \frac{40.5 - 40}{0.625} = 0.8\). The standard normal table or a software calculator gives that \(P(-0.8 \leq z \leq 0.8)\) is approximately 0.5764.
05

Find the Probability that \(\bar{x}\) will differ from \(\mu\) by more than 0.7

We will calculate the z-scores for the values 40.7 and 39.3. These z-scores are \(z_1 = \frac{39.3 - 40}{0.625} = -1.12\) and \(z_2 = \frac{40.7 - 40}{0.625} = 1.12\). The probability that \(\bar{x}\) will differ from \(\mu\) by more than 0.7 is \(P(z \leq -1.12) + P(z \geq 1.12) = 2 * (1 - P(z \leq 1.12))\), by symmetry of the normal distribution. The standard normal table lists that \(P(z \leq 1.12)\) is approximately 0.8686. So the desired probability is approximately \(2*(1-0.8686) = 0.2628\).

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Key Concepts

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

Population Mean
The population mean, often represented by the Greek letter \( \mu \), is a key statistical concept that refers to the average value of a set of measurements in the entire population. When we talk about a population, we're referring to the complete collection of data points or individuals that we're interested in studying. For example, if we are looking at the heights of all adult women in a city, the population mean would be the average height of all these women.

To calculate the population mean, we simply sum up all the values and divide by the total number of observations in the population. It's important to note that when we take a random sample from this population and calculate the average, we're seeking an estimate of this population mean. This estimated average is known as the sample mean.
Standard Deviation of the Sample Mean
The standard deviation of the sample mean, sometimes called the standard error, describes how much variation or 'spread' exists from the sample mean to the true population mean. A smaller standard error indicates that the sample mean is likely to be close to the population mean. This concept is critical when we make inferences about the population from our sample.

In mathematical terms, the standard deviation of the sample mean \( \sigma_{\bar{x}} \) is calculated by dividing the population standard deviation \( \sigma \) by the square root of the sample size \( n \) (\( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \) ). The smaller the standard error, the more confident we can be that our sample mean is a good representation of the population mean.
Central Limit Theorem
The Central Limit Theorem is a fundamental principle in statistics that has far-reaching implications. This theorem states that, given a sufficiently large sample size, the sampling distribution of the mean will be approximately normally distributed, irrespective of the shape of the original population distribution.

This theorem proves especially useful when the population distribution is unknown or not normally distributed. As a rule of thumb, a sample size of 30 or more is often considered sufficient for the Central Limit Theorem to hold, which means that it can be reliably used to approximate the sampling distribution with a normal distribution, as seen in our example with a sample size of 64.
Normal Distribution
A normal distribution, also known as a Gaussian or bell curve, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In a perfectly normal distribution, the mean, median, and mode of the distribution are all equal.

The normal distribution is important in statistics due to its recurrence in numerous natural phenomena. For instance, heights, blood pressure, and test scores are often found to be approximately normally distributed. When a distribution is normal, we can fully describe it with just two parameters: the mean and the standard deviation. This makes it simpler to calculate probabilities for the occurrence of certain events within the distribution, as showcased in the calculation of the probabilities for the sample mean being within a specific range from the population mean.

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Most popular questions from this chapter

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a sample of \(n=25\) adult males is to be selected, what is the probability that the sample average distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm}\) ? b. Suppose that a sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample average distance will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm}\) ?

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there is a weight limit of \(2500 \mathrm{lb}\). Assume that the average weight of students, faculty, and staff on campus is \(150 \mathrm{lb}\), that the standard deviation is \(27 \mathrm{lb}\), and that the distribution of weights of individuals on campus is approximately normal. If a random sample of 16 persons from the campus is to be taken: a. What is the mean value of the distribution of the sample mean? b. What is the standard deviation of the sampling distribution of the sample mean weight? c. What average weights for a sample of 16 people will result in the total weight exceeding the weight limit of \(2500 \mathrm{lb}\) ? d. What is the chance that a random sample of 16 persons on the elevator will exceed the weight limit?

Consider the following population: \(\\{2,3,3,4,4\\}\). The value of \(\mu\) is \(3.2\), but suppose that this is not known to an investigator, who therefore wants to estimate \(\mu\) from sample data. Three possible statistics for estimating \(\mu\) are Statistic \(1:\) the sample mean, \(\bar{x}\) Statistic 2: the sample median Statistic 3 : the average of the largest and the smallest values in the sample A random sample of size 3 will be selected without replacement. Provided that we disregard the order in which the observations are selected, there are 10 possible samples that might result (writing 3 and \(3^{*}, 4\) and \(4^{*}\) to distinguish the two 3 's and the two 4 's in the population): \(\begin{array}{rllll}2,3,3^{*} & 2,3,4 & 2,3,4^{*} & 2,3^{*}, 4 & 2,3^{*}, 4^{*} \\ 2,4,4^{*} & 3,3^{*}, 4 & 3,3^{*}, 4^{*} & 3,4,4^{*} & 3^{*}, 4,4^{*}\end{array}\) For each of these 10 samples, compute Statistics 1,2, and 3. Construct the sampling distribution of each of these statistics. Which statistic would you recommend for estimating \(\mu\) and why?

Suppose that \(20 \%\) of the subscribers of a cable television company watch the shopping channel at least once a week. The cable company is trying to decide whether to replace this channel with a new local station. A survey of 100 subscribers will be undertaken. The cable company has decided to keep the shopping channel if the sample proportion is greater than \(.25\). What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only .20?

Newsweek (November 23,1992 ) reported that \(40 \%\) of all U.S. employees participate in "self-insurance" health plans \((\pi=.40)\). a. In a random sample of 100 employees, what is the approximate probability that at least half of those in the sample participate in such a plan? b. Suppose you were told that at least 60 of the \(100 \mathrm{em}-\) ployees in a sample from your state participated in such a plan. Would you think \(\pi=.40\) for your state? Explain.

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