/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Let \(x_{1}, x_{2}, \ldots, x_{1... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(x_{1}, x_{2}, \ldots, x_{100}\) denote the actual net weights (in pounds) of 100 randomly selected bags of fertilizer. Suppose that the weight of a randomly selected bag has a distribution with mean 50 pounds and variance 1 pound \(^{2}\). Let \(\bar{x}\) be the sample mean weight \((n=100)\). a. Describe the sampling distribution of \(\bar{x}\). b. What is the probability that the sample mean is between \(49.75\) pounds and \(50.25\) pounds? c. What is the probability that the sample mean is less than 50 pounds?

Short Answer

Expert verified
a. The sampling distribution of \(\bar{x}\) is approximately normal with mean 50 pounds and standard deviation 0.1 pounds.\n b. The probability that the sample mean is between 49.75 and 50.25 pounds can be found by looking up the corresponding z-scores of -2.5 and 2.5 in a standard normal probability table or using a calculator.\n c. The probability that the sample mean is less than 50 pounds is 0.5 or 50%.

Step by step solution

01

Describe the sampling distribution of \(\bar{x}\)

According to the central limit theorem (CLT), the sampling distribution of \(\bar{x}\) will approach normal distribution as the number of samples increase. Given that in this problem, the number of samples is 100 which is greater than 30 (typical threshold), we can assume the distribution to be approximately normal. The mean of the distribution of \(\bar{x}\), denoted by \(\mu_{\bar{x}}\), is equal to the mean of the population, \(\mu = 50\) pounds. The standard deviation of the distribution of \(\bar{x}\), denoted by \(\sigma_{\bar{x}}\), is the standard deviation of the population (\(\sigma = 1\) pound) divided by the square root of the number of samples (n = 100), i.e., \(\sigma_{\bar{x}} = \sigma/ \sqrt{n} = 1/ \sqrt{100} = 0.1\) pounds.
02

Find the probability that the sample mean is between 49.75 pounds and 50.25 pounds

We can find this probability by using the z-score formula which relates a raw score, the mean, and the standard deviation in a standard normal distribution. First, calculate the z-scores for weights 49.75 pounds and 50.25 pounds. The z-score is given by the formula: \( Z = (X - \mu) / \sigma \). For \( X = 49.75 \) pounds, the Z score \( Z_{49.75} = (49.75 - 50) / 0.1 = -2.5 \). For \( X = 50.25 \) pounds, the z score \( Z_{50.25} = (50.25 - 50) / 0.1 = 2.5 \). To find the probability that the sample mean is between these two weights, look these z-scores up in a standard normal probability table or use a calculator that can calculate normal probabilities.
03

Find the probability that the sample mean is less than 50 pounds

Similarly to step 2, we need to find the z-score for weight 50 pounds: \( Z_{50} = (50 - 50) / 0.1 = 0 \). Then, look this up in a standard normal probability table or use a calculator. However, since the standard normal distribution is symmetric about the mean, we know that the probability of getting a z-score less than 0 is 0.5 or 50%.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that explains how the distribution of the sample mean approaches a normal distribution as the sample size becomes large, typically over 30. This is incredibly useful because it allows statisticians to make inferences about a population, even if the original data is not normally distributed.

For example, in the problem regarding the weights of fertilizer bags, we have a sample size of 100, which makes it possible to apply the CLT. The theorem ensures that the distribution of the sample mean is approximately normal, with the same mean as the population and a standard deviation reduced by a factor of the square root of the sample size. This normality holds true regardless of the shape of the population distribution, allowing us to predict probabilities related to mean weights.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values, commonly represented by \( \sigma \). A lower standard deviation means that the values tend to be close to the mean (also called the expected value), while a higher standard deviation means that the values are spread out over a wider range.

In the context of the fertilizer bag weights, the standard deviation of the population is 1 pound. When dealing with the standard deviation of the sample mean (\( \sigma_{\bar{x}} \)), we divide the population standard deviation by the square root of the sample size. This reflects how the distribution of the sample mean is more tightly clustered around the true mean than individual observations would be.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a symmetric, bell-shaped curve that describes the distribution of many types of data. It is determined by two parameters: the mean (\( \mu \)), which locates the center of the bell, and the standard deviation (\( \sigma \)), which scales the width of the bell.

In many practical scenarios, including the fertilizer weight example, we can apply the CLT to approximate the sample means to a normal distribution. This normal distribution can then be used to calculate probabilities, make predictions, and infer characteristics about the population from which the sample was drawn.
Z-Score
A z-score, sometimes called a standard score, is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. The formula to calculate a z-score is given by \( Z = (X - \mu) / \sigma \). Non-standard distributions can be standardized using the z-score formula, allowing comparison across different datasets.

In the provided problem, z-scores help determine probabilities associated with specific sample mean weights. Once we calculate the z-score, we can use a standard normal distribution table or calculator to find out the probabilities of the sample mean falling within certain intervals.
Probability
Probability measures the likelihood of an event occurring, ranging from 0 (the event never occurs) to 1 (the event always occurs). It can be thought of as the proportion of times an event would occur over an infinite number of trials in a probabilistic model.

In statistics, probability allows us to quantify the uncertainty of events. For the weights of the fertilizer bags, it enables us to compute how likely it is for the sample mean weight to fall within a certain range (for example, between 49.75 and 50.25 pounds). Probabilities for a normal distribution can be found using z-scores to transform the normal distribution questions into standardized form.

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Most popular questions from this chapter

A random sample is to be selected from a population that has a proportion of successes \(p=.65 .\) Determine the mean and standard deviation of the sampling distribution of \(\hat{p}\) for each of the following sample sizes:\(8.23\) A random sample is to be selected from a population that has a proportion of successes \(p=.65 .\) Determine the mean and standard deviation of the sampling distribution of \(\hat{p}\) for each of the following sample sizes: a. \(n=10\) b. \(n=20\) c. \(n=30\) d. \(n=50\) e. \(n=100\) f. \(n=200\)

Explain the difference between a population characteristic and a statistic.

The thickness (in millimeters) of the coating applied to disk drives is one characteristic that determines the usefulness of the product. When no unusual circumstances are present, the thickness \((x)\) has a normal distribution with a mean of \(2 \mathrm{~mm}\) and a standard deviation of \(0.05 \mathrm{~mm}\). Suppose that the process will be monitored by selecting a random sample of 16 drives from each shift's production and determining \(\bar{x}\), the mean coating thickness for the sample. a. Describe the sampling distribution of \(\bar{x}\) (for a sample of size 16). b. When no unusual circumstances are present, we expect \(\bar{x}\) to be within \(3 \sigma_{\bar{x}}\) of \(2 \mathrm{~mm}\), the desired value. An \(\bar{x}\) value farther from 2 than \(3 \sigma_{\bar{x}}\) is interpreted as an indication of a problem that needs attention. Compute \(2 \pm 3 \sigma_{\bar{x}} .\) (A plot over time of \(\bar{x}\) values with horizontal lines drawn at the limits \(\mu \pm 3 \sigma_{x}^{-}\) is called a process control chart.) c. Referring to Part (b), what is the probability that a sample mean will be outside \(2 \pm 3 \sigma_{\bar{x}}^{-}\) just by chance (that is, when there are no unusual circumstances)? d. Suppose that a machine used to apply the coating is out of adjustment, resulting in a mean coating thickness of \(2.05 \mathrm{~mm}\). What is the probability that a problem will be detected when the next sample is taken? (Hint: This will occur if \(\bar{x}>2+3 \sigma_{\bar{x}}\) or \(\bar{x}<2-3 \sigma_{\bar{x}}\) when \(\left.\mu=2.05 .\right)\)

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 proportion of all subscribers who watch it is only \(.20\) ?

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. In addition, there is a weight limit of 2500 pounds. Assume that the average weight of students, faculty, and staff on campus is 150 pounds, that the standard deviation is 27 pounds, 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 expected value of the distribution of the sample mean weight? 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 pounds? d. What is the chance that a random sample of 16 people will exceed the weight limit?

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