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Larger Sample, More Accurate Estimate. Suppose that, in fact, the total cholesterol level of all men aged 20-34 follows the Normal distribution with mean \(\mu=182\) milligrams per deciliter (mg/dL) and standard deviation \(\sigma=37 \mathrm{mg} / \mathrm{dL}\). a. Choose an SRS of 100 men from this population. What is the sampling distribution of \(x\) ? What is the probability that \(x\) takes a value between 180 and 184 \(\mathrm{mg} / \mathrm{dL}\) ? This is the probability that \(x\) estimates \(\mu\) within \(\pm 2 \mathrm{mg} / \mathrm{dL}\). b. Choose an SRS of 1000 men from this population. Now what is the probability that \(x\) falls within \(\pm 2 \mathrm{mg} / \mathrm{dL}\) of \(\mu\) ? The larger sample is much more likely to give an accurate estimate of \(\mu\).

Step by step solution

01

Understand the Sampling Distribution

When we select a simple random sample (SRS) of size \(n\) from a population with mean \(\mu = 182\) mg/dL and standard deviation \(\sigma = 37\) mg/dL, the sampling distribution of the sample mean \(\bar{x}\) will be normally distributed with mean \(\mu\) and standard deviation \(\sigma_\bar{x} = \frac{\sigma}{\sqrt{n}}\). This is due to the Central Limit Theorem.

02

Determine Sampling Distribution for n = 100

For an SRS of 100 men, the sample size \(n = 100\). Thus, the standard deviation of the sampling distribution is \(\sigma_\bar{x} = \frac{37}{\sqrt{100}} = 3.7\). So the sampling distribution of \(\bar{x}\) is \(N(182, 3.7)\).

03

Calculate Probability for n = 100

We need to find the probability that \(\bar{x}\) falls between 180 and 184 mg/dL. We standardize using the Z-score formula: \(Z = \frac{\bar{x} - \mu}{\sigma_\bar{x}}\). Thus: \(Z(180) = \frac{180 - 182}{3.7} = -0.54\) and \(Z(184) = \frac{184 - 182}{3.7} = 0.54\). Using standard normal distribution tables or a calculator, the probability \(P(-0.54 < Z < 0.54)\) is approximately 0.4076.

04

Determine Sampling Distribution for n = 1000

For an SRS of 1000 men, the sample size \(n = 1000\). Thus, the standard deviation of the sampling distribution is \(\sigma_\bar{x} = \frac{37}{\sqrt{1000}} = 1.17\). So the sampling distribution of \(\bar{x}\) is \(N(182, 1.17)\).

05

Calculate Probability for n = 1000

We need the probability that \(\bar{x}\) is within \(\pm 2\) mg/dL of 182. Using Z-scores: \(Z(180) = \frac{180 - 182}{1.17} = -1.71\) and \(Z(184) = \frac{184 - 182}{1.17} = 1.71\). The probability \(P(-1.71 < Z < 1.71)\) is approximately 0.9134 using tables or a calculator.

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Key Concepts

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

Normal Distribution

The Normal distribution, often referred to as the bell curve, is a probability distribution that is symmetrical around its mean. In the context of our problem, it describes the distribution of cholesterol levels among men aged 20-34.

• Key feature: The curve is highest at the mean, which is also the median and mode.
• Characterized by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)).
• In our exercise, the total cholesterol level is normally distributed with a mean (\( \mu = 182 \ ext{mg/dL} \)) and a standard deviation (\( \sigma = 37 \ ext{mg/dL} \)).

This distribution tells us that most men's cholesterol levels are around 182 mg/dL, and the spread of these levels is measured by the standard deviation. The Central Limit Theorem informs us that when we take a large enough sample, the sample mean will also follow a normal distribution.

Sampling Distribution

A sampling distribution is the probability distribution of a statistic based on a random sample. Here, we focus on the sampling distribution of the sample mean (\( \bar{x} \)).

• The Central Limit Theorem states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough, regardless of the population's distribution.
• The mean of the sampling distribution is equal to the population mean (\( \mu \)).
• The standard deviation of the sampling distribution is (\( \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}} \)), where (\( n \)) is the sample size.

In our example, when the sample size is 100, (\( \sigma_\bar{x} = 3.7 \)). For a sample size of 1000, (\( \sigma_\bar{x} = 1.17 \)). Thus, larger samples lead to a narrower distribution of the sample mean, making it a more precise estimate of (\( \mu \)).

Standard Deviation

Standard deviation is a key measure of variability in statistics. It tells us how spread out the numbers in a data set are.

• Population standard deviation (\( \sigma \)) measures how much individual data points differ from the mean for the entire population.
• In our exercise, (\( \sigma = 37 \ ext{mg/dL} \)), indicating average variability in cholesterol levels.
• Sample standard deviation, or more specifically the standard error of the mean, (\( \sigma_\bar{x} \)), adjusts the population standard deviation for the size of the sample.

This adjusted standard deviation becomes crucial when assessing how close a sample mean is likely to be to the population mean. In the exercise, a reduction in (\( \sigma_\bar{x} \)) with larger sample sizes shows increased accuracy of the sample mean estimation.

Z-score

The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is expressed in terms of standard deviations.

• Calculating a Z-score involves subtracting the mean from the data point, then dividing by the standard deviation.
• For the exercise, we compute Z-scores to determine how unusual a sample mean is compared to the expected mean.
• For example, for a mean of 180, with standard deviation (\( \sigma_\bar{x} = 3.7 \)), the computation yields a Z-score of -0.54.
• Similarly, for a mean of 184, the Z-score is calculated as 0.54 with (\( n=100 \)). For (\( n=1000 \)), Z-scores of -1.71 and 1.71 reflect the reduced standard error of the mean.

These Z-scores allow us to use standard normal distribution tables or calculators to find probabilities related to our sample means.