Pregunta
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37. There were 5,317 previously owned homes sold in a western city in the year 2000 . The distribution of the sales prices of these homes was strongly right-skewed, with a mean of \( \$ 206,274 \) and a standard deviation of \( \$ 37,881 \). If all possible simple random samples of size 100 are drawn from this population and the mean is computed for each of these samples, which of the following describes the sampling distribution of the sample mean? (A) Approximately normal with mean \( \$ 206,274 \) and standard deviation \( \$ 3,788 \) (B) Approximately normal with mean \( \$ 206,274 \) and standard deviation \( \$ 37,881 \) (C) Approximately normal with mean \( \$ 206,274 \) and standard deviation \( \$ 520 \) (D) Strongly right-skewed with mean \( \$ 206,274 \) and standard deviation \( \$ 3,788 \) (E) Strongly right-skewed with mean \( \$ 206,274 \) and standard deviation \( \$ 37,881 \)

Ask by Clark Ramirez. in the United States
Jan 30,2025

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The sampling distribution of the sample mean is approximately normal with a mean of \$206,274 and a standard deviation of \$3,788.

Solución

To determine the sampling distribution of the sample mean, we'll use the **Central Limit Theorem (CLT)**. The CLT states that, for a sufficiently large sample size, the distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution. Given the details: - **Population Mean (\(\mu\))**: \$206,274 - **Population Standard Deviation (\(\sigma\))**: \$37,881 - **Sample Size (\(n\))**: 100 According to the CLT: 1. **Mean of the Sampling Distribution (\(\mu_{\bar{x}}\))**: This is equal to the population mean. \[ \mu_{\bar{x}} = \mu = \$206,274 \] 2. **Standard Deviation of the Sampling Distribution (\(\sigma_{\bar{x}}\))**: This is the population standard deviation divided by the square root of the sample size. \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{37,881}{\sqrt{100}} = \frac{37,881}{10} = \$3,788.1 \] Since the sample size is large (\(n = 100\)), the sampling distribution of the sample mean will be approximately **normal**, even though the original distribution is strongly right-skewed. **Answer:** **(A) Approximately normal with mean $206,274 and standard deviation $3,788**

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The Deep Dive

When you're working with the means of samples, the Central Limit Theorem kicks in, letting us know that even if the original distribution is skewed, the distribution of the sample means will trend toward normality as the sample size increases. Now, with a sample size of 100, we're in a good place for this to happen! The standard deviation of the sampling distribution, known as the standard error, is calculated by dividing the population standard deviation by the square root of the sample size. So that's \( \$ 37,881 \div \sqrt{100} = \$ 3,788 \). Therefore, the sampling distribution of the sample mean is approximately normal, with a mean of \( \$ 206,274 \) and a standard deviation of \( \$ 3,788 \). So, the answer is (A).

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