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For samples of the specified size from the population described, find the mean and standard deviation of the sample mean $\bar{x}$.



The National Weather Service keeps records of rainfall in valleys. Records indicate that in a certain valley, the annual rainfall has a mean of 77 inches and a standard deviation of 12 inches. Suppose the rainfalls are sampled during randomly picked years, and $\bar{x}$ is the mean amount of rain in these years. For samples of size 36, determine the mean and standard deviation of $\bar{x}$.


A$\mu_{\bar{x}} = 77, \sigma_{\bar{x}} = 2$
B$\mu_{\bar{x}} = 2, \sigma_{\bar{x}} = 77$
C$\mu_{\bar{x}} = 77, \sigma_{\bar{x}} = 12$
D$\mu_{\bar{x}} = 12, \sigma_{\bar{x}} = 77$

Verified Answer :

  • The mean of the sampling distribution of the sample mean is equal to the population mean: \(\mu_{\bar{x}} = \mu = 77\).
  • The standard deviation of the sampling distribution is the standard error: \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2\).
  • The final result is: \(\boxed{\mu_{\bar{x}} = 77, \sigma_{\bar{x}} = 2}\).

Explanation

  1. Analyze the problem data

To understand the behavior of the sample mean \(\bar{x}\), we use the principles of the Central Limit Theorem. We are given the population mean \(\mu = 77\) inches and the population standard deviation \(\sigma = 12\) inches. We are taking samples of size \(n = 36\). The sampling distribution of the sample mean describes how the average rainfall varies across different samples of 36 years.

  1. Calculate the mean of the sample mean (\(\mu_{\bar{x}}\))

The mean of the sample mean, denoted as \(\mu_{\bar{x}}\), is equal to the population mean. Therefore, the calculation is straightforward:
\[\mu_{\bar{x}} = \mu = 77\]
This tells us that, on average, the sample mean will be centered around the population mean of \(77\) inches.

  1. Calculate the standard deviation of the sample mean (\(\sigma_{\bar{x}}\))

The standard deviation of the sample mean, \(\sigma_{\bar{x}}\), also known as the standard error, measures how much the sample mean deviates from the true population mean. It is calculated using the formula:
\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\]
Substituting our known values, where \(\sigma = 12\) and \(n = 36\):
\[\sigma_{\bar{x}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2\]
So, the standard deviation of the sample mean is \(2\) inches.

  1. Final conclusion

Comparing our results with the given options, we find that the mean of the sample mean is \(77\) and the standard deviation of the sample mean is \(2\). This corresponds to the first option: \(\mu_{\bar{x}} = 77, \sigma_{\bar{x}} = 2\). Thus, the correct choice is \(\boxed{\mu_{\bar{x}} = 77, \sigma_{\bar{x}} = 2}\).

Examples

In practice, this is essential for quality control or climate analysis. For example, if you are analyzing the average yield of crops over a 36-year period, this math tells you how much the average yield is expected to vary from the historical average, allowing you to establish confidence intervals and better predict future outcomes. The relationship \(\sigma_{\bar{x}} = \sigma / \sqrt{n}\) is fundamental to understanding that larger sample sizes lead to more precise estimates of the population mean.

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