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Chapter 7: Problem 36

The weights of all people living in a particular town have a distribution that is skewed to the right with a mean of 133 pounds and a standard deviation of 24 pounds. Let \(\bar{x}\) be the mean weight of a random sample of 45 persons selected from this town. Find the mean and standard deviation of \(\bar{x}\) and comment on the shape of its sampling distribution.

Short Answer

Expert verified
The mean of the sample means is 133 pounds and the standard deviation of the sample means is 24/鈭�45 pounds. The distribution of sample means will be approximately normal.

Step by step solution

01

Identify the Population Mean and Standard Deviation

The problem provides that the population mean weight (\( \mu \)) of persons in this town is 133 pounds and the population standard deviation (\( \sigma \)) is 24 pounds.
02

Find the Mean of the Sampling Distribution

The mean of the sample means (\( \mu_{\bar{x}} \)) is equal to the population mean. Hence, \( \mu_{\bar{x}} = \mu = 133 \) pounds.
03

Find the Standard Deviation of the Sampling Distribution

The standard deviation of the sample means (\( \sigma_{\bar{x}} \)) is equal to the population standard deviation divided by the square root of the sample size (n). Given that the sample size (n) is 45, therefore \( \sigma_{\bar{x}} = \sigma / \sqrt{n} = 24 / \sqrt{45} \) pounds.
04

Comment on the Shape of the Sampling Distribution

Because the sample size (45) is greater than 30, we can infer from the Central Limit Theorem that the distribution of sample means is approximately normal, even though the population distribution is skewed to the right. Therefore, the shape of the sampling distribution is about normal (bell-shaped curve).

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