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Question

# A normal population has a mean of 75 and a standard deviation of 5. You choose a 40-person sample. Calculate the likelihood that the sample mean is less than 74.

Solution

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The population mean $\mu$ is a parameter that indicates the average of the entire population. Its estimator is the sample proportion $\overline X$ and the estimate, or its particular value, is $\overline x$. Since $\overline X$ is a random variable whose value depends on the chosen random sample, then we need to discuss its sampling distribution.

The given population is normal. As such, the sampling distribution of the sample mean, regardless of the size, will also be normal. Therefore, we can convert the normal distribution $\overline x$ to its standard normal distribution $z$ using the formula below

$z = \dfrac{\overline x - \mu}{ \sigma / \sqrt n }, \tag 1$

where $\overline x$ is the estimate, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size.

We can then determine the likelihood of the estimate $\overline x$ by looking at the table for the areas under the curve of the z-values.

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