## Related questions with answers

An IRS research found that the average time for taxpayers to prepare, copy, and electronically file a 1040 tax form is 330 minutes. The standard deviation for this time distribution, which is normal, is 80 minutes. A group of 40 taxpayers are chosen at random by a consumer protection organisation. What is the likelihood the sample mean is greater than 320 minutes?

Solution

VerifiedThe 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 has a **normal distribution**. 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

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

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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