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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 between 320 and 350 minutes?

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The population mean μ\mu is a parameter that indicates the average of the entire population. Its estimator is the sample proportion X\overline X and the estimate, or its particular value, is x\overline x. Since X\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 x\overline x to its standard normal distribution zz using the formula below

z=xμσ/n=xμσx(1)z = \dfrac{\overline x - \mu}{ \sigma / \sqrt n } =\dfrac{\overline x - \mu}{\sigma _{\overline x} }\tag 1

where x\overline x is the estimate, μ\mu is the population mean, σ\sigma is the population standard deviation, and nn is the sample size.

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

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