According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepa

Question

According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. This distribution of times follows the normal distribution and the standard deviation is 80 minutes. A consumer watchdog agency selects a random sample of 40 taxpayers. What is the standard error of the mean in this example? (Round your answer to 3 decimal places.) What is the likelihood the sample mean is greater than 320 minutes? (Round z value to 2 decimal places and final answer to 4 decimal places.) What is the likelihood the sample mean is between 320 and 350 minutes? (Round z value to 2 decimal places and final answer to 4 decimal places.) What is the likelihood the sample mean is greater than 350 minutes? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Explanation / Answer

Solution....

What assumption or assumptions do you need to make about the shape of the population?
What is the standard error of the mean in this example?
s = 80/sqrt(40)
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What is the likelihood the sample mean is greater than 320 minutes?
z(320)= (320-330)/(80/sqrt(40) = -0.79057
P(z>-0.79057)= 0.7854
P(sample mean > 320) = 0.7854

What is the likelihood the sample mean is between 320 and 350 minutes?
Find z(350); Find z(320)=-0.79057
Find P(z(350)-z(320)) = 0.7285
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What is the likelihood the sample mean is greater than 350 minutes?
Find P(z>z(350)) = 0.0569

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