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. a. What is the standard error of the mean in this example? b. What is the likelihood the sample mean is greater than 360 minutes?

Answer:78.52% probability that the sample mean is greater than 320 minutesStep-by-step explanation:To solve this question, we need to understand the normal probability distribution and the central limit theorem.Normal probability distributionProblems of normally distributed samples are solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.Central Limit TheoremThe Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.In this problem, we have that:[tex]\mu = 330, \sigma = 80, n = 40, s = \frac{80}{\sqrt{40}} = 12.65[/tex]What is the likelihood the sample mean is greater than 320 minutes?This is 1 subtracted by the pvalue of Z when X = 320. So[tex]Z = \frac{X - \mu}{\sigma}[/tex]By the Central Limit Theorem[tex]Z = \frac{X - \mu}{s}[/tex][tex]Z = \frac{320 - 330}{12.65}[/tex][tex]Z = -0.79[/tex][tex]Z = -0.79[/tex] has a pvalue of 0.21481 - 0.2148 = 0.785278.52% probability that the sample mean is greater than 320 minutes