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Probability Q? Incomes in a certain town are strongly right-skewed with mean $36,000 and standard deviation $7000. A random sample of 75 households is taken. What is the probability the sample mean is greater than $37,000?
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I think the fact that it's trongly right-skewed doesn't matter. The sample size is large enough (greater than 30) to have the distribution of sample means be a normal distribution so basically the xbar distribution is approximately normal
what you want to find is P( X > 37000) where the mean of the distribution is 36000 and the standard deviation is 7000/sqrt(75) = 808.290376865
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