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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? 0.1075 0 0.4432

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Since the sample is greater than 10, we can approximate this binomial problem with a normal distribution.

First, calculate the z-score:

z = (x -Ā Ī¼) /Ā Ļƒ = (37000 - 36000) / 7000 = 0.143

The probability P(x > 37000$) = 1 - P(x < 37000$),Ā 
therefore we need to look up at a normal distribution table in order to findĀ 
P(z < 0.143) = 0.55567Ā 
AndĀ 
P(x > 37000$) = 1 -Ā 0.55567 = 0.44433

Hence, there is a 44.4% probability thatĀ the sample mean is greater than $37,000.

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