Question

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

Answer 1

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

First, calculate the z-score:

z = (x - μ) / σ = (37000 - 36000) / 7000 = 0.143

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

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