The mean per capita income is 24,787 dollars per annum with a variance of 169,744. What is the probability that the sample mean would differ from the true mean by less than 42 dollars if a sample of 412 persons is randomly selected?

Question

I know what formula to use for this question (z: x-μ/(σ/√n)) and I know that the standard deviation would be σ= √169 744 = 142. But, I'm just confused about what this statement means: "probability that the sample mean would differ from the true mean by less than 42 dollars"?

Answer 1

Z = (24,745-24,787)/(142/20) = ?

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Since you want to know above AND below the mean, double that value.