A random sample of customer incomes yields a mean income of $35000 with a standard deviation of $4500. Assume that the distribution of the incomes in normal. What income range would include 90% of the most popular incomes?

Answer 1

Given:
μ=35000
σ=4500
X~N(35000,4500^2)

Look for the two tail probability in the normal distribution table for 90% = 0.9, or
Z=±1.645
So the range that contains 90% would be between μ±1.645*σ