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?

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

To determine the income range that includes 90% of the most popular incomes, we can use the concept of a confidence interval. In this case, we want to find the range that includes 90% of the incomes, so we need to find the boundaries of a 90% confidence interval.

First, let's find the margin of error, which is a measure of how much variation we expect in our sample mean. The formula for the margin of error is:

Margin of Error = Z * (standard deviation / sqrt(sample size))

Since the distribution of incomes is assumed to be normal, we can use the Z-score to find the corresponding confidence level. For a 90% confidence level, the Z-score is 1.645 (you can find this value in a Z-table or use a calculator).

Given that the standard deviation (σ) is $4500 and assuming we have a large enough sample size, let's calculate the margin of error:

Margin of Error = 1.645 * ($4500 / sqrt(sample size))

Since we don't have information about the sample size, we can calculate the margin of error for different sample sizes to get an idea of the income range.

For example, for a sample size of 100, the margin of error would be:

Margin of Error = 1.645 * ($4500 / sqrt(100)) = 1.645 * ($4500 / 10) = $743.85

To determine the income range that includes 90% of the most popular incomes, we need to add and subtract the margin of error from the sample mean.

Lower Bound = Mean - Margin of Error = $35000 - $743.85 = $34256.15

Upper Bound = Mean + Margin of Error = $35000 + $743.85 = $35743.85

Therefore, the income range that includes 90% of the most popular incomes would be approximately $34256.15 to $35743.85.