Question

Last year, the revenue for utility companies had a mean of 60 million dollars with a standard deviation of 17 million. Find the percentage of companies with revenue less than 21 million or more than*99 million dollars. Assume that the distribution is normal. Round your answer to the nearest hundredth.

Answer 1

To find the percentage of companies with revenue less than 21 million, we first need to find the z-score associated with 21 million:

z = (X - μ) / σ
z = (21 - 60) / 17
z = -39 / 17
z = -2.29

Using a standard normal distribution table, we find that the percentage of companies with revenue less than 21 million is approximately 0.0117, or 1.17%.

Next, we find the percentage of companies with revenue more than 99 million. Again, we first need to find the z-score associated with 99 million:

z = (X - μ) / σ
z = (99 - 60) / 17
z = 39 / 17
z = 2.29

Using a standard normal distribution table, we find that the percentage of companies with revenue more than 99 million is approximately 0.0117, or 1.17%.

To find the percentage of companies with revenue less than 21 million or more than 99 million, we add the two percentages together:

1.17% + 1.17% = 2.34%

Therefore, the percentage of companies with revenue less than 21 million or more than 99 million is approximately 2.34%.