Question

Last year, the revenue for utility companies had a mean of 80 million dollars with a standard deviation of 21 million. Find the percentage of companies with revenue less than 79 million dollars. Assume that the distribution is normal. Round your answer to the nearest hundredth.

Answer 1

The percentage of utility companies with revenue less than $79 million is approximately 48.19%, calculated using the z-score from the normal distribution.

Step-by-step explanation:

To find the percentage of utility companies with revenue less than $79 million, you need to use the z-score formula since we are dealing with a normal distribution. The z-score formula is z = (X - μ) / σ, where X is the value in question, μ is the mean, and σ is the standard deviation. Here, the mean (μ) is $80 million and the standard deviation (σ) is $21 million.

First, calculate the z-score for $79 million:

z = (79 - 80) / 21

z = -1 / 21

z ≈ -0.0476

Now, we look up this z-score in a standard normal distribution table, or use a calculator with normal distribution functions, to find the percentage of values less than this z-score.

The normal distribution table shows that the percentage of companies with a z-score less than -0.0476 is approximately 48.19%. Thus, the percentage of utility companies with revenue less than $79 million is about 48.19%, rounded to the nearest hundredth.