Question

Last year, the revenue for medical equipment companies had a mean of 90 million dollars with a standard deviation of 19 million. Find the percentage of companies with revenue less than 75 million or more than 105 million dollars. Assume that the distribution is normal. Round your answer to the nearest hundredth.

Answer 1

[AI-generated answer — Bot GPT 3.5]

To solve this problem, we first need to find the z-scores for 75 million and 105 million dollars.

For 75 million:
z = (75 - 90) / 19 = -0.79

For 105 million:
z = (105 - 90) / 19 = 0.79

Next, we use a standard normal distribution table or a calculator to find the percentage of companies with a z-score less than -0.79 or greater than 0.79.

Looking at the standard normal distribution table, the percentage of companies with a z-score less than -0.79 is approximately 0.2149, and the percentage of companies with a z-score greater than 0.79 is also approximately 0.2149.

Therefore, the total percentage of companies with revenue less than 75 million or more than 105 million dollars is:
0.2149 (for z < -0.79) + 0.2149 (for z > 0.79) = 0.4298

Multiplying by 100% to convert to percentage, we get:
0.4298 * 100% = 42.98%

So, approximately 42.98% of companies have revenue less than 75 million or more than 105 million dollars.