Question

Suppose a Normal model describes the fuel efficiency of cars registered in California. The mean is 24 mpg, with a standard deviation of 6. What percent of cars get more than 32 mpg?

A) 16.67%
B) 50%
C) 83.33%
D) 95%

Answer 1

To find the percent of cars that get more than 32 mpg, we need to find the area under the Normal curve to the right of 32. Using the z-score formula, we can calculate the z-score for 32 mpg and look it up on a standard Normal distribution table to find the corresponding area. The percent of cars that get more than 32 mpg is approximately 9.18%.

Step-by-step explanation:

To find the percent of cars that get more than 32 mpg, we need to find the area under the Normal curve to the right of 32. We can use the z-score formula to transform the value 32 into a standard score.

z = (x - μ) / σ

where x is the value of interest, μ is the mean, and σ is the standard deviation.

In this case, x = 32, μ = 24, and σ = 6. Plugging these values into the formula, we get:

z = (32 - 24) / 6 = 1.33

Looking up the z-score in a standard Normal distribution table, we find that the area to the right of z = 1.33 is approximately 0.0918, which corresponds to 9.18%.

Therefore, the percent of cars that get more than 32 mpg is approximately 9.18%. The correct answer is not available in the options provided.