Question

Suppose that the heights of adult women in the United States are normally distributed with a mean of 64.5 inches and a standard deviation of 2.4 inches. Jennifer is taller than 70% of the population of U.S. women. How tall is Jennifer? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.

Answer 1

Jennifer's height is approximately 65.8 inches.

Step-by-step explanation:

To find Jennifer's height, we need to find the value of x that corresponds to the 70th percentile in a normal distribution with a mean of 64.5 inches and a standard deviation of 2.4 inches.

First, we need to find the corresponding z-score for the 70th percentile. We can use the z-score formula: z = (x - mean) / standard deviation. Rearranging the formula, we get: x = mean + (z * standard deviation).

Using a z-table or a calculator, we find that the z-score for the 70th percentile is approximately 0.5244. Plugging this value into the formula, we get:

x = 64.5 + (0.5244 * 2.4) = 64.5 + 1.2586 = 65.7586

Rounding to one decimal place, Jennifer's height is approximately 65.8 inches.