Assuming that the heights of college women are normally distributed with mean 62 inches and standard deviation 3.5 inches, what percentage of women are between 58.5 inches and 72.5 inches? (Points: 5)

34.1%
84.0%
15.7%
13.6%
97.6%

Z = (score - mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion between the two Z scores.

gruff

To find the percentage of women with heights between 58.5 inches and 72.5 inches, we need to calculate the proportion of women within this range.

First, we need to convert these heights into z-scores. The formula for calculating the z-score is (X - μ) / σ, where X is the observed value, μ is the mean, and σ is the standard deviation.

For 58.5 inches:
z = (58.5 - 62) / 3.5 = -1

For 72.5 inches:
z = (72.5 - 62) / 3.5 = 3

Next, we can use a standard normal distribution table or a calculator to find the area/proportion between these z-scores.

Looking up these z-scores in a standard normal distribution table, the area to the left of z = -1 is 0.1587, and the area to the left of z = 3 is 0.9987.

So, the proportion of women with heights between 58.5 inches and 72.5 inches is 0.9987 - 0.1587 = 0.84, or 84%.

Therefore, the correct answer is 84.0%.