The heights of women are normally distributed with a mean of 65 inches and a standard deviation of 2.5 inches. The heights of men are also normal with a mean of 70 inches. Approximately what percent of women are taller than a man of average height?

Step-by-step explanation:

Since the heights of women are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = heights of women.

µ = mean height

σ = standard deviation

From the information given,

µ = 65 inches

σ = 2.5 inches

The probability that a woman is taller than 70 inches is expressed as

P(x > 70) = 1 - P(x ≤ 70)

For x = 70

z = (70 - 65)/2.5 = 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.977

Therefore,

P(x > 70) = 1 - 0.977 = 0.023

The percent of women that are taller than a man of average height is 0.023 × 100 = 2.3℅

whatvare the choices

step-by-step explanation:

0.0076, if you meant 4.125 divided by 0.03125 it would be 132

step-by-step explanation:

first you have to move the decimal in 0.03125 over until its behind the 5 which is 5 places over. now the number is 3125

then you have to move the decimal in 4.125, 5 places over. now the number is 4,112,500.

now you divide like you would normally would and get 0.0076