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℅

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)