The distribution of height of adult american women is approximately normal with a mean of 65.5 inches and a standard deviation of 2.5 inches what percentage of women are taller than 70.5 inches?

2.275%

Step-by-step explanation:

We start by calculating the z-score

Mathematically;

z-score = (x-mean)SD

here, x = 70.5

mean = 65.5

SD = 2.5

Substituting these values;

z = (70.5-65.5)/2.5

= 5/2.5 = 2

So we want to calculate the probability that;

P( z> 2)

We check the standard normal distribution table for this

That will be

0.02275

In percentage, this is 2.275 %

the answer

b. they are parallel and will never intersect.