1. ## normal probability distributions

heights are normally distributed with mean 63.6 in and standard deviation 2.5 in

What percentage of women are at least 70 in?

How would I go about doing this?

2. Originally Posted by maria13always
heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in.

What percentage of women are at least 70 in?

How would I go about doing this?
Convert to Standard Normal:

$\displaystyle \mathbb{P}\!\left(X\geq 70\right)=\mathbb{P}\!\left(\frac{X-\mu}{\sigma}\geq \frac{70-63.6}{2.5}\right)=\mathbb{P}\!\left(Z\geq 2.56\right)$

Note that $\displaystyle \mathbb{P}\!\left(Z\geq 2.56\right)=1-\mathbb{P}\!\left(Z<2.56\right)\approx.00523$. This means that .523% of women are 70in or taller.

Does this make sense?

3. They tell us the distribution, so we know what particular distribution to use. You need to find out how many standard deviations 70 inches is from 63.6 inches. Do you know how to do this:

$\displaystyle \frac{\overline{x}-x}{\sigma_{x}}=z$

Do you know how to read a z-chart? Also carefully read the sentence. They are asking for what percentage of women are at LEAST ($\displaystyle P(X\geq70)$) 70 inches.