normal probability distributions

maria13always · #1 $Old$ October 25th, 2009, 03:48 PM

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?

Chris L T521 · #2 $Old$ October 25th, 2009, 06:35 PM

Quote:

Originally Posted by maria13always $View Post$

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:

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

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

Does this make sense?

ANDS! · #3 $Old$ October 25th, 2009, 07:52 PM

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:

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

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 ( $P(X\geq70)$ ) 70 inches.