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Baxter-Dunn1 · Apr 30, 2009, 12:28 AM

How do you figure out this question? Women make up 23% of the workforce. In a random sample of 392 employees, what is the probability that more than 108 are women. The probability that more than 108 are women is what percentage?

galactus · Apr 30, 2009, 11:44 AM

Since we are dealing with large numbers, we could use the binomial distribution.

That is, $np={\mu}\Rightarrow 392(.23)=90.16, \;\ \sqrt{npq}={\sigma}\Rightarrow 8.332$

and use the continuity correction.

$z=\frac{108.5-90.16}{8.332}=2.2$

Look up the value for z=2.2 in the z table and subtract from 1 because it states "more than".

Also, if you have a calculator or computer you can use

$1-{\sum_{k=109}^{392}C(392,k)(.23)^{k}(.77)^{392-k}$

The respective results should be close to one another. I checked and they are.

The latter is cumbersome to compute. That is why we use the short method outlined first.