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?

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.