Women Make up......Statistics Question?

Question

women make up 24% of the science and engineering workforce. In a random sample of 400 science and engineering employees, what is the probability that more than 120 are women?

please put it in words i can understand!!!! THANK YOU!!!!

Answer 1

Your question pertains to a Binomial distribution, since your sample is a fixed number, i.e. 400, and the probability is a fixed percentage, i.e. 24%. Also, the calculation in this case is really tedious, since you want the probability over the range n > 120. So, in finding the answer to your problem, I say
that's what computers are for.

The following Web site has a nice applet that you can use to get your answer. Plug in p = .24, N = 400, and range = 121 to 400. When I do this, I get a probability of .0025.

http://www.ruf.rice.edu/~lane/stat_sim/normal_approx/index.html

Answer 2

Repeating the answer I gave to the last person who asked the exact same question:

"The probability that any given member of the sample is a woman is p = 24/100 = 6/25. Therefore, the probability that exactly n women are in the sample is 400!/(n!(400-n)!) * p^n * (1-p)^(400-n), and so the probability that more than 120 women are in the sample is [n=121, 400]∑400!/(n!(400-n)!) * p^n * (1-p)^(400-n). This is an exact value, but very hard to compute. Therefore, we take advantage of the fact that for sufficiently large sample sizes, the binomial distribution is approximately normal. The expected value of the number of women in the sample is 400*(6/25) = 96 women. The variance will be n*p*(1-p), which is 96*(19/25) = 912/25 = 72.96. Therefore, the probability that n>120 is approximately the probability that a standard normal is greater than (120-96)/√72.96. (120-96)/√72.96 ≈ 2.809757. Looking this up in a standard normal table, we see that the probability that z<2.809757 ≈ .997521, so the probability that there are more than 120 women is only 0.002479."