The operation manager of a large production plant wants to estimate the average ammount of time a worker requires to assemble a new electronic component. She observed a number of workers and noted the shortest time was 8 minutes and the longest was 20 minutes. With these observations, how large a sample of workers should she take if she wishes to estimate the mean assembly time within 18 seconds? Additional info: use the confidence interval 95%. The standard deviation is the range of the observed values divided by 4.

How in the hell do you do this?

Use CLT
N = sample of workers
T = vector of times

sqrt(N)*(sum(T[i])/N - mean) ~ Normal(0,3²)

confidence interval 95% :

P(-t(95%)) < sqrt(N)(sum(T[i]/N - mean) < t(95%)) = 0.95, t(95%) being the quantile for N(0,9); t(95%) = 5.88

Rewrite this because only the mean is interesting you:

0.95 = (sum(T[i]/N) - t(95%)/sqrt(N) < mean < t(95%)/sqrt(N) + sum(T[i]/N))
Now you want to estimate the mean within 18 seconds, that is to say: 2*t(95%)/sqrt(N) = 18/60 = 0.3
solve for N, N = 1536

Hope i'm right

Thanks for trying but you're wrong. I googled my problem and I found a powerpoint from Lamar University that had an identical example as this one, they saved my ass because this is on my take home test and he never talked about this!). For anyone who wants to know how to do it, here's what you do. The short hand version of how to do this is using the formula n=(z*standard deviation/w)all squared. In this case z= 100%-95%=5/2=2.5%, then do 100%-2.5%=97.5%=.975 then look up that z score in the back of the book. The standard deviation they told you how to calculate, it is the range which is 22 minutes-10 minutes=12minutes/ 4=3 minutes which is 180 seconds(to keep units the same). W is what they told us they wanted the mean to be within, in this case 20 seconds. So z=1.96, Standard deviation=180, and w =20. (1.96 * 180/20) squared equals 311.16 but we have to round up to 312 because there's no such thing as .16 of a person. So the answer is that 312 work samples are needed to estimate mean assembly time within 20 seconds.