Posted by Gift on .
The time required by workers to complete an assembly job has a mean of 50 minutes and a standard deviation of 8 minutes. To spot check the workers' progress on a particular day, their supervisor intends to record the times 60 workers take to complete one assembly job apiece. What is the probability that the sample mean will be more than 52 minutes?

statistics 
PsyDAG,
Z = (mean1  mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.