Stats 330 HELP
posted by John .
I need some help on this question it has two parts but i can't figure it out. Can someone help me.
The NDSU library checks out an average of 320 books per day, with a standard deviation of 75 books. Suppose a simple random sample of 30 days is selected for observation.
What is the probability that the sample mean for the 30 days will be between 300 and 340 books?
Answer
0.4279
0.2128
0.8558
0.1064
0.9279
There is a 90% chance that the sample mean will fall below how many books?
Answer
323.5
416.0
342.5
337.5
443.4

Stats 330 HELP 
PsyDAG
Z = (mean1  mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n1)
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 proportions related to the Z scores.
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Does anyone have any idea on how to do this?