Posted by **John** on Thursday, March 10, 2011 at 2:20pm.

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**, Friday, March 11, 2011 at 11:29am
Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

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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