Asked 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
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
Answers
Answered by
PsyDAG
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.