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Computing Probability and Deciding Extreme Samples

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Please explain how to solve. I have attempted to answer, please let me know if I'm right or wrong. Thank you!

a random sample of n=9 scores is selected from a normal distribution with u= 80 and o=12. what is the probability that the sample mean will be between 76 and 84?
* 0.9974
*0.3830
* 0.2586
* 0.6426 ( is this correct?)

a random sample of n= 16 scores is selected from a normal distribution with u= 500 and o= 200. for this sample which of the following is true?
* p(450<M<550)=0.95
*p(425<M<575)=0.95
*p(402<M<598)= 0.95
*p(490<M<510)=0.95

A sample is selected from a normal population with u= 50 and o= 12. which of the following samples would be considered extreme and unrepresentative for this population?
*M=53 and n=4
*M=53 and n= 16
*M=56 and n=4 ( is this correct?)
*M=56 and n=16

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Solution Summary

The solution gives detailed steps on computing probability of normally distributed data and also deciding extreme samples.

Solution Preview

1. P(76<X<84)=P((76-80)/(12/sqrt(9))<Z<(84-80)/(12/sqrt(9)))=P(-1<Z<1)=0.6426 from standard normal table

2. We know that P(-1.96<Z<1.96)=0.95. So we ...

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