The ideal (daytime) noise-level for hospitals is 45 decibels with a standard dev

Question

The ideal (daytime) noise-level for hospitals is 45 decibels with a standard deviation of 10 db. A simple random sample of 81 hospitals at a moment during the day gives a mean noise level of 47 db. Assume that the standard deviation of noise level is really 10 db. 1.)Assuming that the average noise level of hospitals is what it's supposed to be, what is the probability of a sample of 81 hospitals producing an average as high as our sample's (p-value)? NOTE: I have asked this question before and nobody seems to have gotten it correctly. The answer is not .07186 or .1171. The question is looking for the probability.

Explanation / Answer

here std error =std deviation/(n)1/2 =10/(81)1/2 =1.1111

hence P(X<47) =P(Z<(47-45)/1.1111)=P(Z<1.8)= 0.9641

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