## Question 824:

1## Answer:

No answer provided yet.This question is asking you about both z-scores for a single value and z-scores for a mean. To find the probability associated with a score or mean, each value is converted into a z-score, then the z-score is looked up using a table of values. The general formula for a z-score is (point-mean)/sd.a. The z-score is (75-78)/6 = -0.5. We look up this value in a normal table or using the excel formula =NORMSDIST(-.5) and we get .309.

b. Now we need to find the z-score for a mean. The only difference is instead of dividing by the standard deviation we divide by the standard error of the mean (SE). The SE is the standard deviation divided by the square root of the sample size. For a sample of 50, the SE = 6/SQRT(50) = .848528. The z-score is then (75-78)/.8485 = -3.535. Looking up the area under the normal curve for -3.535 we get the probability of a mean this low as .000203. In other words, it would be extremely rare to have such a low mean given the large sample size.

For more background on these concepts see the tutorials on the standard error of the mean.