Finding the mean score

Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 503 and a standard deviation of 103. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the preparation course has no effect. 

 

If 14 of the men are randomly selected, find the probability that their mean score is at least 547.
P(M > 547) =

 

Answer 1

Since the course has no effect, you would expect the sample mean and standard deviation to be the same as the population.

 

In order to find P(>547) we have to calculate the z score

 

Z = ^{(x - μ)}/_σ 

 

Z = ^547-503/₁₀₃ = 0.4272 which we can round up to 0.43

 

We can use that z score and a z score calculator or chart and find that the percentage is 0.6664

 

That percentage is the percent of students BELOW a z score of 0.43. Since we want the percentage above, we subtract our answer from 1

 

1-0.6664 = 0.3336, or 33.36%