 Quandaries and Queries My name: Gary I'm a student. The level of the question: secondary (10-12). I have the following questions which I can not answer: In a law school class, the entering students averaged about 160 on the last LSAT; the standard deviation was about 8. The historgram of the lLSAT scores follwed the normal curve reasonable well. Q. About what percentage of the class scored below 166? Q. One student was 0.5 above average on the last LSAT, about what percentage of the students had lower scores than he did? Thanks for your help. Regards, Gary Hi Gary, Since the histogram showed approximately a normal distribution you are expected to assume that the distributioin is normal with mean 160 and standard deviation 8. If X is the random variable that is the score of a student on the test then the percentage of the class that scored below 166 is Pr(X < 166) Since you are to assume that X is normal with mean 160 and standard deviation 8 you can convert this to a probability concerning the standard normal variable Z by using the transformation Z = (X - mean)/standard deviation Thus the probability becomes Pr(X < 166) = Pr(Z < (166 - 160)/8) Use the standard normal tables to calculate this probability. For the second part you can use the same method to find Pr(X < (160 + 0.5)) = Pr(X < 160.5) Penny Go to Math Central