1. The problem statement, all variables and given/known data an exam is normally distributed with a mean of 75 and a variance of 64. students with a mark of 90 or above recieve an A+ a) what percentage of students recieve an A+? b) if 12 students write the exam what is the probability 3 get A+? 2. Relevant equations z-score- z=mark-mean/standard deviation my teacher gave us a chart that gives probabilities as area under a curve for normal distributions 3. The attempt at a solution a) since the standard deviation is the square root of the variance: SD= 8 z-score= 90-75/8= 1.88 now using the area under the curve chart, for a z-score of 1.88 it reads 0.9699. from this i subtract 1 because i want the amount above 90 and this value is below 90, so i get: = 1-0.9699 = 0.0301 or 3.01% b) if 12 students write and 3 get an A+ then 9 students must get something else: p(x=3)= 3(0.0301)/9(0.9699)= 0.0103 since the probability of getting over A+ is 0.0301 and anything else is one subtract 0.0301. do my answers seen correct? if not can someone guide me in the right direction?