Question: In 2006 the average combined MCAT score ( physical science, verbal reasoning, biological science) for students in Canada was 24.7. Suppose that approximately 45% of all students took this test, and that 100 students are randomly selected from throughout Canada.
Which of the following random variables has an approximate binomial distribution? If possible, give the values for n and p.
a. The number of students who took the MCAT
b. The scores of the 100 students on the MCAT
c. The number of students who scored above average on the MCAT
d. The amount of time it took each student to complete the MCAT  Edit

Answer: a. The number of students who took the MCAT =n= 45% of 100=45
b. The scores of the 100 students on the MCAT=24.7

c. np=24.7 so, p=24.7/n=24.7/45=0.55
Need to find P(x>24.7),
z= (x-np)/sqrt(npq)=(24.7-24.7)/sqrt(npq)=0

P(x>24.7)=P(z>0)=0.5

n= 24.7/0.5=49.4=50(Ans.)

  Edit