Question

1. The grade point average (GPA) of a large population of college students follows a normal distribution with mean 2.6, and standard deviation of 0.5. Students with GPA higher than 3.5 are considered “exceptional”, 3.0 to 3.5 are considered to be “good”, 2.0 to 3.0 are considered “average”, and below 2.0 are considered to be “poor”.

(a) For a randomly selected student, what is the probability that he has a “good” GPA? 


(b) Suppose 10 students are randomly selected. Let Y be the number of students with “good” GPA. Find the mean and variance of Y . 


(c) Suppose 200 students are randomly selected. Approximate the probability that at most 50 students have “good” or “exceptional” GPA.

Answer 1

(a) Let X denote the grade point average of the randomly selected student.

= 0.9640697 - 0.7881446

= 0.1759

(b) Y is the number of students with good GPA. then, Y has binomial distribution with n=10, p=0.1759

Mean of Y = np = 10*0.1759 = 1.759

Variance of Y = npq = 10*0.1759*(1-0.1759) = 1.45

(c) P(GPA>=3)

= 1 - 0.7881446

=0.2119

Let Y be the number of students with good or exceptional GPA. Then Y follows binomial with n=200, p=0.2119

= 0.90658