Question

11. (20) GPA distribution in UPW university is a normal distribution with an average of 2.68 and a standard deviation of 0.4.

(a) About what proportion of the students have GPA at most 3?

(b) About what proportion of the students’ GPA are between 2 and 3?

(c) The President of the university is establishing a new scholarship, the minimum qualification is that students GPA have to be among top 3%, what is the numerical GPA a student must have in order to qualify?

(d) A students’ club has a minimum GPA requirement of 2.7 or higher. You heard that Kelly is going to attend a club members’ meeting, you are wondering: what is the chance that Kelly’s GPA is lower than 3.3?

(e) If we randomly choose 7 students in the university, what is the chance that at least 2 have GPA over 3.3?

Answer 1

Let X be the GPA distribution

then

a) P( X < 3)

= P( z < 0.8)

= 0.7881 (from z table)

Proportion of students have GPA at most 3 = 0.7881

b) P(2< X < 3)

= P(-1.7 < z < 0.8)

= P(-1.7 < z < 0) +P( 0 < z < 0.8)

= 0.4554 + 0.2881 ( from z table)

= 0.7435  

Proportion of students have GPA between 2 and 3 = 0.7435

c) Given , P( X > c) =0.03

From z table

P( z > 1.88) = 0.03

The numerical value the student must have in order to qualify = 3.432

d) Kelly's GPA is more than 2.7 ( as she is in the club)

P(2.7< X < 3.3)

= P(0.05  < z < 1.55)

= P( z < 1.55) - P( z < 0.05)

= 0.9394  0.5199 ( from z table)

= 0.4195

Probability that Kelly's GPA is lower than 3.3 = 0.4195

e) P( X > 3.3 )

= P( z > 1.55)

= 1- P( z < 1.55)

= 1-0.9394

= 0.0606

Probability that GPA is more than 3.3 is p = 0.0606

Let Y be the number students have GPA over 3.3

Y follow Binomial with n =7 , p =0.0606

The probability mass function of Y

, y=0,1,2,...7

To find P( Y 2)

P( Y 2) = 1- P( Y )

= 1-( P( Y=0) +P( Y=1 ))

= 0.0629

Probability that at least 2 have GPA over 3.3 is 0.0629