Question

A professor of a introductory statistics class has stated that, historically, the distribution of final exam grades in the course resemble a normal distribution with a mean final exam mark of 60% and a standard deviation of 9%.
(a) What is the probability that a randomly chosen final exam mark in this course will be at least 75%?
(b) In order to pass this course, a student must have a final exam mark of at least 50%. What proportion of students will not pass the statistics final exam?
(c) The top 2% of students writing the final exam will receive a letter grade of at least an A in the course. To four decimal places, find the minimum final exam mark needed on the statistics final to earn a letter grade of at least an A in the course.

Answer 1

X: final exam grades in the course

X follows normal distribution with a mean final exam mark of 60% and a standard deviation of 9%.

(a) Probability that a randomly chosen final exam mark in this course will be at least 75% = P(X>75)

P(X>75)=1-P(X 75)

Z-score for 75 = (75-mean)/standard deviation = (75-60)/9 = 15/9 = 1.67

From standard normal tables,

P(Z1.67) = 0.9525

P(X 75) = P(Z1.67) = 0.9525

P(X>75)=1-P(X 75) = 1-0.9525=0.0475

Probability that a randomly chosen final exam mark in this course will be at least 75% = 0.0475

(b)

In order to pass this course,

a student must have a final exam mark of at least 50%.

Proportion of students will not pass the statistics final exam = Probability that a randomly chosen final exam mark in this course will be at least 50%= P(X>50)

P(X>50)=1-P(X 50)

Z-score for 50= (50-mean)/standard deviation = (50-60)/9 = -10/9 = -1.11

From standard normal tables,

P(Z-1.11) = 0.1335

P(X 50) = P(Z-1.11) = 0.1335

P(X>50)=1-P(X 50) = 1-0.1335 =0.8665

Proportion of students will not pass the statistics final exam = 0.8665

(c)

Let the X_A : be the final exam mark needed on the statistics final to earn a letter grade of at least an A in the course

Top 2% of students writing the final exam will receive a letter grade of at least an A in the course i.e

P(X>X_A) = 2/100 =0.02

P(X>X_A) = 1-P(XX_A) =0.02

P(XX_A) = 1-0.02 =0.98

Let Z_A be the Z-score for X_A

P(ZZ_A) = P(XX_A) = 0.98

From standard normal tables,

P(Z2.05) = 0.97980.98

Z_A = 2.05;

Z_A = (X_A - 60)/9

2.05 = (X_A -60)/9

X_A = 60 + 9 x 2.05 = 60 + 18.45 = 78.45

Minimum final exam mark needed on the statistics final to earn a letter grade of at least an A in the course = 78.45