Question

Your friend’s professor gives out reasonably hard exams 70% of the time, and ridicu- lously hard exams 30% of the time. On hard exams, each student’s score on the exam is a normally distributed random variable with μH = 70 and σH = 10. On ridiculously hard exams, each student’s score on the exam is a normally distributed random variable with μR = 50 and σR = 15. Suppose you have four friends in the class, not just one. Let A be the average score of your four friends: A= (F1 +F2 +F3 +F4)/ 4 Where F1 is your first friend’s score, and F2 is your second friend’s score. Find E[A] and V ar(A) if the exam is ridiculously hard. (e) Find E[A] and V ar(A) if the exam is reasonably hard. (f) Since A is the sum of normal random variables, it is itself a normal random variable. Find P (A > 65) if the exam is reasonably hard, and if it is ridiculously hard. (g) If A is greater than 65, what is the probability the exam was ridiculously hard?

Answer 1

(d)

E[A] and V ar(A) if the exam is ridiculously hard

On ridiculously hard exams, each student’s score on the exam is a normally distributed random variable with = 50 and = 15

Let A be the average score of your four friends: A= (F1 +F2 +F3 +F4)/ 4

F1,F2,F3,F4 follow normal distribution with mean   = 50 and standard deviation = 15

A= (F1 +F2 +F3 +F4)/ 4

E[A] = (E[F1] +E[F2] +E[F3}+E[F4])/ 4 = =

Var[A] =( Var[F1]+Var[F2]+Var[F3]+Var[F4]) / 4² =

E[A] = = 50

Var[A] = =

(e)

E[A] and V ar(A) if the exam is reasonably hard

On hard exams, each student’s score on the exam is a normally distributed random variable with = 70 and = 10

Let A be the average score of your four friends: A= (F1 +F2 +F3 +F4)/ 4

F1,F2,F3,F4 follow normal distribution with mean   = 70 and = 10

A= (F1 +F2 +F3 +F4)/ 4

E[A] = (E[F1] +E[F2] +E[F3}+E[F4])/ 4 = =

Var[A] =( Var[F1]+Var[F2]+Var[F3]+Var[F4]) / 4² =

E[A] = = 70

Var[A] = =

(f)

P (A > 65) if the exam is reasonably hard

if the exam is reasonably hard, A follows normal distribution with mean 70 and standard deviation := =5

P (A > 65) = 1 - P(A65)

Z-score for 65 = 65-70/5 = -5/5 = -1

From standard normal tables, P(Z -1) = 0.1587

P(A65) = P(Z -1) = 0.1587

P (A > 65) = 1 - P(A65) = 1-0.1587= 0.8413

P (A > 65) if the exam is reasonably hard = 0.8413

P (A > 65) if the exam is ridiculously hard

if the exam is ridiculously hard, A follows normal distribution with mean 50 and standard deviation := =7.5

P (A > 65) = 1 - P(A65)

Z-score for 65 = 65-50/7.5 = 15/7.5 = 2

From standard normal tables, P(Z 2) = 0.9772

P(A65) = P(Z 2) = 0.9772

P (A > 65) = 1 - P(A65) = 1-0.9772= 0.0228

P (A > 65) if the exam is ridiculously hard = 0.0228

(g)

If A is greater than 65, probability the exam was ridiculously hard

Let X : Event of A greater than 65

H: Exam is reasonably hard

R : exam is ridiculously hard

From the above :

Probability that A > 65 Given that the Exam is reasonably hard P(X|H) = 0.8413

Probability that A > 65 Given that the Exam is ridiculously hard P(X|R) = 0.0228

Professor gives out reasonably hard exams 70% of the time, and ridiculously hard exams 30% of the time

i.e

P(H) =0.7

P(R) = 0.3

If A is greater than 65, probability the exam was ridiculously hard = P(R|X)

From Bayes theorem

If A is greater than 65, probability the exam was ridiculously hard = P(R|X) = 0.011481326

If A is greater than 65, probability the exam was ridiculously hard = 0.011481326