Question

3.2 Our overworked student trudges to his 3rd exam of the day, a true/false exam with 100 questions. Again,he just guesses the answers. Notice that this time he has a 50% chance of getting each particular question correct. (10 Marks: 2 Marks for each of parts a), b), c), d) and e).

a. What is the probability that the student gets at least one question correct?

b. What is the probability that the student gets between 15 and 35 questions (inclusive) correct?

c. What is the probability that the student gets between 40 and 60 questions (inclusive) correct?

d. What is the probability that the student passes the exam (gets 60 or more questions correct)?

e. Why does the student have a better chance of passing the exam in question 3.2 than the one in question 3.1?

Answer 1

Number of questions, n = 100

P(correct answer), p = 0.5

q = 1 - p = 0.5

a) P(the student gets at least one question correct) = 1 - P(student getting all questions wrong)

= 1 - 0.5¹⁰⁰

= 1

b) Normal approximation for binomial distribution

Mean, np = 100 x 0.5

= 50

Standard deviation,

= 5

P(X < A) = P(Z < (A - mean)/standard deviation)

P(the student gets between 15 and 35 questions (inclusive) correct) = P(15 X 35)

= P(X < 35.5) - P(X < 14.5) (continuity correction of 0.5 is applied)

= P(Z < (35.5 - 50)/5) - P(Z < (14.5 - 50)/5)

= P(Z < -2.9) - P(Z < -7.1)

= 0.0019 - 0

= 0.0019

c) P(the student gets between 40 and 60 questions (inclusive) correct) = P(40 X 60)

= P(X < 60.5) - P(X < 39.5) (continuity correction of 0.5 is applied)

= P(Z < (60.5 - 50)/5) - P(Z < (39.5 - 50)/5)

= P(Z < 2.1) - P(Z < -2.1)

= 0.9821 - 0.0179

= 0.9642

d) P(gets 60 or more questions correct) = 1 - P(X < 60)

= 1 - P(X < 59.5) (continuity correction of 0.5 is applied)

= 1 - P(Z < (59.5 - 50)/5)

= 1 - P(Z < 1.9)

= 1 - 0.9713

= 0.0287