Suppose a student in Stat 2160 has not been attending class this semester but de

Question

Suppose a student in Stat 2160 has not been attending class this semester but decides to take the exam anyway. If he randomly guesses on each of the 25 questions, then he has a 1 out of 5 chance of getting a correct answer, since it is a multiple choice exam with choices a, b, c,d, or e.

(a) How many questions should the student expect to get correct on this exam?

(b) What is the probability that the student will get a "C" or better, which is 16 or more correct?

(c) What is the probability that the student will score lower than a "C" (15 or less correct)?

Explanation / Answer

There are 25 questions. Each question has 5 options out of which one is correct.

So the probability of getting a correct answer in each case is 1/5 and that of incorrect answer is 4/5. So, this problem can be taken as B(25 , 1/5).

Note: Binomial Distibution, the pmf is given by B(x; n, p) = nCx * px * (1 - p)n - x  or

P(X=x) = nCx * px * (1 - p)n - x

where,

a) There are 25 questions, and on average 1/5 of the questions will be answered correctly.

So, number of  questions should the student expect to get correct on this exam = Expected value of X, E(X) = n* P(x) = 25 * 1/5 = 5   

So, student should expect 5 correct questions in this exam.

b) The probability that the student will get a "C" or better, which is 16 or more correct is P (X>16) :

P (X>16) = P(X= 16) + P(X= 17) + P(X= 18) + P(X= 19) + P(X= 20) + P(X= 21) + P(X= 22) +

P(X= 23 ) + P(X=24) + P(X= 25)

= 25C16 * (1/5)16 * (1 - 1/5)25 - 16 + 25C17 * (1/5)17 * (1 - 1/5)25 - 17 + 25C18 * (1/5)18 * (1 - 1/5)25 -18

+25C19 * (1/5)19 * (1 - 1/5)25 - 19 +25C20 * (1/5)20 * (1 - 1/5)25 - 20 +25C21 * (1/5)21 * (1 - 1/5)25-21

+25C22 * (1/5)22 * (1 - 1/5)25 - 22 +25C23 * (1/5)23 * (1 - 1/5)25 - 23 + 25C24 * (1/5)24 * (1 -1/5)25-24

+25C25 * (1/5)25 * (1 - 1/5)25 - 25

=  0.00000206392

c) The probability that the student will score lower than a "C" (15 or less correct) is  P (X<15) :

P (X<15) = 1 - P (X>16) = 1 - 0.00000206392 = 0.099999793608

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