A multiple choice test has 25 questions, each with 5 possible choices, exactly o

Question

A multiple choice test has 25 questions, each with 5 possible choices, exactly one of which is correct. The test is marked by giving four points to each correct choice, and subtracting one point for each incorrect choice. No points are either given or subtracted for questions for which no choice is selected.Amanda and Beatrice each know the answer to 12 of the 25 questions, but are unsure about the rest. Amanda leaves the other 13 unanswered, whereas Beatrice answers them randomly, picking one choice for each question with equal probabilities, independently between questions.

Cynthia is also writing the same exam, and also is sure about only 12 questions. For the remaining questions, Cynthia is successfully able to eliminate three of the five choices as being clearly wrong, but she is not sure about which of the remaining two are correct. She therefore randomly and independently picks one of the two answers for these 13 questions.

(a) Find the expectation and variance of Cynthia’s mark.

Explanation / Answer

(a) Find the expectation and variance of Cynthia’s mark.

Cynthia is able to answer 12 questions. So the marks obtained for these 12 quetsions = 12 * 4 = 48 marks

For the remaining 13 questions, as Cynthia is successfully able to eliminate three of the five choices as being clearly wrong, the probability of correctly answering the questions = 1/2 = 0.5

This is the case of a binomial distribution, where the binomial experiment consists of n = 13 trials and results in x successes (correct answers). If the probability of success on an individual trial is P=0.5, then the binomial probability is:

b(x; n, P) = nCx * Px * (1 - P)n - x

So, b(x; 13, 0.5) = 13Cx * 0.5x * (1 - 0.5)13 - x

If x be the number of correct answers, then 13-x be the number of incorrect answers with P = 0.5. So,

b(13-x; 13, 0.5) = 13C13-x * 0.513-x * (1 - 0.5) x

= 13Cx * 0.5x * (1 - 0.5)13 - x = b(x; 13, 0.5)

So, the probability distribution of correct and incorrect anwers is same as b(x; 13, 0.5).

So, the distribution of the marks would be

4 * b(x; 13, 0.5) - 1 * b(x; 13, 0.5) = 3 * b(x; 13, 0.5)

Distribution of Cynthia marks = 48 + 3  * b(x; 13, 0.5)

Expectation of of Cynthia marks = E[48 + 3  * b(x; 13, 0.5) ]

= E[48] + 3* E[b(x; 13, 0.5)]

= 48 + 3 * 13 * 0.5 (Expectation of binomial distribution = nP)

= 67.5

Variance of of Cynthia marks = Var[48 + 3  * b(x; 13, 0.5) ]

= Var[48] + Var[3 * b(x; 13, 0.5)]

= Var[48] + 9* Var[b(x; 13, 0.5)]

= 0 + 9 * 13 * 0.5 * (1 - 0.5) (Varianve of binomial distribution = nP(1-P))

= 29.25

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