You are to take a multiple-choice exam consisting of 64 questions with 5 possible responses to each question. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let x represent the number of correct responses on the test. (a) What is your expected score on the exam? (Hint: Your expected score is the mean value of the x distribution.)(b) Compute the variance and standard deviation of x. Variance =Standard deviation

For each question, there are only two possible outcomes. Either you guesses the answer correctly, or you does not. The probability of guessing the answer of a question correctly is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The variance of the binomial distribution is:

[tex]V(X) = np(1-p)[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

64 questions.

So [tex]n = 64[/tex]

5 possible answers, one correctly, chosen at random:

So [tex]p = \frac{1}{5} = 0.2[/tex]

(a) What is your expected score on the exam?

[tex]E(X) = np = 64*0.2 = 12.8[/tex]

(b) Compute the variance and standard deviation of x. Variance =Standard deviation

[tex]V(X) = np(1-p) = 64*0.2*0.8 = 10.24[/tex]

Variance 10.24

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{64*0.2*0.8} = 3.2[/tex]

Standard deviation 3.2