Consider a multiple-choice examination with 50 questions. Each question has four

Question

Consider a multiple-choice examination with 50 questions. Each question has four possible answers. Assume that a student who has done the homework and attended lectures has a 75% probability of answering any question correctly.

a. A student must answer 43 or more questions correctly to obtain a grade of A. What percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination?

b. A student who answers 35 to 39 questions correctly will receive a grade of C. What percentage of students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination?

c. A student must answer 30 or more questions correctly to pass the examination. What percentage of students who have done their homework and attended lectures will pass the examination?

d. Assume that a student has not attended class and has not done the homework for the course. Furthermore, assume that the student will simply guess at the answer to each question. What is the probability that this student will answer 30 or more questions correctly and pass the examination?

Explanation / Answer

a.)Let Xb be the number of correct answers. Xb has the binomial distribution with n = 50 trials and success probability p = 0.75

In general, if X has the binomial distribution with n trials and a success probability of p then
P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[Xb = x] = 0 for any other value of x.

To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p > 10 and n * (1-p) > 10.

Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker especially if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.

In this case you have:
n * p = 50 * 0.75 = 37.5 expected success
n * (1 - p) = 50 * 0.25 = 12.5 expected failures

We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.

If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean ? = n * p, variance ?

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