Question

Ada Nixon, a student, has just begun a 30-question, multiple-choice exam. For each question, there is exactly one correct answer out of four possible choices. Unfortunately, Ada hasn't prepared well for this exam and has decided to randomly select an answer choice for each question.

(a) (4 points) For a given question, what is the probability Ada picks the correct answer, assuming each answer choice is equally likely to be selected?

(b) Assume the number of questions Ada answers correctly is Binomially distributed.

i. (4 points) What is the average number of questions Ada will answer correctly? Show your work and round your nal answer to 1 decimal place.

ii. (3 points) If Ada answers at least 16 questions correctly, she will receive a passing grade. What is the probability that Ada receives a passing grade? Show your work, including any calculator functions you use, and round your nal answer to 3 decimal places.

(c) (3 points) Suppose Ada comes across a question for which she knows two of the answer choices are certainly wrong, which means the correct answer must be one of the two remaining choices. Assuming Ada will answer every other question using the same random selection procedure as before, will the number of questions she answers correctly remain binomially distributed? State Yes or No and explain why

Answer 1

a) P(correct answer) = 1/4 = 0.25

b)i) n = 30

p = 0.25

Average = n * p = 30 * 0.25 = 7.5 = 8

ii) 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) + P(X = 26) + P(X = 27) + P(X = 28) + P(X = 29) + P(X = 30)

= 30C16 * (0.25)^16 * (0.75)^14 + 30C17 * (0.25)^17 * (0.75)^13 + 30C18 * (0.25)^18 * (0.75)^12 + 30C19 * (0.25)^19 * (0.75)^11 + 30C20 * (0.25)^20 * (0.75)^10 + 30C21 * (0.25)^21 * (0.75)^9 + 30C22 * (0.25)^22 * (0.75)^8 + 30C23 * (0.25)^23 * (0.75)^7 + 30C24 * (0.25)^24 * (0.75)^6 + 30C25 * (0.25)^25 * (0.75)^5 + 30C26 * (0.25)^26 * (0.75)^4 + 30C27 * (0.25)^27 * (0.75)^3 + 30C28 * (0.25)^28 * (0.75)^2 + 30C29 * (0.25)^29 * (0.75)^1 + 30C30 * (0.25)^30 * (0.75)^0 = 0.001

c) No, the number of questions she answers correctly will not be binomially distributed. Because the probability of choosing correct answer for that particular question in which she knows two answer choices are wrong = 1/2 = 0.5

Since the probability of choosing correct answer for all the questions are not same, so it will not be binomially distributed.