Suppose that a student in stat 2160 has not been attending class this semester b

Question

Suppose that 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 change of getting a correct answer, since it is a multiple choice exam with choices a,b,c,d, or e. 1. How many questions should the student expect to get correct on this exam? 2. What is the probability that the student will get a “C” or better, which is 16 or more correct? 3. What is the probability that the student will score lower than “C” (15 or less correct)? Suppose that 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 change of getting a correct answer, since it is a multiple choice exam with choices a,b,c,d, or e. 1. How many questions should the student expect to get correct on this exam? 2. What is the probability that the student will get a “C” or better, which is 16 or more correct? 3. What is the probability that the student will score lower than “C” (15 or less correct)? 1. How many questions should the student expect to get correct on this exam? 2. What is the probability that the student will get a “C” or better, which is 16 or more correct? 3. What is the probability that the student will score lower than “C” (15 or less correct)?

Explanation / Answer

From the information

p : probability of getting the correct answer = 1/5 =0.20

n = number of question in an exam =25

Define the random variable X

X : number of correct answer in an exam.

X follows binomial distribution with parameter n =25 and p=0.20

X ~ B( n=25 ,p=0.20)

P(X=x) = 25 C x 0.20x * 0.80 25 -x ; x= 0, 1,2............25

E(X) = np =25* 0.20 =5

Var(X) =npq =25 *0.2 *0.8 = 4

S.D(X) =2

1) E(X) =expected number of questions should the student expect to get correct = 5

2) Required probability = P ( X >= 16 )

By using Normal approximation

Z =(X-E(X) ) /(S.D(X)) ~ N(0,1)

P(X >=16 ) = P( (X-E(X) ) /(S.D(X)) > ( 16 - 5) /2)

= P(Z > 5.5 )

= 0

3) Required probability = P ( X <= 15) =1- P( X > 15)

by normal approximation

= 1- P( (X-E(X) ) /(S.D(X)) > ( 15 - 5) /2)

= 1 - P( Z > 5)

= 1-0

=1

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