Question

You may need to use the appropriate appendix table or technology to answer this 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 probability of 0.70 of answering any question correctly.

(a)

A student must answer 45 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? Use the normal approximation of the binomial distribution to answer this question.

(b)

A student who answers 34 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? Use the normal approximation of the binomial distribution to answer this question.

(c)

A student must answer 28 or more questions correctly to pass the examination. What percentage of the students who have done their homework and attended lectures will pass the examination? Use the normal approximation of the binomial distribution to answer this question.

(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 28 or more questions correctly and pass the examination? Use the normal approximation of the binomial distribution to answer this question.

Answer 1

a)
p=   0.7
n=   50
E(X) = np = 35
SD(X) = sqrt(npq) = 3.240370349

P( X>45) = 1 - P(X<45) = ?
  
I know that, z = (X-mean)/(sd)  
z1 = (45-35)/3.24037034920393) =   3.0861
  
hence,  
P( X>45)
= 1- P(Z<3.0861)
= 1 - NORMSDIST(3.0861)
= 0.0010

b)
p=   0.7
n=   50
E(X) = np = 35
SD(X) = sqrt(npq) = 3.240370349

P(34 < X < 39) = P(X<39) - P(X<34) = ?
  
I know that, z = (X-mean)/(sd)  
z1 = (34-35)/3.24037034920393) = -0.3086
z2 = (39-35)/3.24037034920393) = 1.2344
  
hence,  
P(34 < X < 39)
= P(Z<1.2344) - P(Z<-0.3086)
= NORMSDIST(1.2344) - NORMSDIST(-0.3086)
= 0.5127

c)
p=   0.7
n=   50
E(X) = np = 35
SD(X) = sqrt(npq) = 3.240370349

P( X>28) = 1 - P(X<28) = ?
  
I know that, z = (X-mean)/(sd)  
z1 = (28-35)/3.24037034920393) = -2.1602
  
hence,  
P( X>28)
= 1- P(Z<-2.1602)
= 1 - NORMSDIST(-2.1602)
= 0.9846

d)
p = 1/4 = 0.25
n = 50
E(X) = np = 12.5
SD(X) = sqrt(npq) = 3.061862178

P( X>28) = 1 - P(X<28) = ?
  
I know that, z = (X-mean)/(sd)  
z1 = (28-12.5)/3.06186217847897) =   5.0623
  
hence,  
P( X>28)
= 1- P(Z<5.0623)
= 1 - NORMSDIST(5.0623)
= 0.0000