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 65% chance of answering any question correctly. (Round your answers to two decimal places.)

(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? Use the normal approximation of the binomial distribution to answer this question.

(b)

A student who answers 36 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 30 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 30 or more questions correctly and pass the examination? Use the normal approximation of the binomial distribution to answer this question.

Answer 1

NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 50 * 0.65 = 32.5
standard deviation ( √npq )= √50*0.65*0.35 = 3.37268439080801
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)

a.
A student must answer 43 or more questions correctly to obtain a grade of A
percentage of the students who have done their homework and attended lectures will obtain a grade of A
on this multiple-choice examination
P(X < 43) = (43-32.5)/3.3727
= 10.5/3.3727= 3.1132
= P ( Z <3.1132) From Standard Normal Table
= 0.9991
P(X > = 43) = (1 - P(X < 43))
= 1 - 0.999074749815058 = 0.0009252
= 0.092%

b.
A student who answers 36 to 39 questions correctly will receive a grade of C.
percentage of students who have done their homework and attended lectures will obtain a grade of C
on this multiple-choice examination
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 36) = (36-32.5)/3.3727
= 3.5/3.3727 = 1.0377
= P ( Z <1.0377) From Standard Normal Table
= 0.85031
P(X < 39) = (39-32.5)/3.3727
= 6.5/3.3727 = 1.9272
= P ( Z <1.9272) From Standard Normal Table
= 0.97303
P(36 < X < 39) = 0.97303-0.85031 = 0.1227
= 12.27%

c.
A student must answer 30 or more questions correctly to pass the examination
percentage of the students who have done their homework and attended lectures will pass the examination
P(X < 30) = (30-32.5)/3.3727
= -2.5/3.3727= -0.7412
= P ( Z <-0.7412) From Standard Normal Table
= 0.2293
P(X > = 30) = (1 - P(X < 30))
= 1 - 0.229272182966224 = 0.77072
= 77.072%

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.
the probability that this student will answer 30 or more questions correctly
and pass the examination
P(X < 30) = (30-32.5)/3.3727
= -2.5/3.3727= -0.7412
= P ( Z <-0.7412) From Standard Normal Table
= 0.2293
P(X > = 30) = (1 - P(X < 30))
= 1 - 0.229272182966224 = 0.7707