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 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? 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

Let

Xb represent the binomial distribution,

Xn represent the normal distribution.

The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution.

So, the normal approximation of the binomial distribution is given as,

P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5))

P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5))

a)

P( Xb ≥ 43) ≈ P( Xn ≥ (43 - 0.5)) ≈ P(X ≥ 42.5)

Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ

P(X ≥ 42.5)=

= P( Z ≥ (42.5 - .65*(50)) / sqrt(50*.65*.35) )

= P( Z ≥ ( 42.5 - 32.5)/ 3.37268 )

= P( Z ≥ 2.96499 )

=0.0015

=0.15%

b)

Student who answers more than 35 questions correctly,

P( Xb ≥x)≈ P( Xn ≥ (x-0.5) ≈ P( Xn ≥34.5)

=P(Z≥ (34.5-32.5) / 3.37268)

=P(Z≥0.59299)

=0.2766

Student who answers less than 39 questions correctly,

P(Xb≤x)= P(Xn≤ (x + 0.5) =P(Xn≤39.5)

=P(Z≤ (39.5-32.5) / 3.37268) =P(Z≤2.07549) =0.981

Students must answer 35 to 39 questions correctly,

= 0.981 – 0.2766

= 0.7044

=70.44%

c)

Student must answer 28 or more questions correctly,

P(Xb≥x)= P(Xn≥ (x-0.5) =P(Xn≥27.5)

=P(Z≥ (29.5-32.5) / 3.37268)

=P(Z≥-1.48249)

=0.9309

=93.09%