ere is a simple probability model for multiple-choice tests. Suppose that each s
ID: 3356179 • Letter: E
Question
ere is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher p than a weak student.) The correctness of answers to different questions are independent. Jodi is a good student for whom p = 0.83. (a) Use the Normal approximation to find the probability that Jodi scores 77% or lower on a 100-question test. (Round your answer to four decimal places.) (b) If the test contains 250 questions, what is the probability that Jodi will score 77% or lower? (Use the normal approximation. Round your answer to four decimal places.) (c) How many questions must the test contain in order to reduce the standard deviation of Jodi's proportion of correct answers to half its value for a 100-item test? questions (d) Laura is a weaker student for whom p = 0.78. Does the answer you gave in (c) for standard deviation of Jodi's score apply to Laura's standard deviation also? Yes, the smaller p for Laura has no effect on the relationship between the number of questions and the standard deviation. No, the smaller p for Laura alters the relationship between the number of questions and the standard deviation.
Explanation / Answer
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.83
a.
for n= 100,
proportion ( p ) = 0.83
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.83*0.17/100)
=0.0376
P(X > 0.77) = (0.77-0.83)/0.0376
= -0.06/0.0376 = -1.5957
= P ( Z >-1.596) From Standard Normal Table
= 0.9447
P(X < = 0.77) = (1 - P(X > 0.77)
= 1 - 0.9447 = 0.0553
b.
for n=250,
proportion ( p ) = 0.83
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.83*0.17/250)
=0.0238
P(X > 0.77) = (0.77-0.83)/0.0238
= -0.06/0.0238 = -2.521
= P ( Z >-2.521) From Standard Normal Table
= 0.9941
P(X < = 0.77) = (1 - P(X > 0.77)
= 1 - 0.9941 = 0.0059
c.
Since std = sqrt(pq/n)
(1/2)sqrt(pq/n) = sqrt(pq/4n)
4*100 = 400
d.
No, the smaller p for Laura alters the relationship between the number
of questions and the standard deviation.
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