Here is a simple probability model for multiple-choice tests. Suppose that each

Question

Here 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 = 0.82. (a) Use the Normal approximation to find the probability that Jodi scores 75% 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 75% 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

Explanation / Answer

Solution:

(a) Mean = np = 100(0.82) = 82
Standard deviation = sqrt [np(1-p)] = sqrt[100(0.82)(0.18)] = 3.8419

We want to find: P( x <= 75)

= 82
= 3.8419
standardize x to z = (x - ) /
P(x < 75) = P( z < (75-82) / 3.8419)
= P(z < -1.8220) = 1P ( Z<1.822 )
= 10.9656
= 0.0344
(From Normal probability table)

(b) Mean = np = 250(0.82) = 205
Standard deviation = sqrt [np(1-p)] = sqrt[250(0.82)(0.18)] = 6.0745

75% of 250 = 187.5
P( x <= 187.5) =

= 205
= 6.0745
standardize x to z = (x - ) /
P(x < 200) = P( z < (187.5-205) / 6.0745)
= P(z < -2.8809) = 1P (Z < 2.8809 )
= 10.998 =0.002
(From Normal probability table)

(c) Since std = sqrt(pq/n) = 0.82*0.18/100 = 0.0384
(1/2)sqrt(pq/n) = sqrt(pq/4n)
0.0192 = sqrt(0.82*0.18/4n)
0.0003686 = 0.82*0.18/4n
0.001474n = 0.1476
n=100.11 or 100

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