Here is a simple probability model for multiple-choice tests. Suppose that each
ID: 3327474 • Letter: H
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 = 0.76. (a) Use the Normal approximation to find the probability that Jodi scores 71% or lower on a 100-question test. (Round your answer to four decimal places.) .1208 Correct: Your answer is correct. (b) If the test contains 250 questions, what is the probability that Jodi will score 71% or lower? (Use the normal approximation. Round your answer to four decimal places.) .0320 Correct: Your answer is correct. (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? (I have done A and B but I can't get C in bold) A is .1208 and B is .0320
Explanation / Answer
NORMAL DISTRIBUTION
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 ) = 0.76
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.76*0.24/100)
=0.0427
a.
P(X < 0.71) = (0.71-0.76)/0.0427
= -0.05/0.0427= -1.171
= P ( Z <-1.171) From Standard Normal Table
= 0.1208
b.
P(X < 0.71) = (0.71-0.76)/0.027
= -0.05/0.027= -1.8519
= P ( Z <-1.8519) From Standard Normal Table
= 0.032
c.
since standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.76*0.24/100)
=0.0427
Since std = sqrt(pq/n)
(1/4)sqrt(pq/n) = sqrt(pq/16n)
0.0427/4 = sqrt ( 0.76 * 0.24 ) / 16n
0.010675 = sqrt ( 0.76 * 0.24 ) / 16n
0.000113955625 = ( 0.76 * 0.24 ) / 16n
16n = ( 0.76 * 0.24 ) / 0.000113955625
16n = 1600.6230
n = 100.038 ~ 100