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

Question

Here is a simple probability model for multiple-choice tests. Suppose 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.) Answers to different questions are independent. (Round your answers to four decimal places.)

(a) Stacey is a good student for whom p = 0.75. Use the Normal approximation to find the probability that Stacey scores between (and including) 70% and 80% on a 100-question test.

(b) If the test contains 250 questions, what is the probability that Stacey will score between (and including) 70% and 80%? You see that Stacey's score on the longer test is more likely to be close to her "true score."

Explanation / Answer

a) Stacey is a good student for whom p = 0.75. Use the Normal approximation to find the probability that Stacey scores between (and including) 70% and 80% on a 100-question test.

mean = n*p = 100*0.75 = 75
std = sqrt(n*p*q) = sqrt(75*0.25) = 4.330
----
Binomial: P(70<= x <=80) ~
z(70) = (70-75)/4.330 = -1.154
z(80) = (80-75)/4.330 = 1.154
------
Ans: p(-1.154 < z < 1.154) = normalcdf(-1.154,1.154) = 0.7498
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(b) If the test contains 250 questions, what is the probability that Stacey will score between (and including) 70% and 80%? You see that Stacey's score on the longer test is more likely to be close to her "true score."

mean = n*p = 250*0.75 = 180
std = sqrt(180*0.25) = 6.8465
Note: 0.7*250 = 175 ; 0.8*250 = 200
--------
z(175) = (175- 0.75(250))/6.8465 = -1.826
z(200) = (200-0.75(250))/6.8465 = 1.826

P( -0.8606 < Z < 3.4426) == normalcdf(--1.826 ,1.826) = 0.9328

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