Question 1054112: 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 = 0.71. 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."
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (a) Stacey is a good student for whom p = 0.71. 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*0.71 = 71
std = sqrt(n*p*q) = sqrt(71*0.29) = 4.5376
----
Binomial: P(70<= x <=80) ~ Normal:P(69.5<= x <=80.5)
z(69.5) = (59.5-71)/4.5376 = -2.5344
z(80.5) = (80.5-71)/4.5376 = 2.0936
------
Ans: p(-2.7344 < z < 2.0936) = normalcdf(-2.7344,2.0936) = 0.9762
-----------------------------------------
(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*0.71 = 177.5
std = sqrt(177.5*0.29) = 7.1746
Note: 0.7*250 = 175 ; 0.8*250 = 200
--------
z(175) = (175-177.5)/7.1746 = -2.5
z(200) = (200-177.5)/7.1746 = 3.1363
---
etc.
I'll let you finish the problem.
Cheers,
Stan H.
-------------
|
|
|
