Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

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

ID: 3225392 • 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 72% or lower on a 100-question test. (Round your answer to four decimal places.) (b) If the test contains 2.50 questions, what is the probability that Jodi will score 72% 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 proportion of correct answers to half Its value for a 100-item test? questions (d) Laura is a weaker student for whom p = 0.71. Does the answer you gave In (c) for standard deviation of Jodi's score apply to Laura's standard deviation also?

Explanation / Answer

(a) Mean = p ( where n = 100, p = 0.76)

so mean = p = 0.76 = 0.76

Std. deviation = sqrt [p(1-p)/n] = sqrt [ 0.76 * 0.24/100 ] = 0.0427

so Pr (p <=0.72; 0.76; 0.0427) = P ( X <0.72 + 0.005 ; 76; 4.271) = P(X < 72.5; 76; 4.271)

(this is done by using binomial correction factorr)

Z - value = (0.725 - 0.76)/ 0.04271 = -0.8195

Pr((X <=72; 76; 4.271)) = 0.2061 [ from Z table]

(ii) Mean = p ( where n = 250, p = 0.76)

so mean = p = 0.76 = 0.76

Std. deviation = sqrt [p(1-p)/n] = sqrt [ 0.76 * 0.24/250 ] = 0.027

so Pr (p <=0.72; 0.76; 0.0427) = P ( X <0.72 + 0.005 ; 76; 4.271) = P(X < 72.5; 76; 4.271)

(this is done by using binomial correction factorr)

Z - value = (0.725 - 0.76)/ 0.027 =-1.2963

Pr((X <=72; 76; 4.271)) = 0.1093 [ from Z table]

(c) Standar deviaition is inversely proportion of square root of sample size

s100 / sn = 2/1 = sqrt (n/100)

n = 400

(d) Yes, the answer given is applicable in the case of Laura also.Laura' standard deviation of score will also get half when we will increase number of question in a test to 4 times.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote