11. 1 points MintroStat6 5.E.033. My Notes Ask Your Teacher Here is a simple pro

Question

11. 1 points MintroStat6 5.E.033. My Notes Ask Your Teacher 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 = 0.79. (a) Use the Normal approximation to find the probability that Jodi scores 73% 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 73% 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 (d) Laura is a weaker student for whom p = 0.74. Does the answer you gave in (c) for standard deviation of Jodi's score apply to Laura's standard deviation also? O Yes, the smaller for Laura has no effect on the relationship between the number of questions and the standard deviation O No, the smaller p for Laura alters the relationship between the number of questions and the standard deviation

Explanation / Answer

Pr(Jodi) = 0.79

(A) Here if we do normal approximation than E(X) = 100 * 0.79 = 79

Standard error of proportion sep = sqrt [100 * 0.79 * 0.21] = 4.073

Here, we will do continuity correction also

Pr(X 73) = NORMAL (x < 73.5) = NORMAL (x < 73.5 ; 79 ; 4.5376)

Z = (73.5 - 79)/4.0.73 -1.35

Pr(X 73) = NORMAL (x < 73.5) = NORMAL (x < 73.5 ; 79 ; 4.5376) = Pr(Z < -1.35) = 0.0885

(b) Here n = 250

Here if we do normal approximation than E(X) = 250 * 0.79 = 197.5

Standard error of proportion sep = sqrt [250 * 0.79 * 0.21] = 6.44

so here the number of maks when she scores 73% = 250 * 0.73= 182.5

Here, we will do continuity correction also

Pr(X 182.5) = NORMAL (x < 182.5) = NORMAL (x < 182.5 ; 250; 6.44)

Z = (182.5 - 197.5)/6.44 = -2.329

Pr(X 182.5) = NORMAL (x < 182.5) = NORMAL (x < 182.5 ; 250; 6.44) = Pr(Z < -2.329) = 0.01

(c) Here as the standard deviation is inversely proportion of square root of sample size so to reduce the sample size into half, we have to increase the sample size by 4 times so new sample size would be

n = 100 question.

(d) Yes, the answer applies to Laura also as smaller p for laura doesn't effect the ration in which standard deviation is getting reduced or increased.

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