Here Is a simple probability model for multiple-choice tests. Suppose that each
ID: 3225392 • Letter: H
Question
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.
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