(Normal approximation to Binomial) Suppose that each student has probability p o
ID: 3331445 • Letter: #
Question
(Normal approximation to Binomial) Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. The correctness of an answer to a question is independent of the correctness of answers to other questions.
(a)Roxi is a good student for whom p = 0.85. Use Normal approximation to find the probability that Roxi will score 90% or better on a 100 question test. (
b)If the test contains 200 questions what is the probability that Roxi will score 90% or better?
(c)Zulekha is a weaker student for whom p = 0.65. Suppose the passing grade for the test is 70%. What is the probability that Zulekha will pass a 200 question test? Suppose she is given a choice to write a 100 question test instead. Should she write the shorter test?
Explanation / Answer
a) p=0.85 ; n=100
here std error =(p(1-p)/n)1/2 =0.0357
probability that Roxi will score 90% or better =P(X>0.90)=1-P(X<0.90)=1-P(Z<(0.9-0.85)/0.0357)=1-P(Z<1.4003)
=1-0.9193 =0.0807
b) for n=200; std error =(p(1-p)/n)1/2 = 0.0252
probability that Roxi will score 90% or better =P(X>0.90)=1-P(X<0.90)=1-P(Z<(0.9-0.85)/0.0252)=1-P(Z<1.9803)
=1-0.9762 =0.0238
c) for Zulekha p=0.65 ; n=200
std error =(p(1-p)/n)1/2 = 0.0337
probability that Zulekha will pass a 200 question test=P(X>0.70)=1-P(X<0.70)=1-P(Z<1.4825)=1-0.9309=0.0691
for 100 question test:
std error =(p(1-p)/n)1/2 = 0.0477
probability that Zulekha will pass a 100 question test=P(X>0.70)=1-P(X<0.70)=1-P(Z<1.0483)=1-0.8527=0.1473
therefore in 100 question test probability of her passing the test increases threfore she should write the test,
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