Normal approximation to Binomial) Suppose that each student has probability p of
ID: 2929138 • Letter: N
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) here mean proportion =0.85
and n=100
therefore std error =(p(1-p)/n)1/2 =0.0357
hence 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)
n=200
therefore std error =(p(1-p)/n)1/2 =0.0252
hence 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)
mean proportion =0.65
and n=200
therefore std error =(p(1-p)/n)1/2 =0.0337
hence probability thatprobability that Zulekha will pass =P(X>0.70)=1-P(X<0.70)=1-P(Z<(0.7-0.65)/0.0337)
=1-P(Z<1.4825)=1-0.9309 =0.0691
for n=100
therefore std error =(p(1-p)/n)1/2 =0.0477
hence probability thatprobability that Zulekha will pass =P(X>0.70)=1-P(X<0.70)=1-P(Z<(0.7-0.65)/0.0477)
=1-P(Z<1.0483)=1-0.8527 =0.1473
as probability of her passing is higher for a 100 question test ; she should write the shortest test,
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