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Unformatted text preview: ng evidence that the prior estimate of a 24% infection rate was too
low? Same question for if 75 ﬁsh were found to be infected.
SOLUTION: Let X indicate the number of ﬁsh infected by parasite, i.e. X ∼ Bin(n =
300, p = 0.24), with µ = 300(0.24) = 72 and σ 2 = 300(0.24)(0.76) = 54.72.
a. Since np = 72 and n(1 − p) = 228 are both greater than 10, normal approximation to
binomial with continuity correction is adopted for approximation.
60 − 1 − 72
84 + 1 − 72
√2
≤Z≤ √ 2
)
54.72
54.72
= P(−1.69 ≤ Z ≤ 1.69) = 2P(Z ≤ 1.69) − 1 = 2(0.9545) − 1 = 0.909. P(60 ≤ X ≤ 84) ≈ P( 85− 1 −72 2
b. P(X ≥ 85) ≈ P(Z ≥ √54.72 ) = P(Z ≥ 1.69) = 1 − 0.9545 = 0.0455. Again, observing
such an unusual event (prob < 0.05) would provide strong evidence against the claim. The
infection rate is unlikely to be as low as 24%. 75− 1 −72 2
c. P(X ≥ 75) ≈ P(Z ≥ √54.72 ) = P(Z ≥ 0.34) = 1 − 0.6331 = 0.3669. An event with
probability 0.37 of happening is not unusual under the claim.
Note that the reason why we look at the righttail probability is because 75 is above the
mean (72)....
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This document was uploaded on 02/15/2014 for the course MATH 231 at Lehigh University .
 Fall '11
 DANIELCONUS
 Statistics, Probability, Standard Deviation

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