# E x binn 300 p 024 with 300024 72 and 2

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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 right-tail 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 .

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