Chapter 8.1 hw

pg 462 #34 pg 461 #29 5C0 (.65)0 (.35)5 5C1 (.65)1 (.35)4 5C2 (.65)2 (.35)3 5C3 (.65)3 (.35)2 5C4 (.65)4 (.35)1 5C5 (.65)5 (.35)0 P(X = 0) P(X = 1) P(X = 2) P(X = 3) P(X = 4) P(X = 5) = (a) n = 5 and p = .65. This is a binomial distribution B(5, .65) (b) X can be any value from 0 to 5. (c) the chart and histogram follow: (d) mean = μ= 5 * .65 = 3.25 standard deviation = σ = √(5 * .65 * .35) = 1.067 Hint: to get all the answers at the same time, use binompdf(n,p) If you want to round the answers, say, to 3 decimal places, use round(binompdf(5,.65), 3) and this will give you the answers in the last column (try it!) 1. There are only 2 possibilities - detect antibodies or not. J 2. There are only 20 test samples, a fixed number. 3. Each sample is independent from the others. 4. The test has the same probability of detecting antibodies for each sample. (a) Is this distribution a binomial distribution? Test the 4 conditions. 20C20 (.99)20 (.01)0 (b) P(X = 20) = P(at least 1 unit is not detected) = 1 - .818 = .182 (c) mean = μ= 20 * .99 = standard deviation = σ = √(20 * .99 * .01) = Therefore, this is a binomial distribution B(20, .99) 0.00 0.00 00.00 00.00 00.00 0.00 0.00 00.00 0.00 0.00 0.00E-00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 00.00 0.00 0.00 0.00 0.00 0.00 00.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00