SUPP 2.7 In a certain population 27% of the residents are overweight. Choose three residents at random. Let X: the number of overweight residents in the sample. a) Find the probability all three residents in the sample are overweight. P(X=3) = ? If you are stuck, look at the table below which has some values filled in. b) Find the probability none of the three residents in the sample are overweight. P(X=0)= ? Find the probability that all three are not overweight. How is this value related to P(X=0)? c) What are the valid values for X? (I) There are eight possible arrangements for the sample of three where 0=overweight, N=not overweight. For example (0,0,0) means all three are overweight. (N,0,0) means that the first person is not overweight, but the second and third residents are overweight. List all eight arrangements. e) Find the probability of each arrangement (they are not all equally likely), and check that the probabilities add to one. f) Find the P(X=0), P(X=1), P(X=2), P(X=3) using your answers above. You may find it helpful to write your answers in this table. P(X=x) (1-0.27)*(1-0.27)*(1-0.27) (0.73*0.73*0.27)*3 0.73*0.27*0.27*3= 0.27*0.27*0.27 =0.73*0.73*0.73 =D.431649 ? =0.019683 =0.389017 g) Graph (pa per and pencil are great) the probability distribution of X. (X on x-axis and probability on the y-axis). You don't need to submit this for the hw since it is awkward just graph it for yourself. h) Find the E(X). COMMENT: WE WILL SEE IN WEEK 3 THAT THIS IS A BINOMIAL PROBLEM, AND THERE IS A COMPUTER FUNCTION FOR IT. WE WILL SEE IN WEEKS 4 AND 5 THAT THESE PROBABILITIES CAN BE VIEWED AS P-VALUES AND ARE CENTRAL T0 INFERENCE . . .50 THIS IS HOW THIS ALL FITS TOGETHER! !!! SUPP 2.8 A sample of three patients with the flu is selected. Let X: number in the sample that are fever-free after 3 days on Tamiflu. The distribution of X is given in this table. X=Nu mber 0 1 2 3 fever-free Prob 0.2746 0.4436 0.2389 0.0429 Find the mean (= E(X))