951生物统计学作业5.

1、951 生物統計學作業 5第六章間斷型機率分布各題機率值請至少取到小數 3 位以上！一、 假設屏東縣居民贊成公投的機率為 0.65如果從此縣居民中隨機抽出15人, 讓X為其中”贊成公投”的人數，1.列出 X 的機率分布函數式。(只要數學式!)P X =x =f(x) C150.65X0.3515-X，X=0152.列出 X 的機率分布表。(針對每個 x 值算出 f(x)xf(x)運算方式00.0000C05汉0.65汉0.3515-0=15!汉0.65帳0.3515= 0.00000!(15-0)10.0000C；5X0.651X0.3515-1=15!x0.65q 0.3514= 0.000

2、01!(15-1 !20.000115215-215!213C2工0.65疋0.35=工0.65疋0.35= 0.00012!(15-2J30.000515315 315!312C3疋0.65 X0.35-=-:X0.65 0.35 = 0.000533!(15-340.0028C；5汉0.654X0.3515-4=15X0.654X0.3511=0.0028 4!(15-4 !50.0124C；5汉0.655汉0.3515-5=15X0.655汉0.3510=0.0124 5! (15-5)60.0422C；5沢0.656X0.3515-6=15X0.656X0.359=0.04226!(1

3、5-670.1132C；5X0.657X0.3515-7=15x0.657x0.358= 0.11327!(15-7)80.1452C；5汇0.658X0.3515-8=15X0.658X0.357=0.1452 8!(15-8!90.4357C；5汉0.659x0.3515-9 15乂0.659x0.356 0.4357 9!(15-9100.6481C；0汉0.6510汉0.35=-汇0.65汉0.355= 0.648110!(15-10)110.8273Cl：汶0.6511x0.35151115!乂0.6511疋0.354 0.827311!(15-11!120.9382C；5汉0.65

4、12汇0.3515-12/5!严0.6512汉0.3530.938212!(15-12 !130.9858C；5汉0.6513汇0.3515-1315!汇0.6513汉0.352 0.985813!(15-13)140.9984151415-1415!141C；5沢0.65汉0.351-汽0.65汉0.35 - 0.998414!(15-14 !151.00001515015!150C；5汇0.65汇0.35=；- -汉0.65汇0.35= 1.000015!(15-15!3.15 人中，有 8 到 12 人贊成的機率為何?PX =4)=1-f(3)=1-0.9662 =0.0338(POISSON)3.What is the probability that at most two cases of teta nus will be reported during a give n week? (Suppose that there are 4 weeks in a give n mon th) 平均=124=0.3,P(X=2)=f(2)= 0.9964(POISSON)4.What is the probability that at least eight cases will be reported