951生物統(tǒng)計(jì)學(xué)作業(yè)5.

1、951 生物統(tǒng)計(jì)學(xué)作業(yè) 5第六章間斷型機(jī)率分布各題機(jī)率值請(qǐng)至少取到小數(shù) 3 位以上！一、 假設(shè)屏東縣居民贊成公投的機(jī)率為 0.65如果從此縣居民中隨機(jī)抽出15人, 讓X為其中”贊成公投”的人數(shù)，1.列出 X 的機(jī)率分布函數(shù)式。(只要數(shù)學(xué)式!)P X =x =f(x) C150.65X0.3515-X，X=0152.列出 X 的機(jī)率分布表。(針對(duì)每個(gè) x 值算出 f(x)xf(x)運(yùn)算方式00.0000C05漢0.65漢0.3515-0=15!漢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漢0.654X0.3515-4=15X0.654X0.3511=0.0028 4!(15-4 !50.0124C；5漢0.655漢0.3515-5=15X0.655漢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匯0.658X0.3515-8=15X0.658X0.357=0.1452 8!(15-8!90.4357C；5漢0.659x0.3515-9 15乂0.659x0.356 0.4357 9!(15-9100.6481C；0漢0.6510漢0.35=-匯0.65漢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漢0.65

4、12匯0.3515-12/5!嚴(yán)0.6512漢0.3530.938212!(15-12 !130.9858C；5漢0.6513匯0.3515-1315!匯0.6513漢0.352 0.985813!(15-13)140.9984151415-1415!141C；5沢0.65漢0.351-汽0.65漢0.35 - 0.998414!(15-14 !151.00001515015!150C；5匯0.65匯0.35=；- -漢0.65匯0.35= 1.000015!(15-15!3.15 人中，有 8 到 12 人贊成的機(jī)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