解ï¼3é¢ï¼(1)âµéæºäºä»¶åççæ¦çä¹å为1ï¼å³âP(x=i)=1ï¼â´3a+1/6+3a+a+11/30=1ï¼â´a=1/15ã(2)å½X=-2ã-1ã0ã1ã3ï¼Y=X^2-1对åºçå¼åå«æ¯3ã0ã-1ã0ã8ï¼èP(Y=0)=P(X=1)+P(X=-1)=1/6+1/15=7/30ï¼â´Yçåå¸å¾ä¸ºï¼Y=-1ã0ã3ã8ï¼å¯¹åºæ¦çåå«æ¯1/5ã7/30ã1/5ã11/30ã(3)P(0â¤Yâ¤3)=7/30+11/30=3/5ã
4é¢ï¼(1)âµâ«(-â,â)f(x)dx=1ï¼â´Aâ«(-â,â)dx/[e^x+e^(-x)]=1ãèâ«(-â,â)dx/[e^x+e^(-x)]=2â«(0,â)dx/[e^x+e^(-x)]=2arctan(e^x)丨(x=0,â)=Ï/2ï¼â´A=2/Ïã(2)P{0<x<(ln3)/2}=â«(0,(ln3)/2)f(x)dx=(2/Ï)arctan(e^x)丨(x=0,(ln3)/2)=1/6ã
5é¢ï¼âµX~U(2,6)ï¼â´f(x)=1/(6-2)=1/4ã2<x<6ï¼f(x)=0ï¼xåå
¶å®å¼ãâ´p=P(X>3)=â«(3,6)f(x)dx=â«(3,6)dx/4=3/4ãç¬ç«è§å¯X3次ï¼è³å°2次åºç°pçæ¦çï¼ç¬¦åäºé¡¹åå¸B(3,p)çæ¡ä»¶ï¼â´P{ç¬ç«è§å¯X3次ï¼è³å°2次åºç°pçæ¦ç}=C(3,2)(p^2)(1-p)=3*(1-3/4)(3/4)^2=(3/4)^3ã
ä¾åèã
温馨提示:答案为网友推荐,仅供参考