山东大学概率论与数理统计(第二版刘建亚)习题解答——第2章

概率论与数理统计（第二版.刘建亚）习题解答第二章1解：不能。因为P(X=-1)=-O.5vO;P(X=x)=0.850。12i2解：P1/103/1056/10#3解：取法：n=C4,X的取值：0,1,2,3。所以5P(X=k)=CkC4-k312C415(k=0,1,2,3)，分布列为3X012P33/9144/9166/4554/455#4解：由概率的规范性性质P(X=炉1得:(1) 邋P(X=k)=上=a=1;a=1Nk=1k=1、錨(2) 邋P(X=k)=a=1;a=12kk=1k=15解：P(X=k)=據尹p(x=2n)=M(k1,2,)(n1,2,)#P(X=偶数)=邋P(X=2n)=313?44-鹽4k=1k=126解：X23456789101112P1/362/363/364/365/366/365/364/363/362/361/3611P(X?4)P(7#X10)-o6227解：n重贝努利试验，XB(20,0.1)解法一：(1) P(X=3)=C3p3(1-p)17=0.1901；20(2) P(X?3)1-P(X?2)1-P(X=0)-P(X=1)-P(X=2)=0.3231；(3) 最可能值：k=(n+1)?0.12；P(X=2)=0.2852。lk解法二：利用泊松定理，P(X=k)蛔一e-1(k0,1,),l=np=20?0.12(1) P(X=3)=e-2=0.1804；引(2) P(X?3)1-P(X?2)1-P(X=0)-P(X=1)-P(X=2)=0.3233(3) 最可能值：k=(n+1)?0.12；P(X=2)=0.270728解：1XBn,p),n=7301P=,10,p=0.01N)N)=1-P(X；N)1邋Ae-1k!k=0查泊松分布表得N+l=5，即N=4。lke-1k!k=N+10.01211解：XB(50000,0.0001)1=np=51k5k泊松定理知P(X=k)蛔一e-1e-5k0厂1，)nk!k!P(X=0)蛔竺e-56.73810-30!P(X10)=ae-1=0.00284k!k=111k-e-1=0.0297k!k=9213解：(1)由概率的规范性1=蝌+f(x)d=x1-0(2) P(0.3X0.7)=60.72xdx=0.4；0.3(3) 由题意知对0#a1有蝌2xdx=0,2-c得得a2=1-a2a=1沁2#(4) 分布函数定义式：F(x)=6xf(t)dt当x0时，F(x)=0；当0?x1时，F(x)=0+6x2tdt=x22当x31时，F(x)=110x3时，P(X?k)蝌+f(x)dx=k所以，当1#x3时，P(X?k)蝌kx)dx1221Idx+0+369dx+0=+(l-dx+0=-(699f(x)dx=0+k)；#215解：由概率的规范性p2cdt=cpp21=蝌f(x)dx=1Cdxx=sint-?01-x2口Xl?2)pdxp(i+6)=2V216解：(1)当x0时，f(x)=F(x)=e-x；当x=0时，F(x)存在，且F(0)=0,f(0)=0f(xTx0x0(2)P(X?4)F(4)=1-e-4,P(X1)=1-P(X?1)1-F(1)=e-1217218219220解：Xe(l)l=0.1(1)P(X?10)0+0.1e-0.1xdx=e-1-0=0.3678810、(2)P(10X20)=0200.1e-0.1xdx=e-1-e-2=0.232510221解：XN(160,0.062)(0.05?0.12)m2sP(|X-0.050.12)=1-P(X-m2s)=1-P(m-2sXm+2s)=1-0.9545=0.0455222解：XN(160,s2)0P(120X200)200-m)120-m)#s01=0.8=F(40)-F(-巴)=2F(10)-sss4040得s31.250F(40)=0.9，查表得一1.28ss00223224设随机变量XN(3,22)。(1) 求P(2c)=P(Xc)；(3) 设d满足P(Xd)0.9，问d至多为多少？解：P(22)=1-P(|X21一犏鼾斷泞=1-F(-0.5)+F(-2.5)=1-1-F(0.5)+1-F(-2.5)=1+F(0.5)-F(2.5)=1+0.6915-0.99379=0.697712）由条件P(Xc)=P(X)c得P(X?c)P(X?1-c)P(Xc)=1-P(Xc),0.5蓿P(Xc)訂壬0.5(0)7#3=m#3)因XN(3,22)，所以X关于U=3对称，根据性质#P(Xd)=P-32d-3观赵-3-_3=F0.92桫2d0.436225#226*证明：/X服从几何分布，P(X=k)=qk-1p(q=1-p,k=1,2,)P(X=n+kIX?n)P(X=n+k)P(X?n)qn+k-1p1-氏qk-1pk=1qn+k-1pp(1+q+qn-1)n+k-1pq11-qn1-p匸qk-1p=P(X=k)227*略。228解：1)Y=2X+1-3-1P(Y=yi)1/101/51/41/41/5#2)Y=X2014P(Y=yi)1/49/203/10229解：#f(x)彳00x1其它#y=-2lnx,g-1(y)=-(0x1 ye-221)xg-1(y)=ey2,(00Ie2j(y)=12y010230解:9f(x)=i1100x6其它#当0y3时，x-3=鞭yg-1(y)=盶1j(y)idi)163#当y为其它时，j(y)=0，综合得0y3其它231解：1)y=2x2+1?(y1)g-1(y)=?耳g-i(y)2f2(y-当y1时，(y)=占e-v轾轾轾犏|/2(y-1)2Qp(y-1)#当y匚1时j(y)=0,综上得#y1y12)y=|蕹(y0)g(y)y,g-i(y)1y2当0时j(yre-2(1+1)=存当y匚0时j(y)=0,综上得j(y)=iVioyy0另一解法：F(y)=P(Y?y)P(|X|?y)$iP(-y#Xi0y)y0y0而P(-yXy)蝌f(x)dxX-y2pye-2dx-y2迄-ye-2dX#232*解：当k=4n-P(Y=-f(y)=f、fy=!021时，Y=l；当k=4n-2或k=4n时，Y=0；当k=4n+3时，Y=-1。1)=邋P(X=4n-1)=n=11=224n-115n=1124n-n=11=824n-315n=1P(Y=0)=邋P(X=4n-2)+P(X=4n)=n=1P(Y=1)=邋P(X=4n-3)=I#1n=1Y的分布列：-12/151/38/1511#233*略。#