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概率论与数理统计（第二版.刘建亚）习题解答——第四章
4－1 解：E(X) =1′0.25+2′0.4+3′0.2+4′0.1+5′0.05= 2.3
4－2 解： 由D(X) = E(X 2)-[E(X)]2 得
E(X)
E(X2 )
D(X) X 1 50 2501 1 X 2
50
2502
2
∵ D(X 1) < D(X 2)
， 用甲法测定的精度高。
4－3 解：
X 0 1 2 3 P
0.75
0.2045
0.0409
0.0045 E(X)=0.3003，E(X2)=0.4086，D(X)=0.3184，[D(X)]1/2=0.5643。4－4 解：
E(X *) = E X -E(X) = 1 E[X -E(X)]= 1
[E(X)-E(X)]= 0 D(X) D(X) D(X)
2
D(X *) =E(X*)2 -[E(X*)]2 =E(X*)2 =E X-E(X) = 1 E[X-E(X)]2 = 1 D(X) = 1
D(X) D(X) D(X)
4－5 解：
+￥ 1
x
E(X) = -￥ xf (x)dx = -1 p 1-x 2 dx = 0 E(X 2) = -+￥￥ x2 f (x)dx = -11 p 1x-2 x2 dx = 01 p 21x-2 x 2 dx 1
x = sint 1 0p2 2sintdx = 1
p 2
D(X) = E(X 2)-[E(X)]2 = 4－6 解：
+￥ +￥
0p2 (1-cost)dx = 1 p
1 -x
dx = 0
E(X) = -￥ xf (x)dx = -￥ x× 2 e D(X) = E{[X -E(X)]2}
=
-+￥￥
(x-0)2 ×e-xdx
=
+2-x0￥
xedx ；
=-x2e-x+0￥ +2 0+￥xe-xdx =-2xe-x +0 ￥ +2 0+ ￥ e-xdx = 2
a
1 p ，则 =1-p，a = ；
4－7 解：令 p=
1+a
1+a 1- p
k
E(X) = k 0 kP(X = k) = k 0 k× 1
=
=
???1+aa÷÷÷ = k=0 k×(1- p)pk = p(1- p)k= 1 kp k-1
1+a
d d d p
= p(1- p)k= 1 (pk) = p(1- p) ???k=1 pk÷÷÷÷= p(1- p) dp???1-p ÷÷÷÷
dp
dp
d 1 d 1 1p
= p(1- p) dp1-p -1÷=÷÷ p(1- p) dp1-p ÷÷÷= p(1- p)× (1-p)2 = 1-p = a
k
E(X 2) = k 0 k2P(X = k) = k 0 k2 × + 1 ???1+aa÷÷÷ = p(1- p)k= 1 [k(k-1)+k]p k-1
= =
1 a
d2
2
= p(1- p) k(k-1)pk-1 + kpk-1 = p(1- p) p
k=1
k=1
(pk)+ kpk-1
k=2
dp k=1
2
= p2(1- p) dp d22 ????k￥ = 2 pk÷÷÷÷+a = a+ p2(1- p) dpd22 ????1-p2p÷ ÷÷÷
= p2(1- p)×
p)
+a = 2???1-p ÷÷÷÷ +a = 2a2 + a (1-
p
2
D(X) = E(X 2)-[E(X)]2 = 2a2 +a-a2 = a2 +a
4－8 证明：设 X 为连续型随机变量，其概率密度函数为 f (x) 。
+￥
+￥
+ ￥
（1）E(aX +b) = -￥ (ax+b) f (x)dx = a -￥ xf (x)dx+b - ￥ f (x)dx = aE(X)+b （2）D(cX) = E(c2X 2)-[E(cX)]2 =c2E(X 2)-c2[E(X)]2 =c2D(X)。 4－9 证明：
D(X) = E[(X -E(X)]2 = E{(X -C)-[E(X)-C]}2
= E{(X -C)2}-2E{(X -C)[E(X)-C]}+E{[E(X)-C]2} = E(X -C)2 -2[E(X)-C]2 +[E(X)-C]2
= E(X -C)2 -[E(X)-C]2 ￡ E(X -C)2
4－10 解：
X
X ~ N(m, s2) ，已知：m=143.10, s2 = 5.672 ，则 U = -m ~ N(0,1) ，由双侧分位点知：[-u , u ]内的概率为
a 2 a 2 s 1-a = 0.95, a = 0.05,1-ua2= 0.975，查表得 u∴ m sua2143.10
5.67
1.96
a 2
=1.96，
∴ 95％正常范围为[131.99，154.22]。
4－11 证明：
E(X)-c = E(X -c) = -+￥￥(x-c) f (x)dxt = x-c - + ￥ ￥ tf (c+t)dt
0 + ￥
= - ￥ tf (c+t)dt+ 0 tf (c+t)dt
3