# 我们已经介绍了随机变量的数学期望，它体现了随机变量取值的平均水平，是随机变量的一个重要的数字特征 . - PowerPoint PPT Presentation

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.

.

a10X

a

,10

?

.

,.

XE[(X-E(X)]2<

D(X)=E[X-E(X)]2 (1)

X.

X-E(X)

X.

D(X)=E[X-E(X)]2

.

X

X .

D(X)=0,r.v X1 .

X

g(X)=[X-E(X)]2 .

X

P(X=xk)=pk

X

X~f(x)

D(X)=E(X2)-[E(X)]2

D(X)=E[X-E(X)]2

=E{X2-2XE(X)+[E(X)]2}

=E(X2)-2[E(X)]2+[E(X)]2

=E(X2)-[E(X)]2

.

1 r.vX

P(X=k)=p(1-p)k-1, k=1,2,n

0<p<1,D(X)

q=1-p

+E(X)

D(X)=E(X2)-[E(X)]2

X1 X2

D(X1 +X2)=

1. C,D(C)=0;

2. C,D(CX)=C2D(X);

3. X1X2

D(X1+X2)= D(X1)+D(X2);

X1,X2,,Xn,

4.D(X)=0 P(X= C)=1 C=E(X)

P(X= x)

.

i=1,2n

n

2

X~B(n,p), Xn

.

E(Xi)=P(Xi=1)= p,

E(Xi2)= p,

D(Xi)= E(Xi2)-[E(Xi)]2

= p- p2= p(1- p)

D(Xi)= E(Xi2)-[E(Xi)]2 = p- p2= p(1- p)

i=1,2n

X1,X2,,Xn

= np(1- p)

XE(X)

>0,

{|X-E(X)|< }X.

.

r.vX .

r.v XE(X) 3 0.111 .

P(5200 X 9400)

3 7300700 . 5200~9400 .

X

E(X)=7300,D(X)=7002

P(5200 X 9400)

=P(5200-7300 X-7300 9400-7300)

= P(-2100 X-E(X) 2100)

= P{ |X-E(X)| 2100}

P{ |X-E(X)| 2100}

5200~94008/9 .

4 A 0.75, nn, A0.74~0.760.90?

XnA

X~B(n, 0.75)

E(X)=0.75n,

D(X)=0.75*0.25n=0.1875n

n .

n

P(0.74n< X<0.76n )

=P(-0.01n<X-0.75n< 0.01n)

= P{ |X-E(X)| <0.01n}

= P{ |X-E(X)| <0.01n}

n 18750n

, A0.74~0.76

0.90 .

.

.

r.v