王明慈版 概率论与数理统计 习题五[统计学经典理论].doc

文档介绍：习题五
:

计算样本均值、样本标准差、样本方差与样本二阶中心矩。
__ 1 10 1
解: x = ∑ x
= (+++++++++) = ;
i
10 i =1 10
1 10 __ 1
= ∑(
−)2 = [( − )2 + ( − )2 + …+ ( − )2 ] = ;
i
s 9 i =1 x x 9
1 10 __
2 = ∑(
−)2 = = ;
9
s xi x
i =1
∼ 1 10
__ 9
2 = ∑(
−)2 =
= .
i
ó 10
i =1 x x
10 S
100 个样本观测值如下:
观测值
xi
1
2
3
4
5
6
频数
ni
15
21
25
20
12
7
计算样本均值、样本方差与样本二阶中心矩。解:由书上 127 页()( )( )式可知:
___ 1 6 1
x = ∑ x n
= (1×15 + 2 × 21+ 3× 25 + 4× 20 + 5× 12 + 6× 7) = ;
100
i i
i
i =1
100
1 6 ___ 1
2 = ∑( −
)2 = [(1− )2 ×15 + …+ (6 − )2 × 7] = ;
s 99
i =1 x
x ni 99
∼ 1 6 ___ 99
2 = ∑( −
)2 = × = .
ó

100
xi
i =1
x ni
100
n
n 的样本 X1 ,…, X
,设 为任意常数, 为任意正数,作变换
c k
Y = k ( X
− c), i = 1, 2,⋯, .
i i n
证明:( 1) X Y
x
2
= + c; (2) S
k
2
2
X
2
x
= = S y ; 其中 X 及 S
k
分别是 1 ,…,
X
n
的样本均值及样本
方差; 及
Y
2 分别是
S y
Y1 ,…,Yn
的样本均值及样本方差。
1
证明(1)
 = ∑n
, 由 = (
−) 得
= Y= i +
X X i
1
n i =
Yi k X i c
X i c
k
n
1 (Y ) 1 n 1 Y
∴ X =
∑ i + c =
∑Yi + ⋅ nc = + c
n i =1
k k ⋅ n i =1 n k

2 = 1 ∑n (


−= )2 = 1 ∑n ⎡( −)
2
− )⎤
⎣ ⎦
(
S y Yi Y
n i =1
n i =1
2
k X i
c kX kc
(2)
= 1 ∑n
= ) =
2 ⋅= 1 ∑n
( −= )2 =
2 ⋅ 2
n i =1
2
kX i
2
S y
kX k
X i
n i =1
X k Sx
∴ Sx = 2
k
5. 从总体中抽取两组样本,其容量分别为 n1 及 n2 ,设两组的样本均值分别为 X1 及 X 2 ,
样本方差分别为
2 及 2 , 把这两组样本合并为一组容量为
S1 S2
n1 + n2
的联合样本。
证明:( 1).联合样本的样本均值= = n1 X 1 + n2 X 2 ;
X
n1 + n2
( −1)

2 + (

−1) 2

( −)2
S
(2).联合样本的样本方差
2 = = n1 S1 n2 S2 +
n1n2 X1 X 2
S 1 = n1 X1 ,
n1 + n2 −1
S 2 = n2 X 2
(n1 + n2 ) (n1 + n2 −1)
um um
证明:( 1)
Sum 1 + S um 2
X = =
n1 + n2
n1

n1 X 1 + n2 X 2
n1 + n2
∑
n2
∑( X1
− )2 + (
X X 2i
− )2
X
=
2 =1= i =1=
i i