Estadistica General 4
Páginas: 5 (1036 palabras)
Publicado: 3 de junio de 2012
1
PRINCIPALES DISTRIBUCIONES DE PROBABILIDAD
A.-
DISTRIBUCIONES DE VARIABLES DISCRETAS
•
!
$
•
"%
$
&
•
$
"#
'
!
!
&
"
!
!
#$
%
(
n
P ( x) = C x P x (1 − P ) n − x
n!
P x (1 − P ) n − x
(n − x)! x!
Esperanza Matemática : E ( x) = nP
P ( x) =
V ( x) = nP(1 − P)
Variancia :
&'
&
(
!
)
)
'
(
"% %*
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*
.
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)
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%%
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%%
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%%
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%%
% 5%
%%
% 5%
%%
% 5%
%%
% 5%
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34
% %%5
% %.
% %.
%# 5
% %.
%# 5
%# 5
% 6#
)
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00
0
00
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000
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)
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*
1
1
+ , -'
'
2
". 2
"% %
,- +. .*.) 2
3
4
*
)
*
1
)
1.
&
8
,
3
P ( x) = C x (0.2) x (1 − 0.2) 3− x
+
3
P ( x = 0) = C 0 (0.2) 0 (0.8) 3 = 0.512
P ( x = 1) = C13 (0.2)1 (0.8) 2 = 0.384
*
3
P ( x = 2) = C 2 (0.2) 2 (0.8)1 = 0.096
)
3
P ( x = 3) = C 3 (0.2) 3 (0.8) 0 = 0.008
1
+
P(X) 0.6
0.512
0.5
0.384
0.4
0.3
0.2
0.096
0.1
0.008
0
0
1
2
3
X: Días que vende pasajes aItalia
$
•
38 4"9
•
;
•
@A
(
; 38 4"9
"
*
3
P ( x = 3) = C 3 (0.20) 3 (0.80) 0 =
•
@A
".3% 4"% :
n
x =2
1
*
3#< 4".3% 43% 54"% ?5
)
1
.
.
*B
C
4!
(0.20) 3 (0.80) 0 = 0.008
( 4 − 3)!3!
)
*
P ( x ≥ 2) =
3 4"
)
.
C
n
C x P x (1 − P ) n − x = Pr( x = 2) + Pr( x = 3) = 0.096 + 0.008 = 0.104
*B
(
3
*(
•
7
/3
•
•
4
&
&
/ +. . *. ). 0.1
•
3
4
$
#
”
•
&
7
&
!
)
!
P ( x) =
λ x e −λ
Donde e = 2.71828
x!
Esperanza Matemática : E ( x) = λ = nP
V ( x ) = λ = nP
Variancia :
&'
/
!
*
)
,- +. .*.).01$ 07
P( x) =
λ xe−λ
x!
)
=
2 x e −2
;
x!
*
)
0
1
+
3
3
3
3
3
$
Lamda λ = 2 $
P (x) =
+
$
"%4"
"#4"
" 4"
".4"
"?4"
%<
2 x e −2
x!
D%E #.6.?
"%
D#E
"% F%:F
<
DE
"% F%:F
.<
D.E #5%?6
"%
?<
D?E %G %
"%
HHHHHH
#
#<
*
4
I
(
0.30
P(x)
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
Fallas (x)
$
•
•
;
•
@A
)
P ( x = 3) =
•
−2
0−2
2e
3!
)
2e
0!
@A
)
P ( x) =
P ( x = 3) =
C
(
C
$
C
9 3 4" 3 "%4K 3 "#4K 3 " 4K 3 ".4KH " #
P ( x > 1) = 1 − [P ( x = 0) + P ( x = 1)] = 1 −
•
$
)
)
.
= 0.13534
@A
A
$
= 0.18044
@A
P ( x = 0) =
•
38 4"J"
" J"
; 38 4"9
3
)
$
λ x e −λ
x!
13 e −1
= 0.06138
3!
=
1x e −1
;
x!
2 0 e −2 2 1 e−2
+
= 1 − [0.13534 + 0.27067] = 0.59399
0!
1!
$
.
Lamda λ = 2 / 2 = 1 $
C
5
)
2
& 3&
&
2
•
!
$
$
"%
$
•
"#
$
L
B
344
•'
(
'
$
)
"M !
5
7
(
P( x = k ) =
A N−A
k n−k
N
n
=
B&
N
C kA C n −− A
k
N
Cn
nA
N
nA
A
V ( x) =
1−
N
N
Esperanza Matemática : E ( x) =
Variancia :
&'
/
%
'
$?
#6
(
(
0" %
"#6
0< "6
"? 7
/ +. .*.).0
/
$
$
/
/
(
/
$
,
+
*
)
0
+
N −n
N −1
P ( x) =
3
3
3
3
3
5
C 15 C 4 − x
x
2
C4 0
"%4"A%#6A?6DA? %"% %%#%.
"#4" A##6A.6DA? %"% %.%G :
" 4" A #6A 6DA? %"% #:F
".4" A.#6A#6DA? %"% ?:G 6:
"?4" A?#6A%6DA? %"% 5#F.
#
6
I 1'+ N O - P
P(x)
0.5
0.46956
0.4
0.28173
0.3
0.216720.2
0.1
0.00103
0
0.03096
0
1
2
3
4
Clientes insatisfechos
nA 4(15)
=
= 3 Clientes insatisfechos
N
20
nA
A N −n
4(15)
15 20 − 4
La Variancia : V ( x) =
=
1−
1−
= 0.63158
N
N N −1
20
20 20 − 1
Esperanza Matemática : E ( x) =
2
•
(
Q
.
$
Pr( x = 3) =
•
15!
5!
3!12! 1!4!
=
= 0.46956
20!
4!16!
.
$
Pr( x ≥ 3) =...
PRINCIPALES DISTRIBUCIONES DE PROBABILIDAD
A.-
DISTRIBUCIONES DE VARIABLES DISCRETAS
•
!
$
•
"%
$
&
•
$
"#
'
!
!
&
"
!
!
#$
%
(
n
P ( x) = C x P x (1 − P ) n − x
n!
P x (1 − P ) n − x
(n − x)! x!
Esperanza Matemática : E ( x) = nP
P ( x) =
V ( x) = nP(1 − P)
Variancia :
&'
&
(
!
)
)
'
(
"% %*
(
*
.
(
)
'"'
2
#
2
%%
2
%%
% 5%
0
%%
%5
0
% %5 0
'
.
%%
% 5%
%%
% 5%
%%
% 5%
%%
% 5%
/0"0
0
0
0
0
*
34
% %%5
% %.
% %.
%# 5
% %.
%# 5
%# 5
% 6#
)
*
*
0
0
00
0
00
00
000
*
+
$
07
)
(
*
1
1
+ , -'
'
2
". 2
"% %
,- +. .*.) 2
3
4
*
)
*
1
)
1.
&
8
,
3
P ( x) = C x (0.2) x (1 − 0.2) 3− x
+
3
P ( x = 0) = C 0 (0.2) 0 (0.8) 3 = 0.512
P ( x = 1) = C13 (0.2)1 (0.8) 2 = 0.384
*
3
P ( x = 2) = C 2 (0.2) 2 (0.8)1 = 0.096
)
3
P ( x = 3) = C 3 (0.2) 3 (0.8) 0 = 0.008
1
+
P(X) 0.6
0.512
0.5
0.384
0.4
0.3
0.2
0.096
0.1
0.008
0
0
1
2
3
X: Días que vende pasajes aItalia
$
•
38 4"9
•
;
•
@A
(
; 38 4"9
"
*
3
P ( x = 3) = C 3 (0.20) 3 (0.80) 0 =
•
@A
".3% 4"% :
n
x =2
1
*
3#< 4".3% 43% 54"% ?5
)
1
.
.
*B
C
4!
(0.20) 3 (0.80) 0 = 0.008
( 4 − 3)!3!
)
*
P ( x ≥ 2) =
3 4"
)
.
C
n
C x P x (1 − P ) n − x = Pr( x = 2) + Pr( x = 3) = 0.096 + 0.008 = 0.104
*B
(
3
*(
•
7
/3
•
•
4
&
&
/ +. . *. ). 0.1
•
3
4
$
#
”
•
&
7
&
!
)
!
P ( x) =
λ x e −λ
Donde e = 2.71828
x!
Esperanza Matemática : E ( x) = λ = nP
V ( x ) = λ = nP
Variancia :
&'
/
!
*
)
,- +. .*.).01$ 07
P( x) =
λ xe−λ
x!
)
=
2 x e −2
;
x!
*
)
0
1
+
3
3
3
3
3
$
Lamda λ = 2 $
P (x) =
+
$
"%4"
"#4"
" 4"
".4"
"?4"
%<
2 x e −2
x!
D%E #.6.?
"%
D#E
"% F%:F
<
DE
"% F%:F
.<
D.E #5%?6
"%
?<
D?E %G %
"%
HHHHHH
#
#<
*
4
I
(
0.30
P(x)
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
Fallas (x)
$
•
•
;
•
@A
)
P ( x = 3) =
•
−2
0−2
2e
3!
)
2e
0!
@A
)
P ( x) =
P ( x = 3) =
C
(
C
$
C
9 3 4" 3 "%4K 3 "#4K 3 " 4K 3 ".4KH " #
P ( x > 1) = 1 − [P ( x = 0) + P ( x = 1)] = 1 −
•
$
)
)
.
= 0.13534
@A
A
$
= 0.18044
@A
P ( x = 0) =
•
38 4"J"
" J"
; 38 4"9
3
)
$
λ x e −λ
x!
13 e −1
= 0.06138
3!
=
1x e −1
;
x!
2 0 e −2 2 1 e−2
+
= 1 − [0.13534 + 0.27067] = 0.59399
0!
1!
$
.
Lamda λ = 2 / 2 = 1 $
C
5
)
2
& 3&
&
2
•
!
$
$
"%
$
•
"#
$
L
B
344
•'
(
'
$
)
"M !
5
7
(
P( x = k ) =
A N−A
k n−k
N
n
=
B&
N
C kA C n −− A
k
N
Cn
nA
N
nA
A
V ( x) =
1−
N
N
Esperanza Matemática : E ( x) =
Variancia :
&'
/
%
'
$?
#6
(
(
0" %
"#6
0< "6
"? 7
/ +. .*.).0
/
$
$
/
/
(
/
$
,
+
*
)
0
+
N −n
N −1
P ( x) =
3
3
3
3
3
5
C 15 C 4 − x
x
2
C4 0
"%4"A%#6A?6DA? %"% %%#%.
"#4" A##6A.6DA? %"% %.%G :
" 4" A #6A 6DA? %"% #:F
".4" A.#6A#6DA? %"% ?:G 6:
"?4" A?#6A%6DA? %"% 5#F.
#
6
I 1'+ N O - P
P(x)
0.5
0.46956
0.4
0.28173
0.3
0.216720.2
0.1
0.00103
0
0.03096
0
1
2
3
4
Clientes insatisfechos
nA 4(15)
=
= 3 Clientes insatisfechos
N
20
nA
A N −n
4(15)
15 20 − 4
La Variancia : V ( x) =
=
1−
1−
= 0.63158
N
N N −1
20
20 20 − 1
Esperanza Matemática : E ( x) =
2
•
(
Q
.
$
Pr( x = 3) =
•
15!
5!
3!12! 1!4!
=
= 0.46956
20!
4!16!
.
$
Pr( x ≥ 3) =...
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