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Tabla 2 - Función de distribución
 
p( ) P(X )
x x
1
1
( )
k
i
i
p x




1
1
( )
k
i
i
p x



 
( ) ( )
F a P X a

 
( ) ( )
i
i
x a
F a p x

𝑃(𝑎 < 𝑥 ≤ 𝑏) = 𝐹(𝑏) − 𝐹(𝑎) = ∑ 𝑝(𝑥𝑖)
𝑎<𝑥≤𝑏

F(x ) =1
k
1 0
( )
F x 
 0 y entero.
h
 
1
0

 

si el cliente compra una notebook
si el cliente compra una computadora de escritorio
Y


 
  

0,80 si x=0
( ) 0,20 si x= 1
0 si x 0 o x 1
p x
0 04 0 1 2 3 4 5
0
 

 


, con , , , ,
( )
otro valor de x
K x x y
p x
( )
p x
1
1
( )
k
i
i
p x



0 1 2 3 4 5 1
0 04 0 08 0 12 0 16 0 20 1
6 0 60 1
1 0 60
6
0 06667
( ) ( ) ( ) ( ) ( ) ( )
( , ) ( , ) ( , ) ( , ) ( , )
,
,
,
p p p p p p
K K K K K K
K
K
K
     
          
 




2 4
2 4 2 3 4
2 4 0 147 0 187 0 227 0 56
 
     
     

( ) ( ) ( ) ( ) ( )
( ) , , , ,
i
i
x
P X p x p p p
P X
2 4 4 1
2 4 0 735 0 174
2 4 0 56
   
   
  
( ) ( ) ( )
( ) , ,
( ) ,
P X F F
P X
P X

3 4 4 5 0 227 0 267 0 49
4 1 3 1 0 508
4 0 49
       
    
 
( ) ( ) ( ) ( ) , , ,
( ) ( ) ,
( ) ,
P X P X p p
P X F
P X
𝑝(𝑥) =
{
0,30 𝑠𝑖 𝑥 = 2
0,10 𝑠𝑖 𝑥 = 4
0,05 𝑠𝑖 𝑥 = 6
0,15 𝑠𝑖 𝑥 = 8
0,15 𝑠𝑖 𝑥 = 10
0,25 𝑠𝑖 𝑥 = 12
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
6 2 4 0 30 0 10 0 40
6 4 0 40
    
 
( < ) ( ) ( ) , , ,
( < ) ( ) ,
P X p p
P X F
a- Comprobar que es una función de cuantía.
b- Graficar p(x).
c- Graficar F(x).
d- Calcular P( X  2).
e- Calcular P(X  5).
f- Calcular P( 3  X  9).
𝑝(𝑥) = {𝑘𝑥2
𝑝𝑎𝑟𝑎 𝑥 = 1,2 3,4,5
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
 
 
𝑝(𝑥) = {
𝑥 + 2
15
𝑝𝑎𝑟𝑎 𝑥 = −1,0,1,2 3
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
Variable Clase LI LS MC FA FR
Gastos 1 272,00 522,75 397,38 7 0,02
Gastos 2 522,75 773,50 648,13 30 0,08
Gastos 3 773,50 1024,25 898,88 51 0,13
Gastos 4 1024,25 1275,00 1149,63 102 0,26
Gastos 5 1275,00 1525,75 1400,38 105 0,26
Gastos 6 1525,75 1776,50 1651,13 65 0,16
Gastos 7 1776,50 2027,25 1901,88 36 0,09
Gastos 8 2027,25 2278,00 2152,63 4 0,01
146,63 597,98 1049,33 1500,68 1952,03 2403,38
Gastos
0,00
0,07
0,14
0,21
0,28
frecuencia
relativa
( )
f x
0
( )
f x para todo x

1
( )
f x dx


 
( , )
 
205,13
740,07
1275,00
1809,93
2344,87
Gastos
0,00
0,04
0,08
0,13
0,17
frecuencia
relativa
104,83
494,89
884,94
1275,00
1665,06
2055,11
2445,17
Gastos
0,00
0,05
0,10
0,15
0,20
frecuencia
relativa
( ) ( )
b
a
P a X b f x dx
   
P(a X b) P(a X b) P(a X b) P(a X b)
          
500
250
250 500
( ) ( )
P X f x dx
   
La función de distribución es la que proporciona la probabilidad acu-
mulada para los diferentes valores de la variable. Para las variables continuas,
será el área bajo la curva de la función de densidad a la izquierda de un valor
particular x, o sea acumulada desde el menor valor hasta un valor determina-
do de la variable. Simbólicamente:
'
' '
( ) ( ) ( )
x
F x P X x f x dx

   
( )
( ) '( )
dF x
f x F x
dx
 
( )
F x ( )
f x
( )
F  1
( )
f x dx





1
lim ( )
x
F x


0
lim ( )
x
F x


 0
h
 
( ) ( ) ( )
( ) ( ) ( )
b a
P a X b F b F a
P a X b f x dx f x dx
 
   
   
 
250 500 500 250
( ) ( ) ( )
P X F F
   
𝑓(𝑦) = {
1
18
𝑦 𝑝𝑎𝑟𝑎 0 < 𝑦 < 6
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦
( )
f y
0
( )
f y 
6
0
1
( )
f y dy 

6
6
2
0
0
2
1 1
18 36
1
6
36
1
ydy y




( )
f y
2
0
1 1
18 36
( )
u
F u ydy u
 

146,63 710,81 1275,00 1839,19 2403,38
Gastos
0,00
0,25
0,50
0,75
1,00
frec.
rel.
acumulada
𝑃(𝑦 ≥ 5) = ∫
1
18
𝑦 𝑑𝑦
6
5
=
1
36
(62
− 52)
= 0,3056
2
5 1 5
1
1 5
36
0 3056
( ) ( )
( )
,
P y F
  
 

𝑃(3 ≤ 𝑦 ≤ 4,5) = ∫
1
18
𝑦 𝑑𝑦
4,5
3
=
1
36
(4,52
− 32)
= 0,3125
2 2
3 4 5 4 5 3
1 1
4 5 3
36 36
0 3125
( , ) ( , ) ( )
= ( , ) ( )
= ,
P y F F
   

y
6
5
4
3
2
1
0
0.4
0.3
0.2
0.1
0
y
6
5
4
3
2
1
0
1
0.8
0.6
0.4
0.2
0
y
P(y) F(y)
y
Figura 7
(a) (b)
En la Figura que sigue se muestra la zona sombreada correspondiente a
la probabilidad calculada anteriormente.
𝑓(𝑥) = {𝑘(1 − 𝑥2 ) 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
1
0
1
( )
f x dx 

1
2
0
1 1
( )
k x dx
 

1 1
2 2
0 0
1
3
0
3
1 1
3
1
1
3
( ) ( )
= ( )
= ( )
k x dx k x dx
x
k x
k
  


 
3
1
1 1
3
2
1
3
3
2
( )
( )
k
k
k
 


f(Y))
f(Y)
Y
Y
Y
𝑓(𝑥) = {
3
2
(1 − 𝑥2 ) 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
0 0
0 1
1 1
4
( ) x
x
F x x
x



  

 

3
4
'( )
F x x

𝑓(𝑥) = { 4𝑥3
𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
4
0 40 1 0 40
1 0 40
0 974
( , ) ( , )
( , )
,
P x F
  
 

0
4
0
4
0
0
0 50
0 50
0 50
0 841
( ) ,
,
,
,
F x
x
x
x




𝑓(𝑦) = {
3
8
𝑦 𝑝𝑎𝑟𝑎 0 ≤ 𝑦 ≤ 2
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦
0
( )
f y 
2
0
1
( )
f y dy 

2 2
0 0
2
2
0
3
8
3
8 2
6
8
( )
f y dy ydy
y

 
  
 

 
8 3
6 8
1
2
( ) .
f y y
y


𝑓(𝑦) = {
1
2
𝑦 𝑝𝑎𝑟𝑎 0 ≤ 𝑦 ≤ 2
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦
𝑓(𝑥) = {
1
2
𝑝𝑎𝑟𝑎 1 ≤ 𝑥 ≤ 3
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥







1
0
)
( 2
kx
x
F
x  0
0  x  3
x  3
1


𝑓(𝑦) = {
𝑦5
8
𝑦 𝑝𝑎𝑟𝑎 0 ≤ 𝑦 ≤ 1
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦

1


  
[ ] ( )
k
i i
i
E X x p x
  
𝐸(𝑥) = 𝜇 = 0(0,067) + 1(0,107) + 2(0,147) + 3(0,187) + 4(0,227) + 5(0,267)
= 3,205



  
[ ] ( )
E X x f x dx
𝑓(𝑦) = {
1
18
𝑦 𝑝𝑎𝑟𝑎 0 < 𝑦 < 6
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦
6
0
6
2
0
6
3
0
3
1
18
1
18
1
18 3
1 6
18 3
4
E Y y y dy
y dy
y

 
 
 






𝑓(𝑦) = {
3𝑥2
125
𝑝𝑎𝑟𝑎 0 ≤ 𝑥 < 5
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
5 2
0
5
0
4
5
0
3
125
3
125
3
125 4
3 75
( )
,
x
E x
x

 
 





3
x dx
x dx
𝑓(𝑥) = {
3
2
(1 − 𝑥2
) 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
 
1
2
0
1
0
2 4
1 1
0 0
3
1
2
3
2
3
2 2 4
0 375
( )
( ) ( )
,
E x x
x x
 
 
 
 
  
 
 
 
 
 



3
x dx
x - x dx
X f(X)
2 0,30
4 0,10
6 0,05
8 0,15
10 0,15
12 0,25
Si consideramos que la probabilidad de que el proyecto sea aproba-
do es de 0,30 ¿Cuál sería la utilidad neta esperada?
P X c 1
E c c
E c X c E X
E c.X c.E X
E X 0
𝑓(𝑥) = {
𝑥
16
𝑝𝑎𝑟𝑎 2 ≤ 𝑥 ≤ 6
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
6
6 3
2 2
4 333
16 48
,
x x
E X x dx
  
 
  
0 20
0 20
0 20 4 333 0 8667
( , )
, ( )
, , ,
E Y E x
E x

 
 

  
1 1 10
1 1 10
1 4 7667 5 7667
( , )
, ( )
, ,
E W E x
E x
 
 
 
 
  
2
2
( ) X
V X E X
 
  
 
 
2
1


 
 
 

( ) ( )
k
i i
i
V X x p x
2



 
 
 

( ) ( )
V X x f x dx
μ
2
2
 
   
 
 
( )
V X E X E X
( ) ( ) X
DE X V X 
 
X
P(X)
2
2
2
3 1 5
0 75
( )
( , )
,
V X E x E x
 
   
 
 
 

𝑓(𝑦) = {
1
18
𝑦 𝑝𝑎𝑟𝑎 0 < 𝑦 < 6
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦
6
2 2
0
6
2 2
0
6
3 2 2
0
1
18
1
2
18
1 1 1
18 9 18
( )
( )
( )
y ydy
y y ydy
y y y dy
 
 
 
 
  
  



6 6 6
4 3 2
2
0 0 0
4 3 2
2
1 1 1
18 4 9 3 18 2
1 6 1 6 1 6
4 4
18 4 9 3 18 2
18 32 16
2
y y y
 
  
  
  

2
 
 
E Y
6
2 2
0
6
3
0
6
4
0
4
1
18
1
18
1
18 4
1 6
18 4
18

 
 






E Y y ydy
y dy
y
2
2 2
18 16
2
( )   
    
 
 
 

y
V Y E Y E Y
2
1 4142
( )
,

 

y
DS Y
( )
( )
x
CV X
E X


V X 0
2
V c E c c
= 0
2
V c.X c .V X
V c X V X
6
6 4
2 2
2 2
2
2
20
16 64
20 4 33
1 22
1 22 1 1054
( )
( ) ,
( ) , %
( ) , , %
x x
E X x dx
V X
V X
DS X
  
 

 

1 1054
0 2551
4 333
( ) ,
( ) ,
( ) ,
  
DS X
CV X
E X
2
2
0 20
0 20
0 04 1 22 0 0488
0 0488 0 2211
0 2211
0 2551
0 8667
( ) ( , )
( ) ( , ) ( )
( ) , , , %
( ) , , %
( ) ,
( ) ,
( ) ,
V y V x
V y V x
V y
DS Y
DS Y
CV Y
E Y


  
 
  
2
1 1 10
1 10
1 21 1 22 1 4762
1 4762 1 215
1 215
0 2107
5 7667
( ) ( , )
( ) ( , ) ( )
( ) , , ,
( ) , ,
( ) ,
( ) ,
( ) ,
 

  
 
  
V W V x
V W V x
V W
DS W
DS W
CV W
E W
𝐸(𝑍) =
1
𝜎
[𝐸(𝑥) − 𝜇] =
1
𝜎
(𝜇 − 𝜇) = 0
𝑉(𝑍) =
1
𝜎2
[𝑉(𝑥) − 0] =
1
𝜎2
(𝜎2
− 0) = 1




x
X
Z . X 10
 
2
 
X
, Z.
Dada una variable aleatoria X, cualquiera sea su distribución de probabi-
lidad, con Esperanza ( )
E x y Varianza ( )
V x , puede afirmarse que la proba-
bilidad de que se presenten valores de la variable que se alejen de su Espe-
ranza en más de una distancia d , no supera a
𝑉(𝑥)
𝑑2 .
  2
  
( )
Pr ( )
V x
X E x d
d
(1)
De otra manera, también se puede decir que la probabilidad que se presen-
ten valores de la variable que se alejen de su Esperanza en menos de una
distancia d , es mayor a 1 −
𝑉(𝑥)
𝑑2
  2
1
   
( )
Pr ( )
V x
X E x d
d
(2)
   
, ,
E X d E X d
   
    
   
   
,
E X d E X d
 
 
 
E(X)-d E(X) E(X)+d
d
d
1 2
2
2 2
( ) ( )
( ) ( ) ( )
R R
V X x f x dx
V X x f x dx x f x dx

 


 
 
 
   
   
   

 
1
2
( ) > ( )
R
V X x f x dx


 
 

x 

 
 
   
1
1
2
2
2
2
( ) > ( )
( ) > ( )
( )
( ) > Pr - > Pr - > <
R
R
V X d f x dx
V X d f x dx
V X
V X d X d X d
d
 



 
 
 
2
49
100 10 0 49
100
( )
Pr
Pr ,
V x
X d
d
X

  
   
 

 
 
2
1
49
100 20 1 0 88
400

   
    
( )
Pr
Pr ,
V x
X d
d
X
d
sólo se requiere conocer
d
d k

 
 
2
2 2
2
2
2 2
2
1
1
1
1

 


 

  

   
 
Pr
Pr
X k
k
k
X k
k
k
4 03
1 35
,
,




   
 
  1
𝑓(𝑦) = {
1
18
𝑦 𝑝𝑎𝑟𝑎 0 < 𝑦 < 6
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦
𝜇 = 4 𝑦 𝜎 = 1,41
   
 
 
 
𝑓(𝑥) = {20𝑥3
(1 − 𝑥) 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1
0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
y=4x+50
2
2
2
2
2
Pr d X d 1
d
Pr X d 1
d
1
Pr X k 1
k
2
2
2
Pr X d
d
1
Pr X k
k
Ω= ( C; C) , ( S; C ) , ( C; S ) . ( S; S )
0
0,05
0,1
0,15
0,2
0,25
0,3
0,35
0,4
1 2 3 4 5 6 7
≤
≥
≤ ≤ ≤
   
   
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1
0 1 2 3 4 5 6 7 8
F(x)
0,00
0,05
0,10
0,15
0,20
0,25
0,30
0,35
0,40
0,45
0,50
1 2 3 4 5
0,00
0,10
0,20
0,30
0,40
0,50
0,60
0,70
0,80
0,90
1,00
0 1 2 3 4 5 6
F(x)
xi p( xi)
-1 0,07
0 0,13
1 0,20
2 0,27
3 0,33
Total 1,00
0,00
0,05
0,10
0,15
0,20
0,25
0,30
0,35
1 2 3 4 5
∫
1
2
3
1
𝑑𝑥 =
1
2
(3 − 1) = 1
≤ ≤
∫
1
2
𝑢
1
𝑑𝑥 =
1
2
(𝑢 − 1)
 
∫ 1/2𝑑𝑥
2
1
∫ 1/2𝑑𝑥
2
1,5
a) Donde f(x) = F´(x)
= 2kx
3
0
2 1
kxdx 

 

≥ ∫ 2/9𝑥 𝑑𝑥
3
1
≤
≤
1 5
0
1
2
y
dy 

≠
≤ ≤

≤
∫ 6 𝑦5
𝑑𝑦 = 6
𝑦6
6
| =
0
𝑢
𝑢
0
𝑦6
∫ 6 𝑦5
𝑑𝑦
1
0,8
Actividad Esperanza E(X2
) Varianza
3 4,03 18,07 1,83
4 4,091 17,8 1,06
5 1,667 4,333 1,56
6 73 10550 5221
7 2 4,333 0,33
8 2 4.5 0.5
9 0,857 0,75 0,02
 
 
2
2 2
2
2
2 2
2
1
1
1
1
Pr
Pr
X k
k
k
X k
k
k

 


 

  

   
 
≤ ≤ ≤ μ ≤ ≤ μ ≤  μ  ≤
 μ 
 μ 
 μ ≤
≤ |𝑋 − 𝜇|≤ ≤ |𝑥 − 𝜇|≤

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Estadistica U4

  • 3.   Tabla 2 - Función de distribución   p( ) P(X ) x x
  • 4. 1 1 ( ) k i i p x     1 1 ( ) k i i p x      ( ) ( ) F a P X a    ( ) ( ) i i x a F a p x  𝑃(𝑎 < 𝑥 ≤ 𝑏) = 𝐹(𝑏) − 𝐹(𝑎) = ∑ 𝑝(𝑥𝑖) 𝑎<𝑥≤𝑏  F(x ) =1 k 1 0 ( ) F x   0 y entero. h  
  • 5. 1 0     si el cliente compra una notebook si el cliente compra una computadora de escritorio Y         0,80 si x=0 ( ) 0,20 si x= 1 0 si x 0 o x 1 p x 0 04 0 1 2 3 4 5 0        , con , , , , ( ) otro valor de x K x x y p x ( ) p x 1 1 ( ) k i i p x    0 1 2 3 4 5 1 0 04 0 08 0 12 0 16 0 20 1 6 0 60 1 1 0 60 6 0 06667 ( ) ( ) ( ) ( ) ( ) ( ) ( , ) ( , ) ( , ) ( , ) ( , ) , , , p p p p p p K K K K K K K K K                      
  • 6.  2 4 2 4 2 3 4 2 4 0 147 0 187 0 227 0 56                ( ) ( ) ( ) ( ) ( ) ( ) , , , , i i x P X p x p p p P X 2 4 4 1 2 4 0 735 0 174 2 4 0 56            ( ) ( ) ( ) ( ) , , ( ) , P X F F P X P X  3 4 4 5 0 227 0 267 0 49 4 1 3 1 0 508 4 0 49                ( ) ( ) ( ) ( ) , , , ( ) ( ) , ( ) , P X P X p p P X F P X
  • 7. 𝑝(𝑥) = { 0,30 𝑠𝑖 𝑥 = 2 0,10 𝑠𝑖 𝑥 = 4 0,05 𝑠𝑖 𝑥 = 6 0,15 𝑠𝑖 𝑥 = 8 0,15 𝑠𝑖 𝑥 = 10 0,25 𝑠𝑖 𝑥 = 12 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥 6 2 4 0 30 0 10 0 40 6 4 0 40        ( < ) ( ) ( ) , , , ( < ) ( ) , P X p p P X F
  • 8. a- Comprobar que es una función de cuantía. b- Graficar p(x). c- Graficar F(x). d- Calcular P( X  2). e- Calcular P(X  5). f- Calcular P( 3  X  9). 𝑝(𝑥) = {𝑘𝑥2 𝑝𝑎𝑟𝑎 𝑥 = 1,2 3,4,5 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥     𝑝(𝑥) = { 𝑥 + 2 15 𝑝𝑎𝑟𝑎 𝑥 = −1,0,1,2 3 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥
  • 9. Variable Clase LI LS MC FA FR Gastos 1 272,00 522,75 397,38 7 0,02 Gastos 2 522,75 773,50 648,13 30 0,08 Gastos 3 773,50 1024,25 898,88 51 0,13 Gastos 4 1024,25 1275,00 1149,63 102 0,26 Gastos 5 1275,00 1525,75 1400,38 105 0,26 Gastos 6 1525,75 1776,50 1651,13 65 0,16 Gastos 7 1776,50 2027,25 1901,88 36 0,09 Gastos 8 2027,25 2278,00 2152,63 4 0,01
  • 10. 146,63 597,98 1049,33 1500,68 1952,03 2403,38 Gastos 0,00 0,07 0,14 0,21 0,28 frecuencia relativa ( ) f x 0 ( ) f x para todo x  1 ( ) f x dx     ( , )   205,13 740,07 1275,00 1809,93 2344,87 Gastos 0,00 0,04 0,08 0,13 0,17 frecuencia relativa 104,83 494,89 884,94 1275,00 1665,06 2055,11 2445,17 Gastos 0,00 0,05 0,10 0,15 0,20 frecuencia relativa
  • 11. ( ) ( ) b a P a X b f x dx     P(a X b) P(a X b) P(a X b) P(a X b)            500 250 250 500 ( ) ( ) P X f x dx     La función de distribución es la que proporciona la probabilidad acu- mulada para los diferentes valores de la variable. Para las variables continuas, será el área bajo la curva de la función de densidad a la izquierda de un valor particular x, o sea acumulada desde el menor valor hasta un valor determina- do de la variable. Simbólicamente: ' ' ' ( ) ( ) ( ) x F x P X x f x dx     
  • 12. ( ) ( ) '( ) dF x f x F x dx   ( ) F x ( ) f x ( ) F  1 ( ) f x dx      1 lim ( ) x F x   0 lim ( ) x F x    0 h   ( ) ( ) ( ) ( ) ( ) ( ) b a P a X b F b F a P a X b f x dx f x dx            
  • 13. 250 500 500 250 ( ) ( ) ( ) P X F F     𝑓(𝑦) = { 1 18 𝑦 𝑝𝑎𝑟𝑎 0 < 𝑦 < 6 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦 ( ) f y 0 ( ) f y  6 0 1 ( ) f y dy   6 6 2 0 0 2 1 1 18 36 1 6 36 1 ydy y     ( ) f y 2 0 1 1 18 36 ( ) u F u ydy u    146,63 710,81 1275,00 1839,19 2403,38 Gastos 0,00 0,25 0,50 0,75 1,00 frec. rel. acumulada
  • 14. 𝑃(𝑦 ≥ 5) = ∫ 1 18 𝑦 𝑑𝑦 6 5 = 1 36 (62 − 52) = 0,3056 2 5 1 5 1 1 5 36 0 3056 ( ) ( ) ( ) , P y F       𝑃(3 ≤ 𝑦 ≤ 4,5) = ∫ 1 18 𝑦 𝑑𝑦 4,5 3 = 1 36 (4,52 − 32) = 0,3125 2 2 3 4 5 4 5 3 1 1 4 5 3 36 36 0 3125 ( , ) ( , ) ( ) = ( , ) ( ) = , P y F F      y 6 5 4 3 2 1 0 0.4 0.3 0.2 0.1 0 y 6 5 4 3 2 1 0 1 0.8 0.6 0.4 0.2 0 y P(y) F(y) y Figura 7 (a) (b)
  • 15. En la Figura que sigue se muestra la zona sombreada correspondiente a la probabilidad calculada anteriormente. 𝑓(𝑥) = {𝑘(1 − 𝑥2 ) 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥 1 0 1 ( ) f x dx   1 2 0 1 1 ( ) k x dx    1 1 2 2 0 0 1 3 0 3 1 1 3 1 1 3 ( ) ( ) = ( ) = ( ) k x dx k x dx x k x k        3 1 1 1 3 2 1 3 3 2 ( ) ( ) k k k     f(Y)) f(Y) Y Y Y
  • 16. 𝑓(𝑥) = { 3 2 (1 − 𝑥2 ) 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥 0 0 0 1 1 1 4 ( ) x x F x x x           3 4 '( ) F x x  𝑓(𝑥) = { 4𝑥3 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥 4 0 40 1 0 40 1 0 40 0 974 ( , ) ( , ) ( , ) , P x F       0 4 0 4 0 0 0 50 0 50 0 50 0 841 ( ) , , , , F x x x x     𝑓(𝑦) = { 3 8 𝑦 𝑝𝑎𝑟𝑎 0 ≤ 𝑦 ≤ 2 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦
  • 17. 0 ( ) f y  2 0 1 ( ) f y dy   2 2 0 0 2 2 0 3 8 3 8 2 6 8 ( ) f y dy ydy y            8 3 6 8 1 2 ( ) . f y y y   𝑓(𝑦) = { 1 2 𝑦 𝑝𝑎𝑟𝑎 0 ≤ 𝑦 ≤ 2 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦 𝑓(𝑥) = { 1 2 𝑝𝑎𝑟𝑎 1 ≤ 𝑥 ≤ 3 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥        1 0 ) ( 2 kx x F x  0 0  x  3 x  3
  • 18. 1   𝑓(𝑦) = { 𝑦5 8 𝑦 𝑝𝑎𝑟𝑎 0 ≤ 𝑦 ≤ 1 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦 
  • 19. 1      [ ] ( ) k i i i E X x p x   
  • 20. 𝐸(𝑥) = 𝜇 = 0(0,067) + 1(0,107) + 2(0,147) + 3(0,187) + 4(0,227) + 5(0,267) = 3,205
  • 21.       [ ] ( ) E X x f x dx 𝑓(𝑦) = { 1 18 𝑦 𝑝𝑎𝑟𝑎 0 < 𝑦 < 6 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦 6 0 6 2 0 6 3 0 3 1 18 1 18 1 18 3 1 6 18 3 4 E Y y y dy y dy y              𝑓(𝑦) = { 3𝑥2 125 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 < 5 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥 5 2 0 5 0 4 5 0 3 125 3 125 3 125 4 3 75 ( ) , x E x x           3 x dx x dx
  • 22. 𝑓(𝑥) = { 3 2 (1 − 𝑥2 ) 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥   1 2 0 1 0 2 4 1 1 0 0 3 1 2 3 2 3 2 2 4 0 375 ( ) ( ) ( ) , E x x x x                         3 x dx x - x dx X f(X) 2 0,30 4 0,10 6 0,05 8 0,15 10 0,15 12 0,25
  • 23. Si consideramos que la probabilidad de que el proyecto sea aproba- do es de 0,30 ¿Cuál sería la utilidad neta esperada? P X c 1 E c c E c X c E X
  • 24. E c.X c.E X E X 0
  • 25. 𝑓(𝑥) = { 𝑥 16 𝑝𝑎𝑟𝑎 2 ≤ 𝑥 ≤ 6 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥 6 6 3 2 2 4 333 16 48 , x x E X x dx         0 20 0 20 0 20 4 333 0 8667 ( , ) , ( ) , , , E Y E x E x         
  • 26. 1 1 10 1 1 10 1 4 7667 5 7667 ( , ) , ( ) , , E W E x E x            2 2 ( ) X V X E X         
  • 27. 2 1          ( ) ( ) k i i i V X x p x 2           ( ) ( ) V X x f x dx μ 2 2           ( ) V X E X E X ( ) ( ) X DE X V X    X P(X)
  • 28. 2 2 2 3 1 5 0 75 ( ) ( , ) , V X E x E x              𝑓(𝑦) = { 1 18 𝑦 𝑝𝑎𝑟𝑎 0 < 𝑦 < 6 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦 6 2 2 0 6 2 2 0 6 3 2 2 0 1 18 1 2 18 1 1 1 18 9 18 ( ) ( ) ( ) y ydy y y ydy y y y dy                  6 6 6 4 3 2 2 0 0 0 4 3 2 2 1 1 1 18 4 9 3 18 2 1 6 1 6 1 6 4 4 18 4 9 3 18 2 18 32 16 2 y y y             2     E Y 6 2 2 0 6 3 0 6 4 0 4 1 18 1 18 1 18 4 1 6 18 4 18            E Y y ydy y dy y
  • 29. 2 2 2 18 16 2 ( )                y V Y E Y E Y 2 1 4142 ( ) ,     y DS Y ( ) ( ) x CV X E X  
  • 30. V X 0 2 V c E c c = 0 2 V c.X c .V X V c X V X
  • 31. 6 6 4 2 2 2 2 2 2 20 16 64 20 4 33 1 22 1 22 1 1054 ( ) ( ) , ( ) , % ( ) , , % x x E X x dx V X V X DS X          1 1054 0 2551 4 333 ( ) , ( ) , ( ) ,    DS X CV X E X 2 2 0 20 0 20 0 04 1 22 0 0488 0 0488 0 2211 0 2211 0 2551 0 8667 ( ) ( , ) ( ) ( , ) ( ) ( ) , , , % ( ) , , % ( ) , ( ) , ( ) , V y V x V y V x V y DS Y DS Y CV Y E Y           2 1 1 10 1 10 1 21 1 22 1 4762 1 4762 1 215 1 215 0 2107 5 7667 ( ) ( , ) ( ) ( , ) ( ) ( ) , , , ( ) , , ( ) , ( ) , ( ) ,            V W V x V W V x V W DS W DS W CV W E W
  • 32. 𝐸(𝑍) = 1 𝜎 [𝐸(𝑥) − 𝜇] = 1 𝜎 (𝜇 − 𝜇) = 0 𝑉(𝑍) = 1 𝜎2 [𝑉(𝑥) − 0] = 1 𝜎2 (𝜎2 − 0) = 1     x X Z . X 10   2   X , Z.
  • 33. Dada una variable aleatoria X, cualquiera sea su distribución de probabi- lidad, con Esperanza ( ) E x y Varianza ( ) V x , puede afirmarse que la proba- bilidad de que se presenten valores de la variable que se alejen de su Espe- ranza en más de una distancia d , no supera a 𝑉(𝑥) 𝑑2 .   2    ( ) Pr ( ) V x X E x d d (1) De otra manera, también se puede decir que la probabilidad que se presen- ten valores de la variable que se alejen de su Esperanza en menos de una distancia d , es mayor a 1 − 𝑉(𝑥) 𝑑2   2 1     ( ) Pr ( ) V x X E x d d (2)     , , E X d E X d                  , E X d E X d       E(X)-d E(X) E(X)+d d d
  • 34. 1 2 2 2 2 ( ) ( ) ( ) ( ) ( ) R R V X x f x dx V X x f x dx x f x dx                           1 2 ( ) > ( ) R V X x f x dx        x           1 1 2 2 2 2 ( ) > ( ) ( ) > ( ) ( ) ( ) > Pr - > Pr - > < R R V X d f x dx V X d f x dx V X V X d X d X d d            2 49 100 10 0 49 100 ( ) Pr Pr , V x X d d X                2 1 49 100 20 1 0 88 400           ( ) Pr Pr , V x X d d X d
  • 35. sólo se requiere conocer d d k      2 2 2 2 2 2 2 2 1 1 1 1                   Pr Pr X k k k X k k k 4 03 1 35 , ,        
  • 36.     1 𝑓(𝑦) = { 1 18 𝑦 𝑝𝑎𝑟𝑎 0 < 𝑦 < 6 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦 𝜇 = 4 𝑦 𝜎 = 1,41          
  • 37. 𝑓(𝑥) = {20𝑥3 (1 − 𝑥) 𝑝𝑎𝑟𝑎 0 ≤ 𝑥 ≤ 1 0 ∀ 𝑜𝑡𝑟𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥 y=4x+50
  • 38. 2 2 2 2 2 Pr d X d 1 d Pr X d 1 d 1 Pr X k 1 k 2 2 2 Pr X d d 1 Pr X k k
  • 39.
  • 40. Ω= ( C; C) , ( S; C ) , ( C; S ) . ( S; S )
  • 42. ≤ ≥ ≤ ≤ ≤         0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 0 1 2 3 4 5 6 7 8 F(x)
  • 43. 0,00 0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50 1 2 3 4 5 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00 0 1 2 3 4 5 6 F(x)
  • 44. xi p( xi) -1 0,07 0 0,13 1 0,20 2 0,27 3 0,33 Total 1,00 0,00 0,05 0,10 0,15 0,20 0,25 0,30 0,35 1 2 3 4 5
  • 45. ∫ 1 2 3 1 𝑑𝑥 = 1 2 (3 − 1) = 1 ≤ ≤ ∫ 1 2 𝑢 1 𝑑𝑥 = 1 2 (𝑢 − 1)   ∫ 1/2𝑑𝑥 2 1 ∫ 1/2𝑑𝑥 2 1,5 a) Donde f(x) = F´(x) = 2kx 3 0 2 1 kxdx     
  • 46. ≥ ∫ 2/9𝑥 𝑑𝑥 3 1 ≤ ≤ 1 5 0 1 2 y dy   ≠ ≤ ≤  ≤ ∫ 6 𝑦5 𝑑𝑦 = 6 𝑦6 6 | = 0 𝑢 𝑢 0 𝑦6
  • 48.
  • 49. Actividad Esperanza E(X2 ) Varianza 3 4,03 18,07 1,83 4 4,091 17,8 1,06 5 1,667 4,333 1,56 6 73 10550 5221 7 2 4,333 0,33 8 2 4.5 0.5 9 0,857 0,75 0,02
  • 50.
  • 51.     2 2 2 2 2 2 2 2 1 1 1 1 Pr Pr X k k k X k k k                   ≤ ≤ ≤ μ ≤ ≤ μ ≤  μ  ≤  μ   μ   μ ≤ ≤ |𝑋 − 𝜇|≤ ≤ |𝑥 − 𝜇|≤