TRABAJO DE PROBABILIDAD

Trabajo de probabilidad

3.

N=15 y P=25% (Binomial)

f(x;p,n)= (n¦x) p^x q^(n-x) □(⇒┬ ) f(x;0.25,15)= (15¦x) 〖0.25〗^x• 〖0.75〗^(15-x)

De 3 a 6

P(3≤x ≤6 )=1- ∑_(i=3)^6▒〖(15¦x) 〖0.25〗^x• 〖0.75〗^(15-x) 〗

P(3≤x ≤6 )=0.225+ 0.225+0.165+0.092=0.707

P(3≤x ≤6 )=70.7%

Menos de 4

P( x <4 )=1- ∑_(i=0)^4▒〖(15¦x) 〖0.25〗^x• 〖0.75〗^(15-x) 〗

P( x <4 )=0.013+0.067+0.156+0.225

P( x <4 )=0.461= 46.1%

De 3 a 6

P( x >6 )=1-P( x ≤6 ) = 1- ∑_(i=0)^5▒〖(15¦x) 〖0.25〗^x• 〖0.75〗^(15-x) 〗

P( x >6 )=1-[0.013+0.067+0.156+0.225+0.225+0.165+0.092]

P( x >6 )=1-0.942= 0.057=5.70%

6.

f(x;p)=( 1- 〖p )〗^(x-1) p (Geométrica)

P=0.04

P( x=4 )= ( 1- 〖0.04 )〗^(4-1)• 0.04=0.035=3.5%

P( x=4 )= ∑_(x=1)^2▒〖( 1- 〖0.04 )〗^(x-1)• 0.04=0.04+0.078=0.078=7.8%〗

8.

Distribución normal

μ_x=800 horas y σ_x^2=40 horas

Más de 829 horas

P(x >829)=P ( z > (x- μ_x )/( σ_x^2 )) □(⇒┬ ) P ( z > (829-800 )/( 40)) □(⇒┬ ) P ( z>0.725)

P(x >829)=1-[P ( z ≤0.73 )]=1-[ 0.76730 ]

P(x >829)=0.2327=23.27%

Máximo 820 horas

P(x ≤820)=P ( z ≤ (x- μ_x )/( σ_x^2 )) □(⇒┬ ) P ( z ≤ (820-800 )/( 40)) □(⇒┬ ) P ( z≤0.5)

P(x ≤820)=0.69146

P(x ≤820)=69.146%

Entre 778 y 834 horas

P( 778 ≤x ≤834)=P ((778- μ_x )/( σ_x^2 ) z ≤ (834- μ_x )/( σ_x^2 )) □

P( 778 ≤x ≤834)= P (-0.55 ≤ z ≤0.85 )

P( 778 ≤x ≤834)=0.80234-[ 1-0.70884 ]

P( 778 ≤x ≤834)=0.51118=51.118%

5.

P=87% (Binomial negativa)

f(x;p,r)= ((n-1)¦(r-1))( 1- 〖p )〗^(x-1) p^r

r=2 y x=4

f(x;0.87,2)= ((4-1)¦(2-1))( 1- 〖0.87 )〗^(4-1) 〖0.87〗^2 □(⇒┬ ) f(x;0.87,2)= (3¦2) ( 1- 〖0.87 )〗^2 〖0.87〗^2

f(x;0.87,2)=0.038=3.83%

r=5 y x=10

f(x;0.87,5)= ((10-1)¦(5-1))( 1- 〖0.87 )〗^(5-1) 〖0.87〗^5 □(⇒┬ ) f(x;0.87,2)= (9¦4) ( 1- 〖0.87 )〗^5 〖0.87〗^5

f(x;0.87,2)=2.331 x 〖10〗^(-3)=0.233%

7. distribución poisson

P = 2 errores por página; λ=2

f(x;λ)= (e^(-λ) λ^x)/x!

P( x ≤3 )

P( x ≤3 )=∑_(x=0)^3▒(e^(-2) λ^x)/x!

P( x ≤3 )=0.135+ 0.271+0.271+0.180

P( x ≤3 )=0.857=85.7%

P( x=0 )

P( x ≤3 )= (e^(-2) λ^0)/0!

P( x ≤3 )=0.135=13.5%

P( x ≤3 )

P( x ≥3 )=1-∑_(x=0)^3▒(e^(-2) λ^x)/x!

P( x ≥3 )=1- [ 0.135+ 0.271+0.271 ]

P( x ≥3 )=0.677=67.7%

1. función de probabilidad (c = café, v= verde)

P( vv )=4/11*3/10=0.109=10.9%

P( cv )=7/11*4/10=0.255=25.5%

P( cv)=7/11*4/10=0.255=25.5%

P( cc )=7/11*6/10=0.381=38.1%

μ_x= E_x= [0*0.109]+ [1*0.510]+ [2*0.381]

E_x=1,272

Se espera sacar como mínimo un calcetín café

〖〖σ_x〗^2 = V〗_(x )= ∑_(x=0)^2▒〖( x-μ_(x ) )^2 f_x (x) 〗

〖 V〗_(x )= [(0-1,272)^2*0.109 ]+ [(1-1,272)^2*0.510]+ [(2-1,272)^2*0.381]

〖 V〗_(x )=0.416

〖σ_x= S〗_(x )= √(〖σ_x〗^2 )= √0.416=0.644

4.

a) probabilidad de pasar

f(x;p)=( 1- 〖p )〗^(x-1)

...