• Author / Uploaded
• Jhonatan Espinoza

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distribucion binomial (muestreo con reposicion) presentaan a… p q 0.001 0.999

 n P( X  x)    p x q n x  x

x  0 , 1, 2, ........n

E(X) = n p

V(X) = n p q a) X:Nº de personas morosas p=0.2 q=0.8 n=10

y: N de personas al dia p=0.8 q=0.2

𝑃(𝑦=8)=(█(10@8)) 〖 0.8 〗 ^8 〖 0.2 〗 ^2=0.301989 𝑥=0,1…..10

𝑃(𝑋=𝑥)=(█(10@𝑥)) 〖 0.2 〗 ^𝑥 〖 0.8 〗 ^(10−𝑥) x

P(x)=hi 0 1 2 3 4 5 6 7 8 9 10

### ### ### ### ### ### ### ### ### ### ### ###

total a) P(x=2)= c)P(x>3)

### P(x=4)+….+P(

F(x)=Hi xp(x) 0.10737418 0 0.37580964 0.26843546 0.67779953 0.60397978 0.87912612 0.60397978 0.9672065 0.35232154 0.99363062 0.13212058 0.99913564 0.03303014 0.99992207 0.00550502 0.9999958 0.00058982 0.9999999 3.6864E-05 1 1.024E-06 2

### 0.12087388

d) P(x>=1)

### ### P(x=1)= 𝑃(𝑥≥1∩𝑥=1)=1-P(

E(x)=np=10*0 V(x)=npq=10*

2 1.6

### ###

###

DISTRIBUCION HIPERGEOMETRICA (MUESTREO SIN REPOSICION)

 r  N  r     x  n  x   P( X  x)  N   n 

x  0,1,....... min( n, r )

p=R/N

PROPORCION POBLACIONAL DE ARTICULOS DEFECTUOSOS

E(X)=np

V(x)=npq(N-n)/(N-1)

Ejemplo1 X=Nº de motores defectuosos N=50 n=5 Si X>=1 se revisa todo r=3 a)𝑃(𝑋=𝑥)=(█(3@𝑥))(█(47@5−𝑥))/((█(50@5)) )

x

total

p(x)=hi F(x)=Hi 0 0.72397959 0.72397959 1 0.25255102 0.97653061 2 0.02295918 0.9994898 3 0.0005102 1 1

b)

P(X>=1)=

0.27602041

X: nº DE MOTORES BUENOS r=47 n= 𝑃(𝑋=𝑥)=((█(@)))/

X

P 0 1 2 3 4 5

DISTRIBUCION GEOMETRI qqqq…qp

𝑥=0,….min⁡(𝑛=5, 𝑟=3)

P(X  x)  q x1 p E ( x) 

x  1,2.........

1 p n

qqqqqqp

q p2

V ( x)  q  x 1

x 1

qn 1 p  p. q 1

qqqqp a)

P ( X  5)  (0.99) 4 (0.01)  0.00960596

E(X)=1/p V(X)=q/p2 𝜎

100 9900 99.4987437

𝜇−2𝜎

-98.9974874 298.997487

𝜇+2𝜎

n

q  x 1

x 1

p

qn 1 q 1

0.94996338 p.

c) P(µ-2α