Probabilidad

U J A T

D A I A

william alexis felix

PROF: M. C. RENE SEBASTIAN MORA ORTIZ

GRUPO: K

“TAREA # 3 DE PROBABILIDAD Y ESTADISTICA”CUNDUACAN, TAB. A NOVIEMBRE DEL 2009

Suponga que la variable aleatoria x tiene una distribución geométrica con p= 0.5, calcule las probabilidades siguientes:

a) P( x = 1) = (1 – 0.5)1-1 (0.5 ) =0.5

b) P( x = 4) =(1 – 0.5)4-1 (0.5 ) = 0.0625

c) P( x = 8) = (1 – 0.5)8-1 (0.5 ) = 0.00390

d) P( x ≤ 2) = P( x = 1) + P( x = 2)
= [ (1 – 0.5)1-1 (0.5 )] + [ (1 – 0.5)2-1(0.5 )] = 0.5 + 0.25 = 0.75

e) P( x > 2) = P( x = 3) + P( x = 4) + P( x = 5) + P( x = 6) + P( x = 7) + P( x = 8)

= [ (1 – 0.5)3-1 (0.5 )] + [ (1 – 0.5)4-1 (0.5 )] + [(1 – 0.5)5-1 (0.5 )] +

[ (1 – 0.5)6-1 (0.5 )] + [ (1 – 0.5)7-1 (0.5 )] + [ (1 – 0.5)8-1 (0.5 )]

= 0.125 + 0.0625 + 0.03125 + 0.0156 + 0.00781 + 0.00390= 0.246

Suponga que X es una variable aleatoria binomial negativa con p = 0.2 y r = 4, calcule lo siguiente:

a) E (X) = 4 = 20
0.2


20 - 1
4 - 1

b)P( x = 20) = (1 – 0.2)20-1 (0.2 )4

= (969)(0.02814)(0.0016) = 0.04362

19 - 1
4 - 1

c) P( x = 19) = (1 – 0.2)19-1 (0.2 )4

=(816)(0.03518)(0.0016) = 0.04593

21 - 1
4 - 1

d) P( x = 21) = (1 – 0.2)21-1 (0.2 )4

= (1140)(0.0225)(0.0016) = 0.04107

Suponga que X tiene unadistribución hipergeométrica con N = 100, n = 4 y K = 20, calcule lo siguiente:

100 - 20
4 - 1
20
1

a) P( x = 1) = = (20) (82,160.0520) =0.41905
100
4
3, 921,225.1360

20
6
100 - 20
4 - 6

b) P( x = 6) = =
100
4


100 - 20
4 -...