Probabolidad

Probabilidad y Estadística
Soluciones Ejercicios Cartilla Nro. 1

UNIDAD 3: Modelos de Distribución de Probabilidad

1) X es v. a. discreta; Y es v. a. continua; M es v. a. continua; N es v. a.discreta; Q es v. a. continua; R es v. a. discreta.

2) a) i) pX(x) ( 0 para todo x, entonces c ( 0.
ii) [pic]

b) i) fX(x) ( 0 para todo x ( (, entonces c ( 0.
ii) [pic]Para encontrar FX(x):

x < 0 FX(x) = 0
0 ( x < 1 [pic]
x ( 1 FX(x) = 1

Por lo tanto:
[pic]

Luego, P (0.3 < X < 0.6) = FX(0.6) –FX(0.3) = 0.6 3/2 – 0.3 3/2 = 0.465 – 0.164 = 0.301

3) a) Para encontrar FX(x):
x < 0 FX(x) = 0

0 ( x < 1 [pic]
1 ( x < 2 [pic]
2 ( x < 3 [pic]

3 ( x< 4 [pic]
x ( 4 FX(x) = 1

Por lo tanto:
|1 | | | | | |
|0.99 | | | | | |
|0.94 | | | | | |
|0.78| | | | | |
| | | | | | |
| | | | | | |
| | | | | | |
|0.41 | | | | ||
| | | | | | |
| | | | | | |
| | | | | | |
| | | | | | |
|0 |1 |2 |3 |4|5 |

[pic]

b) P(X > 0) =1 – P(X < 1) = 1 – P(X = 0) = 1 – 0.41 = 0.59
c) P(X < 2) = P(X ( 1) = FX (1) = 0.78

4) a) Transformación: Y = X2
Recorrido: RY = {0, 1, 4, 9}Función de masa:
pY(0) = P(Y = 0) = P(X2 = 0) = pX(0) = [pic]= 0.216
pY(1) = P(Y = 1) = P(X2 = 1) = pX(1) = [pic]= 0.432
pY(4) = P(Y = 4) = P(X2 = 4) = pX(2) =[pic]= 0.288
pY(9) = P(Y = 9) = P(X2 = 9) = pX(3) = [pic]= 0.064
Luego:
|y |0 |1 |4 |9 |
|pY(y) |0.216 |0.432 |0.288 |0.064 |...