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The bivariate distribution of X and Y is described below: X Y 1 2 1 0.21 0.47 2 0.14 0.18 A. Find the marginal probability distribution of X. 1: .35 2: .65 B. Find the marginal probability distribution of Y. 1: .68 2: .32 C. Compute the mean and variance of X. Mean = .35+1.3 Variance = (.35+2.6)-(.35+1.3)^2 C. Compute the mean and variance of Y.

Respuesta :

Answer:

Step-by-step explanation:

Given that the bivariate distribution of X and Y is described below:

x 1 2  

y    

1 0.21 0.14 0.35

0 0.47 0.18 0.65

Pdf of Y is

Y     1     0

p    0.68  0.32

C) Pdf of X is

x      1       2

p  0.35  0.65

Mean of X = [tex]1(0.35)+2(0.65)=1.65[/tex]

Var(x) = E(x^2)-Mean ^2 = [tex]1(0.35)+4(0.65)-1.65^2\\=2.95-2.7225\\=0.2275[/tex]

C) Mean of Y = 1(0.68)+0 = 0.68

Var(y) = 1(0.68)-0.68^2

= 0.2176

The marginal probability of distribution X is 0.36 and 0.64 respectively.

How to calculate the probability?

The marginal probability distribution of X will be:

P(x = 1) = 0.24 + 0.12 = 0.36

P(x = 2) = 0.47 + 0.17 = 0.64

The marginal probability distribution of Y will be:

P(y = 1) = 0.24 + 0.47 = 0.71

P(y = 2) = 0.12 + 0.17 = 0.29

The expected value for X will be:

= (1 × 0.36) + (2 × 0.64)

= 1.64

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