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USPshnik [31]
3 years ago
6

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 distr

ibution 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.
Mathematics
1 answer:
jasenka [17]3 years ago
8 0

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 = 1(0.35)+2(0.65)=1.65

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

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

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

= 0.2176

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Step-by-step explanation:

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Answer with explanation:

→→→Function 1

f(x)= - x²+ 8 x -15

Differentiating once , to obtain Maximum or minimum of the function

f'(x)= - 2 x + 8

Put,f'(x)=0

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Double differentiating the function

f"(x)= -2, which is negative.

Showing that function attains maximum at ,x=4.

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          = -31 +32

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→→→Function 2:

f(x) = −x² + 2 x − 3

Differentiating once , to obtain Maximum or minimum of the function

f'(x)= -2 x +2

Put,f'(x)=0

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Dividing both sides by , 2, we get

x=1

Double differentiating the function,gives

f"(x)= -2 ,which is negative.

Showing that function attains maximum at ,x=1.

f(1)= -1²+2 ×1 -3

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⇒⇒⇒Function 1  has the larger maximum.

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