Answer to Question #122279 in Statistics and Probability for Joseph Se

Question

Determine the value of c that makes the function, f(x,y) = cxy a joint probability density function over the range 0 < x < 3 and 0 < y <x.

c = ______ Round your answer to four decimal places (e.g. 98.7654).

Determine the following:

a. P (X < 1.4, Y< 2.2) = ___ Round your answer to three decimal places (e.g. 98.765).
b. P ( 1.4 X < 2.2) = ____ Round your answer to three decimal places (e.g. 98.765).
c. P (Y > 1) =___ Round your answer to three decimal places (e.g. 98.765).
D. P (X < 2, Y < 2) = ____ Round your answer to three decimal places (e.g. 98.765).
E. E(X) = ____ Round your answer to three decimal places (e.g. 98.765).
F. E(Y) = ____ Round your answer to three decimal places (e.g. 98.765).

Answer 1

$$f(x, y)=cxy, 0<x<3, 0<y<x$$

$$\displaystyle\int_{0}^3\displaystyle\int_{0}^xcxydydx=c\displaystyle\int_{0}^3x\big[{y^2\over 2}\big]\begin{matrix} x \\ 0 \end{matrix}dx={c\over 2}\displaystyle\int_{0}^3x^3dx=$$

$$={c\over 8}[x^4]\begin{matrix} 3 \\ 0 \end{matrix}={81\over 8}c=1$$

Then $c=\dfrac{8}{81}\approx0.0988.$

The joint probability density function then becomes

$$f(x, y)=0.0988xy, 0<x<3, 0<y<x$$

Determine the following:

a.

$$P(X<1.4, Y<2.2)=\displaystyle\int_{0}^{1.4}\displaystyle\int_{0}^{x}0.0988xydydx=$$

$$=0.0494\displaystyle\int_{0}^{1.4}x[y^2]\begin{matrix} x\\ 0 \end{matrix}dx=0.01235[x^4]\begin{matrix} 1.4 \\ 0 \end{matrix}\approx0.047$$

b.

$$P(1.4<X<2.2)=P(1.4<X<2.2, 0<Y<x)=$$

$$=\displaystyle\int_{1.4}^{2.2}\displaystyle\int_{0}^{x}0.0988xydydx=0.0494\displaystyle\int_{1.4}^{2.2}x[y^2]\begin{matrix} x \\ 0 \end{matrix}dx=$$

$$=0.0494\displaystyle\int_{1.4}^{2.2}x^3dx=0.01235[x^4]\begin{matrix} 2.2 \\ 1.4 \end{matrix}\approx0.242$$

c.

$$P(Y>1)=P(1<X<3, 1<Y<x)=$$

$$=\displaystyle\int_{1}^{3}\displaystyle\int_{1}^{x}0.0988xydydx=0.0494\displaystyle\int_{1}^{3}x[y^2]\begin{matrix} x \\ 1 \end{matrix}dx=$$

$$=0.0494\displaystyle\int_{1}^{3}(x^3-x)dx=0.0494[0.25x^4-0.5x^2]\begin{matrix} 3 \\ 1 \end{matrix}\approx$$

$$\approx0.790$$

d.

$$P(X<2, Y<2)=\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{x}0.0988xydydx=$$

$$=0.0494\displaystyle\int_{0}^{2}x[y^2]\begin{matrix} x\\ 0 \end{matrix}dx=0.01235[x^4]\begin{matrix} 2\\ 0 \end{matrix}\approx0.198$$

e.

$$E(X)=\displaystyle\int_{0}^{3}\displaystyle\int_{0}^{x}x\cdot0.0988xydydx=$$

$$=0.0494\displaystyle\int_{0}^{3}x^2[y^2]\begin{matrix} x\\ 0 \end{matrix}dx=0.00988[x^5]\begin{matrix} 3 \\ 0 \end{matrix}\approx2.401$$

f.

$$E(Y)=\displaystyle\int_{0}^{3}\displaystyle\int_{0}^{x}y\cdot0.0988xydydx=$$

$$={0.0988\over 3}\displaystyle\int_{0}^{3}x[y^3]\begin{matrix} x\\ 0 \end{matrix}dx={0.01976\over 3}[x^5]\begin{matrix} 3 \\ 0 \end{matrix}\approx1.601$$