Answer:
-764.28
Step-by-step explanation:
Given the joint cumulative distribution of X and Y as
[tex]F(x,y) = \frac{xy(x+y)}{2000000}\ \ \ \, \ 0\leq x100, 0\leq y\leq 100[/tex]
#First find [tex]F_x[/tex] and probability distribution function ,[tex]f_x(x)[/tex]:
[tex]F_x(x)=F(x,100)\\\\\\=\frac{100x(x+100)}{2000000}\\\\\\\\=\frac{100x^2+10000x}{2000000}\\\\\\=>f_x(x)=\frac{x}{10000}+\frac{1}{200}[/tex]
#Have determined the probability distribution unction ,[tex]f_x(x)[/tex], we calculate the Expectation of the random variable X:
[tex]E(X)=\int\limits^{100}_0 \frac{x^2}{10000}+\frac{x}{200} dx \\\\\\\\=|\frac{x^3}{30000}+\frac{x^2}{400}|\limits^{100}_0\\\\=58.33\\\\[/tex]
#We then calculate [tex]E(X^2)[/tex]:
[tex]E(X^2)=\int\limits^{100}_0 \frac{x^3}{10000}+\frac{x^2}{200}\ dx\\\\=\frac{x^4}{40000}+\frac{x^3}{600}|\limits^{100}_0=4166.67\\\\Var(X)=E(X^2)-(E(X))^2=4166.67-58.33^2\\\\Var(X)=764.28[/tex]
Hence, the Var(X) is 764.28 Â
