Answer:
Step-by-step explanation:
Given [tex]f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y <2 [/tex]
a)
we know that [tex]\int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} {f(x,y)} \, dxdy=1[/tex]
therefore [tex]\int\limits^{40}_{-0}\int\limits^2_{0} {c(4x+2y+1)} \, dxdy=1[/tex]
on integrating we get
c=(1/6640)
b)
[tex]P(X>20, Y>=1)=\int\limits^{40}_{20}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy[/tex]
on doing the integration we get
=0.37349
c)
marginal density of X is
[tex]f(x)=\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dy[/tex]
on doing integration we get
f(x)=(4x+3)/3320 ; 0<x<40
marginal density of Y is
[tex]f(y)=\int\limits^{40}_{0} {\frca{1}{6640}(4x+2y+1)} \, dx[/tex]
on doing integration we get
[tex]f(y)=\frac{(y+40.5)}{83}[/tex]
d)
[tex]P(0<X<40, 2>Y>1)=\int\limits^{40}_{0}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy[/tex]
solve the above integration we get the answer
e)
[tex]P(X>20, 0<Y<2)=\int\limits^{40}_{20}\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dxdy[/tex]
solve the above integration we get the answer
f)
Two variables are said to be independent if there jointprobability density function is equal to the product of theirmarginal density functions.
we know f(x,y)
In the (c) bit we got f(x) and f(y)
f(x,y)cramster-equation-2006112927536330036287f(x).f(y)
therefore X and Y are not independent
