Solution to Question 41
Take the marginal density of X to be; fx(x) This implies that fxx=x1+x2xdy……………………eqn (i)
Integrating eqn (i) above and solving appropriately, we have;
=2xx+1-2x2………………….eqn (ii)
=2x
Considering the below condition where
0<x<1
& fxx=0, We can see that fy(y|X=x) =fX,Y(X,Y)fx(X)=1, x<y<x+10, Otherwise eqn…….. (iii)
Looking at eqn (iii), it can be seen that (Y|X=x) becomes uniform under the interval x, x+1
Thus, from the properties of uniform distribution, it can be confidently concluded that;
Mean=x+12 and Vraince=112
Thus;
The conditional variance=112

Solution to question 46
Considering the graph below, it can be seen that;

fT,X=fXx×fTtX=x………..eqn (i)
But
fXx=38x2
And
fTtX=x=12x-x
Considering the constraints 0 ≤x ≤2 and x t 2x and solving eqn (i), we get fT,X=12x-x×38x2 =38x
Hence, for a value of t within the constraints (0<t<2) as it varies from 12t to t , then the following becomes true fTt=12ttxdx Integrating the above equation, we have; fTt=964t2 Varying the same in the interval2<x<4), we have fTt=12tt38xdx Integrating the equation and solving accordingly we get fTt=-3t264+34 Therefore, probability that a randomly chosen claim on this policy is processed in three hours or more;
PrT>3=34(-3t264+34)dt
Integrating the equation and solving accordingly we get;
PrT>3=34-1-2764
=0.17