The upper quartile of the PDF f(x) is the value of F(x) at x= 0.75
Now F(x) is f(x) dx = (1/2)*x
So, F(0.75) = 0.75/2 = 0.375
So option b is correct.
2. We know the total probability must be equal to 1. From the table we can see that this is satisfied.
Now, Z will take the value 2 when both x and y are 1 so the required probability is 0.045
Z will be 3 when x=1 and y=2 or x= 2 and y=1. So the required probability is 0.08+0.08 = 0.16
Z will be 4 when both x and y are 2 or x=3, y=1 or x=1,y=3. So the required probability is 0.14+0.05+0.125 = 0.315
Similarly probability for Z=5 is 0.14+0.075 =0.215
And Probability for Z=6 is 0.27
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