First determine the distribution function F(x) f(x)= x/16 for 0<x<4 1/2-x/16 for 4[tex]\leq[/tex]x<8 0 for elsewhere So I determined this & came up with the function (which is correct) F(x)= 0 for x<o x^2/32 for 0<x<4 x/2-x^2/32-1 for 4[tex]\leq[/tex]x<8 1 for x[tex]\geq[/tex]8 Then there is the question P(X>10) Now, my thoughts is that from the original problem, the probability that x is greater than 8 is zero, & therefore, it should be zero... But, I got the question wrong & the professor stated that P(X>10) is 1. Which doesn't make a whole lot of sense to me, but she explained it that since it doesn't have an upper bound, it must be 1. Any ideas here? We had a homework question that stated a similar P(11<x<12) but here, it was zero because she stated it was bounded outside of the range.