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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.