Probability given a cumlative funcion

[kuahji]

First determine the distribution function F(x)

f(x)= x/16 for 0<x<4
1/2-x/16 for 4\leqx<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\leqx<8
1 for x\geq8

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.

 

[Dick]

I'm with you. P(X>10) is 0. P(X<10) is 1. I don't see how you could interpret the problem otherwise.

 

[kuahji]

Do you think given that it is a big X & not a little x has anything to do with it? I could see if it was F(X>10) perhaps... the notation always ends up getting the best of me.

 

[HallsofIvy]

Could you state the problem exactly it is given? Yes, P(x> 10) is 0 but you titled this "cumulative" function and have F(x)= 1 for x\ge 8.

 

[kuahji]

It is stated "Determine the distribution function, F(x) of the random variable X whose probability density is..." what I outlined first.

It then says "Use the information about F(x) in the previous question to determine" (which is the cumulative function I outlined)
a) P(X<6) which I did & showed the work 6/2-36/32-1 = 7/8 (which was correct)
b) P(X>10) which I showed the work 1-1=0 (which was marked wrong)
in class she specifically said b) was suppose to be 1. 1-0=1

 

[Dick]

I think this is all a misunderstanding. If P(X<6) is 7/8 then P(X>10) CANNOT be 1. They add to a number larger than 1 and they are mutually exclusive. That would just plain be silly. Please ask your teacher to explain. I can't.