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Given the following CDF

enter image description here

what is

$$P(T > 3)$$

and according to my answer key it's 1-1/2 = 1/2.

Can someone explain to me why it is 1-F(3), and would subtracting F(3) be subtracting 4 as well? Normally I saw stuff like t = 1,2,3,4,5,6...., but I've never seen something where it skips numbers and stuff like 1,3,5,7, ect ect. How does this affect the cdf and such? In short, I'm mainly confused as to how they figured out it was 1 - F(3).

Belphegor
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    $F(3)$ is by definition $P(T \leq 3)$. Using one of the probability axioms, we have $P(T > 3) = 1 - F(3) = 1 - \frac{1}{2} = \frac{1}{2}$. – user217285 Jul 07 '15 at 03:18
  • I see, but how does the interval affect this? I mean like, if I had 3<=t<4, would this still work? This interval thing is just killing me here :( – Belphegor Jul 07 '15 at 03:20
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    The interval does not matter at all. In fact, I didn't even look at the interval. To read a piecewise function, search for the value of $t$ you're looking for and see what the value of $F$ is at that point. – user217285 Jul 07 '15 at 03:23
  • I see, thanks a lot! – Belphegor Jul 07 '15 at 03:25

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Recall that by definition, the cumulative distribution function is the probability of having a value less than or equal to the argument of the function. So $$ P(T>3) = 1-P(T \leq 3)= 1-F(3) = \frac{1}{2} $$ What may be confusing you is that you have a piecewise "step" function. That is, there are probability jumps at $1$, $3$, $5$, and $7$, and no increase in the cumulative distribution (or probability for those values) between those integers.

miradulo
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