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February 5th, 2017, 01:12 PM   #1
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Interpreting probability from a cumulative distribution

The discrete cumulative distribution I have is
F(x)=( 0 for x<1, 1/3 for 1<or=x<4, 1/2 for 4<or=x<6, 5/6 for 6<or=x<10, 1 for x>or=10)
I'm asked to find P(2<X<or=6)
What im confused about is the overlap
I know f(1)=1/3, f(4)=1/6, and f(6)=1/3 but which would be the answer for this problem? I was thinking f(4)+f(6) because they're both within the range specified and 4 and 6 seem to be outputs of the random variable but what about for the values below 4?

Last edited by Rramos2; February 5th, 2017 at 01:15 PM.
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February 5th, 2017, 01:30 PM   #2
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$F(x) = \begin{cases}
0 &x < 1 \\
\dfrac 1 3 &1\leq x < 5 \\
\dfrac 1 2 &4 \leq x < 6\\
\dfrac 5 6 &6 \leq x < 10 \\
1 &10 \leq x
\end{cases}$

Find $P[2 < x \leq 6]$

$P[6]=\dfrac 5 6$

$P[2] = \dfrac 1 3$

$P[2 < x \leq 6] = P[6]-P[2] = \dfrac 5 6 - \dfrac 1 3 = \dfrac 1 2$

You don't need to worry about overlap with the cumulative distribution function. If you were given the probability density function you would have to take care to evaluate each bit with the proper limits but by giving you the CDF they've already taken care of that for you.
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February 5th, 2017, 03:11 PM   #3
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Oh that totally makes sense! You can just subtract by the bounds. Thanks alot bud.
Btw how were you able to write out the function so nicely? Are there tools on this site or do you have a special kind of keyboard app?
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February 5th, 2017, 05:16 PM   #4
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Quote:
Originally Posted by Rramos2 View Post
Oh that totally makes sense! You can just subtract by the bounds. Thanks alot bud.
Btw how were you able to write out the function so nicely? Are there tools on this site or do you have a special kind of keyboard app?
LaTex
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February 5th, 2017, 07:22 PM   #5
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this is kinda neat

Online LaTeX Equation Editor - create, integrate and download
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