# cumulative distribution function

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• Oct 11th 2010, 01:34 PM
DINOCALC09
cumulative distribution function
Question attached.
• Oct 11th 2010, 01:38 PM
Plato
It cannot be done unless you complete that table.
You have given no values for $\displaystyle f(x)$.
• Oct 11th 2010, 01:41 PM
DINOCALC09
Sorry for confusion:

0 = 1/27
1 = 6/27
2 = 12/27
3 = 8/27
• Oct 11th 2010, 01:48 PM
DINOCALC09
So for an answer maybe something along the lines...

F(x) = 0 if x<0
F(x) = 1/27 if 0<=x<1
F(x) = 6/27 if 1<=x<2
F(x) = 12/27 if 2<=x<3
F(x) = 8/27 if 3<=x
• Oct 11th 2010, 01:51 PM
Plato
It will look like this:
$\displaystyle F_X (x) = \left\{ {\begin{array}{rc} {0,} & {x < 0} \\ {\frac{1} {{27}},} & {0 \leqslant x < 1} \\ ?, & {1 \leqslant x < 2} \\ ?, & ? \\ 1, & {3 \leqslant x} \\\end{array} } \right.$
Now you fill in the blanks.
• Oct 11th 2010, 01:56 PM
DINOCALC09
F(x) = 0 if x<0
F(x) = 1/27 if 0<=x<1
F(x) = 6/27 if 1<=x<2
F(x) = 12/27 if 2<=x<3
F(x) = 1 if 3<=x

Okay I see I did a mistake... it should be 1, not 8/27 for the 3<=x
• Oct 11th 2010, 02:21 PM
Plato
No you missed the accumulation.
$\displaystyle F_X (x) = \left\{ {\begin{array}{lc} {0,} & {x < 0} \\ {\frac{1} {{27}},} & {0 \leqslant x < 1} \\ {\frac{7} {{27}}} & {1 \leqslant x < 2} \\ ? & ? \\ 1 & {3 \leqslant x} \\ \end{array} } \right.$
• Oct 11th 2010, 02:33 PM
DINOCALC09
Ok I get it now. ACCUMULATION.... aka adding.

So we get 0+1/27+6/27+12/27=19/27.

Is my 2<=x<3 incorrect though?
• Oct 11th 2010, 02:39 PM
Plato
Quote:

Originally Posted by DINOCALC09
Is my 2<=x<3 incorrect though?

It is wrong also. Add the 'jump'.
When you finish $\displaystyle F_X(x)$ must be continuous on the right.
• Oct 11th 2010, 02:45 PM
DINOCALC09
What do you mean by jump?

I know what continuous is kinda, but not sure how it relates to inequalities.
• Oct 11th 2010, 02:54 PM
Plato
Quote:

Originally Posted by DINOCALC09
What do you mean by jump?

It should go from $\displaystyle \frac{7}{27}$ to $\displaystyle \frac{19}{27}$.

That is a jump of $\displaystyle f(2)=\frac{12}{27}$.
• Oct 11th 2010, 03:11 PM
DINOCALC09
yes, i see that. however, don't see what's wrong with the inequality necessarily. Should 2 not be inclusive?
• Oct 11th 2010, 03:23 PM
Plato
Frankly this is tiresome. It is
$\displaystyle F_X (x) = \left\{ {\begin{array}{lc} {0,} & {x < 0} \\ \\ {\frac{1} {{27}},} & {0 \leqslant x < 1} \\ \\ {\frac{7}{{27}}} & {1 \leqslant x < 2} \\ \\ {\frac{19}{{27}}} & {2 \leqslant x < 3} \\ \\ 1 & {3 \leqslant x} \\ \end{array} } \right.$

I am done with the thread.