Question attached.

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- Oct 11th 2010, 01:34 PMDINOCALC09cumulative distribution function
Question attached.

- Oct 11th 2010, 01:38 PMPlato
It cannot be done unless you complete that table.

You have given no values for $\displaystyle f(x)$. - Oct 11th 2010, 01:41 PMDINOCALC09
Sorry for confusion:

0 = 1/27

1 = 6/27

2 = 12/27

3 = 8/27 - Oct 11th 2010, 01:48 PMDINOCALC09
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 PMPlato
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 PMDINOCALC09
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 PMPlato
No you missed the ac

**cumulation**.

$\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 PMDINOCALC09
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 PMPlato
- Oct 11th 2010, 02:45 PMDINOCALC09
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 PMPlato
- Oct 11th 2010, 03:11 PMDINOCALC09
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 PMPlato
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.