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Math Help - cumulative distribution function

  1. #1
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    cumulative distribution function

    Question attached.
    Attached Thumbnails Attached Thumbnails cumulative distribution function-4.4-statistics.jpg  
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  2. #2
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    It cannot be done unless you complete that table.
    You have given no values for f(x).
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  3. #3
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    Sorry for confusion:

    0 = 1/27
    1 = 6/27
    2 = 12/27
    3 = 8/27
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  4. #4
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    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
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  5. #5
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    It will look like this:
    F_X (x) = \left\{ {\begin{array}{rc}<br />
   {0,} & {x < 0}  \\<br />
   {\frac{1}<br />
{{27}},} & {0 \leqslant x < 1}  \\<br />
   ?, & {1 \leqslant x < 2}  \\<br />
   ?, & ?  \\<br />
   1, & {3 \leqslant x}  \\\end{array} } \right.
    Now you fill in the blanks.
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  6. #6
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    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
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  7. #7
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    No you missed the accumulation.
    <br />
F_X (x) = \left\{ {\begin{array}{lc}<br />
   {0,} & {x < 0}  \\<br />
   {\frac{1}<br />
{{27}},} & {0 \leqslant x < 1}  \\<br />
   {\frac{7}<br />
{{27}}} & {1 \leqslant x < 2}  \\<br />
   ? & ?  \\<br />
   1 & {3 \leqslant x}  \\<br /> <br />
 \end{array} } \right.
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  8. #8
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    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?
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  9. #9
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    Quote Originally Posted by DINOCALC09 View Post
    Is my 2<=x<3 incorrect though?
    It is wrong also. Add the 'jump'.
    When you finish F_X(x) must be continuous on the right.
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  10. #10
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    What do you mean by jump?

    I know what continuous is kinda, but not sure how it relates to inequalities.
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  11. #11
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    Quote Originally Posted by DINOCALC09 View Post
    What do you mean by jump?
    It should go from \frac{7}{27} to \frac{19}{27}.

    That is a jump of f(2)=\frac{12}{27}.
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  12. #12
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    yes, i see that. however, don't see what's wrong with the inequality necessarily. Should 2 not be inclusive?
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  13. #13
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    Frankly this is tiresome. It is
    <br />
F_X (x) = \left\{ {\begin{array}{lc}<br />
   {0,} & {x < 0}  \\ \\<br />
   {\frac{1}<br />
{{27}},} & {0 \leqslant x < 1}  \\ \\<br />
   {\frac{7}{{27}}} & {1 \leqslant x < 2}  \\ \\<br />
   {\frac{19}{{27}}} & {2 \leqslant x < 3}  \\ \\<br />
   1 & {3 \leqslant x}  \\<br /> <br />
 \end{array} } \right.

    I am done with the thread.
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