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$$ x\quad|\quad 0\quad|\quad1\quad|\quad2\quad|\quad4 $$ $$ f_x(x)\quad|\quad1/27\quad|\quad8/27\quad|\quad10/27\quad|\quad8/27 $$

Okay I want to make sure I am doing this correctly: $$ F_x(0) = P(X\leq 0) = P(X=0) = 0.0370 $$ $$ F_x(1) = P(X\leq 1) = P(X=0) + P(X=1) = 0.0370 + 0.2963 = 0.3333 $$ $$ F_x(1) = P(X\leq 2) = P(X=0) + P(X=1) + P(X=2) = 0.0370 + 0.2963 + 0.3704 = 0.7037 $$ $$ F_x(1) = P(X\leq 4) = P(X=0) + P(X=1) + P(X=2) + P(X=4) = 0.0370 + 0.2963 + 0.3704 + 0.2963 = 1 $$ I believe this is how to do the CDF.

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There seem to be some typographical errors in what you have written, but you have the right idea. I would suggest using common fractions instead of decimal fractions, so that you come up with $F_X(0) = \frac{1}{27}$ instead of $F_X(0) = 0.0327$, etc. The pattern of the result will be much clearer that way. Finally, you may be expected to specify $F_X(x)$ not just for integer values of $x$, but for all numbers $x$, so that your answer may need to be expressed as $F_X(x) = \frac{1}{27}$ for all $x$, $0 \leq x < 2$. Watch out for the $\leq$ and $<$ signs when you are writing such things. –  Dilip Sarwate Oct 6 '11 at 1:25
    
To repeat part of the comment by @Dilip Sarwate, you would be expected to write $F_X(x)=0$ if $x<0$; $F_x(x)=\frac{1}{27}$ if $x \le 0<1$; and so on ending up with $F_X(x)=1$ if $x \ge 4$. The cdf is a function defined for all real $x$. –  André Nicolas Oct 6 '11 at 1:50
    
third line..it should be $Fx(2)$ ,fourth line... it should be $Fx(4)$ –  pedja Oct 6 '11 at 3:02
    
Cool, pretty much add a a "basis". –  Tommy Oct 6 '11 at 3:04

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