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If the random variable $X$ has probability function

x          0       1       2       3       4       5

p(x)     0.03    0.06    0.13    0.20    0.31    0.27

,specify the probability function, distribution function, mean and variance of $Y=|X-3|$

callculus42
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user507124
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    What have you tried? Add that to your question (not in a comment). Further welcome to Math.SE. Take a look at both links How to ask a good question at Math.SE and for formatting MathJax. – drhab Nov 26 '17 at 14:25
  • First of all calculate $y=|x-3|$ for all $x=0,1,2,3,4,5$, For instance for $x=0$ we have $y=|0-3|=|-3|=3$ – callculus42 Nov 26 '17 at 14:43
  • As per above the values of y only range between 3,2,1,0,1,2 and if we take the probability of x at each of these points then it doesn't add upto 1. – user507124 Nov 26 '17 at 14:45
  • What "above" you mean? What were your calculations and results? – callculus42 Nov 26 '17 at 14:46
  • Here are the results: x 0 1 2 3 4 5 y 3 2 1 0 1 2 p(y) .20 .13 .06 .03 .06 .13 When I add these then the probability of these doesn't add up to 1. I am not sure if I am solving it right! Apologies for the poor format as I am very new to this. – user507124 Nov 26 '17 at 15:04
  • @user507124 Your values for p(y) are hard to interprete. I have posted a hint. If you have any further questions you can refer to it. – callculus42 Nov 26 '17 at 15:29

1 Answers1

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Hint:

If $x=1$ then $y=2$, If $x=5$ then $y=2$. Therefore for $y=2$ we get $p(y)=0.06+0.27=0.33$

If $x=2$ then $y=1$, If $x=4$ then $y=1$. Therefore for $y=1$ we get $p(y)=0.13+0.31=0.44$

Then the table for $y$ looks like

$$\begin{array}{|c|c|c|c|} \hline y&0&1&2&3 \\ \hline p(y)& ? &0.44&0.33&? \\ \hline \end{array}$$

It is comprehensible? Can you finish the table?

callculus42
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