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Assume the below probability distribution of a dice \begin{array}{c|ccccc} values & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline p(values) & 0.05 & 0.2 & 0.4 & 0.2 & 0.1 & 0.05 \\ \end{array}

When we draw a CDF function we get

\begin{array}{c|ccccc} values & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline p(values) & 0.05 & 0.2 & 0.4 & 0.2 & 0.1 & 0.05 \\ CDF- P(values) & 0.05 & 0.25 & 0.65 & 0.85 & 0.95 & 1 \\ \end{array}

The way I infer is
1. With 5% probability, I can say, at the max one can get value not more than 1
2. With 95% probability, at the max one can get value not more than 5

What does the rest 95% in the first case say? as well as the 5% in the second case says.

-Kamal.

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  • $\begingroup$ Cumulative distribution probabilities are usually interpreted as for example "there is 0.65 probability that one will observe a value less than or equal to three" and since this a discrete case this is the same thing as saying "there is 0.65 probabiliyt of seeing a 1, 2, or a 3" $\endgroup$ – Kamster Apr 2 '15 at 5:22
  • $\begingroup$ Thanks, I understand that but what would the opposite of this? What does the remaining 0.35 convey? $\endgroup$ – Kamal Apr 2 '15 at 5:25
  • $\begingroup$ There is a 0.35 probability that the value is greater than 3. You can see this by observing $p(4) + p(5) + p(6) = 0.35$. $\endgroup$ – William Stagner Apr 2 '15 at 5:31
  • $\begingroup$ Thanks William, with that can I also state the statement 'with $p = 0.35$ the minimum of all would be 3?' In other words can I say with a probability 0.35 at least I will get a value of 3? $\endgroup$ – Kamal Apr 2 '15 at 5:36
  • $\begingroup$ We said with probability of 0.65 the max value or the best value I may get is 3. In reverse to that with 0.35 probability can I say the least of all I may get is 3? $\endgroup$ – Kamal Apr 2 '15 at 5:45

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