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Why nothing is mentioned about the standard deviation of a Bernoulli trial ?

Does it even make sense if I try to visualize it ?

Oleg
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  • With usual terms ($p(1)=p,;p(0)=1-p$) then $\sigma =\sqrt {p(1-p)}$. This comes up a lot! In particular it comes up if you have many trials and would like to approximate with a normal. – lulu Mar 27 '16 at 20:08

2 Answers2

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Why nothing is mentioned about the standard deviation of a Bernoulli trial ?

Where? Standard textbooks will have some mention regarding the variance, at a minimum. Regardless, you (I) can derive it yourself (myself).

If $X$ is a Bernoulli trial with chance $p$, then $$E[X] =\sum_{k = 0}^1 kP(X = k) = 0\cdot(1-p)+1\cdot p = p,$$ and $$E[X^2] = \sum_{k = 0}^1 k^2P(X= k) = 0^2\cdot(1-p)+1^2\cdot p = p.$$

Hence $$\text{Var}(X) = E[X^2]-\{E[X]\}^2 = p-p^2 = p(1-p)$$ and so $$\text{SD}(X) = \sqrt{p(1-p)}.$$

Em.
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If $p$ is the probability of success, then the variance of a Bernoulli random variable is $p(1-p)$, hence the standard deviation is $\sqrt{p(1-p)}$.

carmichael561
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  • yes, but given for e.g the set [10,612,642,3,-64] I can easily see a standard deviation if I were to draw the bell curve but if I am given many 1(success) or 0(failure) of an experiment then it's not clear how spread apart really they are. – Oleg Mar 27 '16 at 20:47
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    Well that's a different random variable. If $X$ counts the number of successes in $n$ trials, then the standard deviation of $X$ is $\sqrt{np(1-p)}$. – carmichael561 Mar 27 '16 at 20:49