Let us assume that(adsbygoogle = window.adsbygoogle || []).push({}); Xhas Bernoulli distribution, with P(X= 1) =pand P(X= 0) =q= 1 -p. Of course, E(X) =pand Var(X) =pq. Now, sincepq< 1, standard deviation is bigger than variance.

I have got the following question:

Does this fact make standard deviations and theorems based on standard deviation (like Chebyshev's inequality) unusable in this case?

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# Standard deviation and Bernoulli distribution

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