Standard deviation and Bernoulli distribution

[Kocur]

Let us assume that X has Bernoulli distribution, with P(X = 1) = p and P(X = 0) = q = 1 - p. Of course, E(X) = p and Var(X) = pq. Now, since pq < 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?

 

[mathman]

I don't see why it should make any difference. Could you elaborate on what's your concern?

 

[EnumaElish]

I don't see why it would. You don't need Bernoulli for this property (sigma < sigma^2), as there is no rule that says the variance of, say, a normal dist. has to be > 1.

 

[Hurkyl]

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

Read the theorem -- it tells you exactly when you can use it.