Thread: Geometric distribution and P( E(X) - 2 < E(X) < E(X) + 2)

If we have some random variable X, and it's distributed geometrically.



and



To get expected value it's just:



But how to get this probability:





thank you for your help.

Originally Posted by Nforce

If we have some random variable X, and it's distributed geometrically.
and
To get expected value it's just:

But how to get this probability:

If this is not a trick question, then it is flawed.
Because for any real number T it is true that

No, it's not a trick question,

I am trying to ask what is the probability, that expected value | E(X) - and a result of real experiment | < 2 ?

we throw a dice and we are looking at just one event: when we throw a 3, for example.

for this is E(X) = 6. In average we need 6 throws to get one 3 on a dice, then we stop.

But in a real experiment maybe we had need it 8 throws to get 3 on a dice.

So what is that probability?

It reads to me as if you are asking for the P that we need 5,6 or 7 goes (?) However your heading is not correct.

I think you mean P(E(X)-2<X<E(X)+2)