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

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

$\displaystyle X = G(p)$

and

$\displaystyle p = \frac{1}{6}$

To get expected value it's just:

$\displaystyle E(X) = \frac{1}{\frac{1}{6}} = 6 $

But how to get this probability:

$\displaystyle P( E(X)-2 < E(X) < E(X)+2 )$

thank you for your help.

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

Quote:

Originally Posted by

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

$\displaystyle X = G(p)$ and $\displaystyle p = \frac{1}{6}$

To get expected value it's just:

$\displaystyle E(X) = \frac{1}{\frac{1}{6}} = 6 $

But how to get this probability: $\displaystyle P( E(X)-2 < E(X) < E(X)+2 )$

If this is not a trick question, then it is flawed.

Because for **any real number** *T* it is true that $\displaystyle \mathcal{P}(T-2<T<T+2)=1.$

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

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?

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

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)