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

• April 15th 2012, 02:41 AM
Nforce
Geometric distribution and P( E(X) - 2 < E(X) < E(X) + 2)
If we have some random variable X, and it's distributed geometrically.

$X = G(p)$

and

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

To get expected value it's just:

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

But how to get this probability:

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

• April 15th 2012, 04:13 AM
Plato
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.
$X = G(p)$ and $p = \frac{1}{6}$
To get expected value it's just:
$E(X) = \frac{1}{\frac{1}{6}} = 6$
But how to get this probability: $P( E(X)-2 < E(X) < E(X)+2 )$

If this is not a trick question, then it is flawed.
Because for any real number T it is true that $\mathcal{P}(T-2
• April 15th 2012, 05:27 AM
Nforce
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?
• April 15th 2012, 10:08 AM
biffboy
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)