# Let X be a random variable with V(X)=25

Let X be a random variable with $\mathbb{V}\left(X\right)=25$
It also satisfies $\mathbb{P}\left(X>20\right)>\frac{1}{4}$
Which of the following is true?
- $\mathbb{E}\left(X\right)>10$
- $5<\mathbb{E}\left(X\right)\le 10$
- $0<\mathbb{E}\left(X\right)\le 5$
- $\mathbb{E}\left(X\right)\le 0$
I tried using Chebyshev's inequality and found out that $\mathbb{E}\left(X\right)>5$ but wasn't sure how to go on about it.
You can still ask an expert for help

## Want to know more about Probability?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

occuffick24
Step 1
As a fairly major hint, you should use Chebyshev's inequality to show
$\mathbb{E}X\ge c-\sqrt{\frac{\mathbb{V}X}{\mathbb{P}\left(X\ge c\right)}}$
Step 2
(provided $\mathbb{P}\left(X\ge c\right)\ne 0$)
###### Did you like this example?
ajanlr
Step 1
There cannot be an upper limit on E(X) - for example there is nothing to stop the distribution being centred around 100 or 1000. So option A is the only possibility.
Step 2
We can check option A by trying to make E(X) as small as possible. Put $p\left(X=20\right)=0.25$ and $p\left(X=a\right)=0.75$. Then
$E\left(X\right)=0.25\ast 20+0.75\ast a=5+0.75a$
$E\left({X}^{2}\right)=0.25\ast {20}^{2}+0.75\ast {a}^{2}=100+0.75{a}^{2}$
$Var\left(X\right)=100+0.75{a}^{2}-\left(5+0.75a{\right)}^{2}=25$
which gives $a\approx 8.45$ and $E\left(X\right)\approx 11.3$. So option A is satisfied.