# Thread: Calculate E[X] and Var[X]

1. ## Calculate E[X] and Var[X]

Question : Calculate E[X] and Var[X] when X represents the outcome when a fair die is rolled.
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$\displaystyle Pr(X) = \frac{1}{6}$

$\displaystyle E[X] = \sum_{x=1}^{6} x f(x)$

$\displaystyle Var[X] = E[(X- \mu)^2]$

2. Originally Posted by zorro
Question : Calculate E[X] and Var[X] when X represents the outcome when a fair die is rolled.
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$\displaystyle Pr(X) = \frac{1}{6}$

$\displaystyle E[X] = \sum_{x=1}^{6} x f(x)$

$\displaystyle Var[X] = E[(X- \mu)^2]$
Substitute f(x) and the values of x into the formulas that you have quoted and do the calculations.

3. I am getting the following results

E[X] = $\displaystyle \frac{10}{3}$

Var[X] = 3.07 .................Is that correct???

4. E(x)=3.5

5. Originally Posted by zorro
Question : Calculate E[X] and Var[X] when X represents the outcome when a fair die is rolled.
---------------------------------------------------------------------

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

$\displaystyle E[X] = \sum_{x=1}^{6} x f(x)$

$\displaystyle Var[X] = E[(X- \mu)^2]$

i think you should try to make a probability
x | f(x)
1 | 1/6
2 | 1/6
: | :
: | :
6 | 1/6

then use the formula:
$\displaystyle E[X] = \sum_{x=1}^{6} x f(x)$
$\displaystyle E[X] = 1. f(1)+ 2. f(2)+.....+ 6. f(6)$

hope it help....

6. Originally Posted by zorro
I am getting the following results

E[X] = $\displaystyle \frac{10}{3}$

Var[X] = 3.07 .................Is that correct???
I think you know now that the first is not correct (I haven't checked the second).

When I say: