# [SOLVED]-aux.09.Probability distribution

#### karush

##### Well-known member
from probability distribution table

$$\displaystyle x\ \ \ P(X=x)$$
$$\displaystyle 1\ \ \ 0.3$$
$$\displaystyle 2\ \ \ 0.15$$
$$\displaystyle 3\ \ \ 0.35$$
$$\displaystyle 4\ \ \ 0.2$$

find $$\displaystyle E[X]$$

I don't know what $$\displaystyle E[X]$$ is

#### MarkFL

Staff member
Re: probability distribution

Wikipedia defines it as:

$$\displaystyle E[X]=\sum_{k=1}^n\left(x_kp_k \right)$$

You can also think of it as a weighted average since we must have:

$$\displaystyle \sum_{k=1}^n\left(p_k \right)=1$$

So, you essentially want to find the product of each pair, and then sum them all up. What do you find?

#### karush

##### Well-known member
Re: probability distribution

Wikipedia defines it as:

$$\displaystyle E[X]=\sum_{k=1}^n\left(x_kp_k \right)$$

You can also think of it as a weighted average since we must have:

$$\displaystyle \sum_{k=1}^n\left(p_k \right)=1$$

So, you essentially want to find the product of each pair, and then sum them all up. What do you find?
$$\displaystyle (1x0.3)+(2x0.15)+(3x0.35)+(4x0.2)=2.45$$

that was easy!