MHB -aux.09.Probability distribution

[karush]

from probability distribution table

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

find $$E[X]$$

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

 

[MarkFL]

Re: probability distribution

Wikipedia defines it as:

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

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

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

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

 

[karush]

Re: probability distribution

Wikipedia defines it as:

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

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

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

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

$$
(1x0.3)+(2x0.15)+(3x0.35)+(4x0.2)=2.45$$

that was easy!

 

[karush]