-aux.09.Probability distribution

[karush]

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]

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]

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!

 

[karush]