MHB -aux.09.Probability distribution

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The discussion focuses on calculating the expected value, E[X], from a given probability distribution table. The probability distribution lists values of x and their corresponding probabilities, P(X=x). The expected value is defined as E[X]=∑(x_k * p_k), which represents a weighted average of the outcomes. Participants confirm that by applying this formula, the expected value is calculated as 2.45. The calculation process is straightforward, demonstrating the concept effectively.
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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
 
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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?
 
Re: probability distribution

MarkFL said:
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!
 
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