Random Variable

Question

Probability	Scores
0.05	0
0.05	3
0.15	5
0.05	7
0.15	9
0.35	11
0.2	13

Find the expected value of the above random variable

Answer 1

Hello Diana,

The table you give is an example of a discrete probability distribution (for a discrete random variable.) The values of the random variable (let's call it X) are the scores in the second column. For each value of X, the corresponding probability P(X) is listed in the first column. The expected value (or mean) of a discrete random variable X (denoted by E[X]) is given by

E[X] = ∑[x·P(x)]

∑ is the usual summation symbol, and the sum is taken over all values of X. In our case,

E[X] = 0·05 + 3·.05 + 5·.15 + 7·.05 + 9·.15 + 11·.35 + 13·.2

E[X] = 0 + .15 + .75 + .35 + 1.35 + 3.85 + 2.6

E[X] = 9.05

Hope that helps! Let me know if you need any further explanation.

William