Consider a six sided standard die with sides labeled 1 through 6. The value of X takes on one of those values.
Let's say you roll the die and record your results in a table. The result will of course be random, but we should expect (roughly) that each face has the same chances of being landed on. Each face has theoretical probability of 1/6 of showing up.
If you roll the die 6000 times, then we should have roughly 1000 occurrences of "1", and 1000 occurrences of "2", and so on. Each "1000" won't be perfectly on the mark, but we'll probably get close.
What we do is multiply the frequency of each face by the face value itself. We'll have 1000*1, then 1000*2, and so on until we get to 1000*6
We get this result:
1000*1+1000*2+1000*3+1000*4+1000*5+1000*6
1000*(1+2+3+4+5+6)
1000*(21)
21000
Then we divide over the 6000 since we had 6000 rolls total to get an average value of 21000/6000 = 3.5 which is the expected value.
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Or you could use the expected value formula of multiplying each P(X) value with the X value and summing
E[X] = expected value
E[X] = sum of X*P(X) values
E[X] = (1/6)*1+(1/6)*2+(1/6)*3+(1/6)*4+(1/6)*5+(1/6)*6
E[X] = (1/6)*(1+2+3+4+5+6)
E[X} = (1/6)*(21)
E[X] = 3.5
This example shows that the expected value is not an integer value, even though X can only take on integer values from 1 to 6. The 3.5 is pretty much a midpoint or center for all of the values to get around. It is possible for E[X] to be equal to one value of X, but we'd have to change the set X around somehow to make this happen.
