A discrete random variable 𝑋 has a uniform probability distribution such that the probability that 𝑋 is equal to 𝑥 is one eleventh, where 𝑥 is an element of the set containing the numbers one, two, three, four, five, six, seven, eight, nine, 10, 11. Determine the expected value of 𝑋.
Notice how each outcome has an equal probability of occurring. This is an example then of a uniform distribution. There is a shortcut to help us find 𝐸 of 𝑋 for uniform distribution. But first, we’ll consider the general formula for the expected value of 𝑥.
The expected value of 𝑋 is found by adding together all of the possible outcomes multiplied by the probability of that outcome occurring. In the case of our probability distribution, the first possible value of 𝑋, the first outcome, is one and the probability of that occurring is one eleventh. So we write one multiplied by one eleventh.
The second possible value of 𝑋 is two, and the probability of that occurring is once again one eleventh. So we write two multiplied by one eleventh. The next possible value of 𝑋 is three, so we repeat this process for three, and then for all the remaining possible values of 𝑋.
Remember, each of these values has a probability of occurring of one eleventh. And if we evaluate the sum of these products, we get that the expected value of 𝑋 is six.
Now we did say that there’s a shortcut to help us find the expected value of 𝑋 for a uniform distribution. For a uniform distribution, where 𝑥 can be any number from one through to 𝑛, the expected value of 𝑋 is given by 𝑛 plus one over two. For our distribution, 𝑥 was any number from one through to 11. So 𝑛 becomes 11, and our expected value of 𝑋 becomes 11 plus one over two, which is once again six.