# Question Video: Finding the Expected Value of a Discrete Random Variable Mathematics • 10th Grade

Let π denote a discrete random variable which can take the values 4, 5, 8, and 10. Given that π(π = 4) = 4/27, π(π = 5) = 5/27, and π(π = 8) = 8/27, find the expected value of π. Give your answer to two decimal places.

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### Video Transcript

Let π denote a discrete random variable which can take the values four, five, eight, and 10. Given that the probability that π is equal to four is four twenty-sevenths, the probability that π is equal to five is five twenty-sevenths, and the probability that π is equal to eight is eight twenty-sevenths, find the expected value of π. Give your answer to two decimal places.

It can be helpful to take the information given and present it in table form. Itβs not necessary though it does make it easier to work with. Weβre given that the random variable π can take the values four, five, eight, and 10 and that the probability π is equal to four is four twenty-sevenths, the probability that π is equal to five is five twenty-sevenths, and the probability that π is equal to eight is eight twenty-sevenths. We arenβt actually told the probability that π is equal to 10. Though, we can work this out.

Remember the sum of all probabilities for our discrete random variable is one. We can find the probability that π is equal to 10 by subtracting the sum of the other probabilities from one whole. Itβs one minus four twenty-sevenths add five twenty-sevenths add eight twenty-sevenths, which is one minus seventeen twenty-sevenths.

Remember one whole is equivalent to twenty-seven twenty-sevenths. So we can subtract seventeen twenty-sevenths from this number to find the probability that π is equal to 10. The probability that π is equal to 10 then is ten twenty-sevenths.

Now that we know all the probabilities, we can apply the formula for the expected value of π. The expected value of π is the sum of each of the possible outcomes multiplied by the probability of this outcome occurring.

Letβs substitute what we have into this formula. π₯ multiplied by the probability of π for the first column is four multiplied by four twenty-sevenths. For the second column, itβs five multiplied by five twenty-sevenths; for the third, eight multiplied by eight twenty-sevenths. And finally, for the fourth column, we have 10 multiplied by ten twenty-sevenths. Thatβs sixteen twenty-sevenths plus twenty-five twenty-sevenths plus sixty-four twenty-sevenths plus one hundred twenty-sevenths. Thatβs a total of two hundred and five twenty-sevenths.

So our expected value of π is two hundred and five twenty-sevenths, which is equal to 7.5925. Correct to two decimal places, the expected value of π is 7.59.

Now, we can look at our table to check whether this answer is likely to be correct. The possible values of π are four, five, eight, and 10. And 7.59 is a little over halfway between four and 10. So 7.59 is likely to be correct for the expected value of this probability distribution.