Video: Finding the Expected Value of a Discrete Random Variable

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

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Video Transcript

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

It can be helpful to take the information given and present it in table form. It’s not necessary though it does make it easier to work with. We’re given that the random variable 𝑋 can take the values four, five, eight, and 10 and that the probability 𝑋 is equal to four is four twenty-sevenths, the probability that 𝑋 is equal to five is five twenty-sevenths, and the probability that 𝑋 is equal to eight is eight twenty-sevenths. We aren’t actually told the probability that 𝑋 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 𝑋 is equal to 10 by subtracting the sum of the other probabilities from one whole. It’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 𝑋 is equal to 10. The probability that 𝑋 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 𝑋. The expected value of 𝑋 is the sum of each of the possible outcomes multiplied by the probability of this outcome occurring.

Let’s substitute what we have into this formula. π‘₯ multiplied by the probability of 𝑋 for the first column is four multiplied by four twenty-sevenths. For the second column, it’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. That’s sixteen twenty-sevenths plus twenty-five twenty-sevenths plus sixty-four twenty-sevenths plus one hundred twenty-sevenths. That’s a total of two hundred and five twenty-sevenths.

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

Now, we can look at our table to check whether this answer is likely to be correct. The possible values of 𝑋 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.

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