Video: Calculating the Expected Value of a Discrete Random Variable

The discrete random variable 𝑋 has the shown probability distribution. Find the value of π‘˜. Hence, determine the expected value of 𝑋.

03:20

Video Transcript

The discrete random variable 𝑋 has the shown probability distribution. Find the value of π‘˜.

To answer this question, we’ll need to recall some information about discrete random variables. We know that the sum of all the probabilities for a discrete random variable must be one. We can therefore form an equation by adding together the probabilities in our table. That gives us π‘˜ over one plus π‘˜ over two plus π‘˜ over three plus π‘˜ over four is equal to one.

Next, we’ll need to evaluate the expression on the left-hand side of our equation. To do that, we’ll need to ensure that all of the fractions have the same denominator. The lowest common denominator we can choose is found by evaluating the lowest common multiple of one, two, three, and four. It’s 12.

To change the denominator of our first fraction to 12, we’ll need to multiply both the numerator and the denominator by 12. That gives us 12π‘˜ over 12. For the second fraction, we’ll multiply both the numerator and the denominator by six, to give us six π‘˜ over 12. For π‘˜ over three, we’ll need to multiply both the numerator and the denominator by four, which gives us four π‘˜ over 12. And we’ll multiply the numerator and the denominator of π‘˜ over four by three, to give us three π‘˜ over 12.

Next, let’s simplify the expression on the left-hand side. Adding together the numerators, 12π‘˜ plus six π‘˜ plus four π‘˜ plus three π‘˜, we get 25π‘˜. So this fraction becomes 25π‘˜ over 12. Then to solve this equation, we’ll first multiply both sides by 12, to get 25π‘˜ equals 12. Finally, we’ll divide both sides by 25, to get π‘˜ is equal to 12 25ths.

Now we can substitute the value of π‘˜ back into our table showing the probability distribution of the discrete random variable. That gives us that the probability π‘₯ is equal to one is 12 25ths, the probability π‘₯ is equal to two is six 25ths, the probability is equal to three is four 25ths, and the probability is equal to four is three 25ths.

Hence, determine the expected value of 𝑋.

Now that we have all the relevant probabilities, we can apply the formula for the expected value of 𝑋. It’s the sum of each of the possible outcomes multiplied by the probability of this outcome occurring. So let’s substitute what we have into this formula.

When π‘₯ is equal to one, it becomes one multiplied by 12 25ths. When π‘₯ is equal to two, π‘₯ multiplied by the probability of π‘₯ is two multiplied by six 25ths. In the third column, it’s three multiplied by four 25ths. And in the fourth and final column, it’s four multiplied by three 25ths. That simplifies to 12 25ths plus 12 25ths plus 12 25ths plus another 12 25ths, which is 48 25ths. The expected value of 𝑋 then is 48 25ths.

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