# Question Video: Finding a Probability for a Continuous Random Variable Mathematics

Let 𝑋 be a continuous random variable with the probability density function 𝑓(𝑥), represented by the following graph. Find the value of 𝑎 that makes 𝑃(5 < 𝑋 < 𝑎) = 1/3.

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### Video Transcript

Let 𝑋 be a continuous random variable with the probability density function 𝑓 of 𝑥 represented by the following graph. Find the value of 𝑎 that makes the probability that 𝑋 is greater than five but less than 𝑎 equal to one-third.

We recall first that for a continuous random variable 𝑋, the probability that 𝑥 lies in a given interval is equivalent to the area under the graph of its probability density function 𝑓 of 𝑥 between the endpoints of that interval. We’re looking to find the probability that 𝑋 is greater than five and less than some value 𝑎. So this will correspond to the orange area shown.

We know what we want the area to be. It’s one-third. We also know the width of this rectangle. From the graph of the probability density function, we can see that the height is one-sixth. The base or length of the rectangle is from five to the unknown value 𝑎. So an expression for the rectangle’s length is 𝑎 minus five. Recalling that the area of a rectangle is found by multiplying its length and width together, we can therefore form an equation. One-sixth multiplied by 𝑎 minus five is equal to one-third. We can then solve this equation to determine the value of 𝑎.

First, we multiply both sides of the equation by six. On the left-hand side, we have 𝑎 minus five, and on the right-hand side, six multiplied by a third or six over three, which is equal to two. We then add five to each side of the equation, giving 𝑎 equals seven.

So by recalling that for a continuous random variable 𝑋, the probability that 𝑋 lies in a given interval is equal to the area under the graph of its probability density function between the endpoints of that interval, we found the value of 𝑎 such that the probability 𝑋 is greater than five but less than 𝑎 is one-third is seven.