### Video Transcript

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

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

We know what we want the area to be. Itβs 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 π. So an expression for the rectangleβs length is π 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 π minus five is equal to one-third. We can then solve this equation to determine the value of π.

First, we multiply both sides of the equation by six. On the left-hand side, we have π 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 π equals seven.

So by recalling that for a continuous random variable π, the probability that π 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 π such that the probability π is greater than five but less than π is one-third is seven.