# Question Video: Finding a Probability for a Continuous Random Variable Using the Graph of Its Probability Density Function Mathematics

Let π be the continuous random variable with the probability density function π(π₯), represented by the graph. Find π(1/2 β€ π β€ 2).

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### Video Transcript

Let π be the continuous random variable with the probability density function π of π₯ represented by the following graph. Find the probability that π is between one-half and two.

In this example, we want to find the probability of an event for a continuous random variable π₯, where the event is that π lies between one-half and two. To do this, we recall that the probability of an event for a continuous random variable is given by the area under the graph of the probability density function over the required interval. We can see this on our graph by shading the region under the graph between π is one-half and π is equal to two. And we note that the area we require is the area of a trapezoid, and thatβs where the area of a trapezoid is given by one over two multiplied by π sub one plus π sub two multiplied by β, where π sub one and π sub two are the lengths of the two parallel sides and β is the height of the trapezoid.

In our case, we see that π sub one has a length of one-third, our height β is two minus one-half, and thatβs 1.5, which can be written as the fraction three over two. And so, to find the area of our trapezoid, we need to find the length π two. And this is the π¦-coordinate of the point on the graph at π is equal to one-half. Now, this point lies on a straight line between the two points with coordinates zero, zero and two, one-third. And since one-half is exactly one-quarter of the way between π is equal to zero and π is equal to two, the π¦-coordinate at π is equal to one-half must lie one-quarter of the way between π¦ is equal to zero and π¦ is one-third. That is, π¦ is equal to one over four multiplied by one over three minus zero, which is one over 12.

And remember, this is our length π two. And so, for our trapezoid, we have the side length π one is equal to one-third. The height is 1.5 or three over two, and π two is one over 12. And so, the area of our trapezoid and hence the probability is equal to one over two multiplied by one-third plus one twelfth multiplied by three over two. And putting our sum over a common denominator of 12, this gives us one over two multiplied by five-twelfths multiplied by three over two. Our area is therefore five over 16 square units.

The probability that continuous random variable π lies between one-half and two is therefore five over 16.