Question Video: Using Probability Density Function of Continuous Random Variable to Find Probabilities | Nagwa Question Video: Using Probability Density Function of Continuous Random Variable to Find Probabilities | Nagwa

# Question Video: Using Probability Density Function of Continuous Random Variable to Find Probabilities Mathematics • Third Year of Secondary School

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Let π be a continuous random variable with the probability density function π(π₯) represented by the following graph. Find π(4 β€ π β€ 5).

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### Video Transcript

Let π be a continuous random variable with the probability density function π of π₯ represented by the following graph. Find the probability that four is less than or equal to π is less than or equal to five.

In this example, we need to find the probability of an event for a continuous random variable, where the event is given by four is less than or equal to π is less than or equal to five. Now we recall that the probability of the event π₯ one is less than or equal to π is less than or equal to π₯ two for a continuous random variable is the area under the probability density function π of π₯ on the interval with boundaries π₯ one and π₯ two. Since our interval is bounded by π₯ is equal to four and π₯ is equal to five, we begin by highlighting the region under the curve over this interval.

To find the probability of our event, we must find the area of the highlighted region, which is a trapezoid. Recalling that the area of a trapezoid is one over two multiplied by the sum of the lengths of the base and top multiplied by the height, so weβll need to find the lengths of our base, top, and height. We can see straight away from the graph that the length of the base is one-quarter. The height is given by five minus four; that is one unit. And it remains to find the length of the top of the trapezoid. And this is the π¦-coordinate of the point on the graph at π₯ is equal to five.

Now this point lies on a straight line between the points with coordinates four, one-quarter and six, zero. Now, since π₯ is equal to five is exactly halfway between π₯ is equal to four and π₯ is equal to six, the π¦-coordinate at π₯ is equal to five must be the average of the π¦-coordinates of the two endpoints. That is the average of one-quarter and zero. And so we have π¦ is equal to one over two times one-quarter plus zero, which is one over eight. And so the length of the top of the trapezoid is one over eight units. And so the base of our trapezoid is one over four units, the top is one over eight units, and the height is one unit.

Now we have everything we need to calculate the area of our trapezoid. And thatβs one over two multiplied by one over four plus one over eight multiplied by one. And this evaluates to three over 16 units squared. And so for the probability density function π of π₯ represented by the graph, the probability that four is less than or equal to π is less than or equal to five is three over 16.

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