Question Video: Using Probability Density Function of Continuous Random Variable to Evaluate an Unknown | Nagwa Question Video: Using Probability Density Function of Continuous Random Variable to Evaluate an Unknown | Nagwa

# Question Video: Using Probability Density Function of Continuous Random Variable to Evaluate an Unknown Mathematics • Third Year of Secondary School

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Let π be a continuous random variable with probability density function π(π₯) = ππ₯ if 1 β€ π₯ β€ 5 and π(π₯) = 0 otherwise. Determine the value of π.

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### Video Transcript

Let π be a continuous random variable with probability density function π of π₯ is equal to ππ₯ if one is less than or equal to π₯ is less than or equal to five and zero otherwise. Determine the value of π.

In this example, weβre given that π of π₯ is a probability density function of a continuous random variable. And weβre asked to find the constant π. To do this, letβs recall the properties of a probability density function. π of π₯ is a probability density function if π of π₯ is greater than or equal to zero for all values of π₯ and the total area under the graph of π¦ is equal to π of π₯ is equal to one. So letβs begin by examining the first condition; thatβs the positivity condition.

We know that π of π₯ is equal to zero outside of the interval one to five. And this means that our first condition is satisfied for π₯ not in the interval one to five. For π₯ inside the interval one to five, we know that π of π₯ is equal to ππ₯. We know that π₯ is greater than zero in this interval. And so π also must be greater than or equal to zero, since π is a probability density function. Our first condition then requires that π is greater than or equal to zero. Now letβs look at our second condition, which tells us that the total area under the graph of π¦ is π of π₯ is one. In order to satisfy this condition, we see that π cannot be equal to zero, since if π is equal to zero, π of π₯ is equal to zero for all π₯. This in turn means that the area would be equal to zero.

Hence, to satisfy the second condition that the total area is equal to one, we cannot have π equal to zero. This means that π must be greater than zero. And since π of π₯ is equal to ππ₯ for π₯ between one and five, the graph of π of π₯ over the interval one to five must be a straight line with a positive slope. Now, from this graph, we see that the area under the graph is a trapezoid. And we recall that the area of a trapezoid is one over two multiplied by the sum of the lengths of the base and the top multiplied by the height. And so to find the area of this trapezoid, thatβs the area under the curve, we must find the lengths of the base, the top, and the height.

To do this, we begin by finding the coordinates of the vertices of the trapezoid. And to do this, we substitute π₯ is equal to one and π₯ is equal to five into our function π of π₯. π of one is equal to π multiplied by one, which is equal to π. And π of five is equal to π multiplied by five, which is five π. And this gives us the coordinates of our vertices, which are one, π and five, five π. And so the top of our trapezoid has a length π and the base has a length five π.

The height of our trapezoid is the length that lies along the π₯-axis. That is five minus one, which is four. And so we have the top π, the base five π, and the height four, which we can now substitute into the formula for the area of the trapezoid. And so we have the area is one over two multiplied by five π plus π multiplied by four. That is two multiplied by six π, which is 12π.

Now to satisfy our second condition, 12π must be equal to one because thatβs the area. Dividing both sides by 12, we can solve for π. And so the value of π is equal to one over 12.

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