Video: Continuous Random Variables

In this video, we will learn how to describe the probability density function of a continuous random variable and use it to find the probability for some event.

17:10

Video Transcript

In this video, we will learn how to describe the probability density function of a continuous random variable and use it to find the probability for some event. We will begin by looking at some key definitions.

A continuous random variable is a random variable where the data can take infinitely many values. For example, the time taken for something to be done is continuous, since there are an infinite number of possible times that can be taken. This means that the probability for a continuous random variable to take a particular value is zero. As a result, we use a probability density function to calculate probabilities when dealing with continuous random variables.

A probability density function or PDF is used to specify the probability of the random variable falling within a particular range of values. This satisfies the following conditions. For any continuous random variable with probability density function 𝑓 of 𝑥, we have that 𝑓 of 𝑥 must be greater than or equal to zero. Also, the integral of 𝑓 of 𝑥 with respect to 𝑥 between the limits negative ∞ and ∞ is equal to one.

As integration calculates the area under a graph and above the 𝑥-axis, when dealing with a probability density function, the area underneath our graph will be equal to one. We can therefore use our probability density function to calculate the probability that a variable 𝑥 lies between two values. As we’re dealing with continuous functions, the probability that 𝑋 is greater than 𝑎 and less than 𝑏 will be the same as the probability that 𝑋 is greater than or equal to 𝑎 and less than or equal to 𝑏. Either way, we can calculate the probability by integrating our probability density function 𝑓 of 𝑥 with respect to 𝑥 between the limits 𝑎 and 𝑏.

In our first example, we will calculate the probability between two limits graphically.

Let 𝑋 be a continuous random variable with the probability density function 𝑓 of 𝑥 represented by the following graph. Find the probability that 𝑋 is greater than or equal to four and less than or equal to five.

We know that when dealing with a continuous random variable, the area between the probability density function graph and the 𝑥-axis is equal to one. We also know that to calculate the probability of our random variable taking a value between 𝑎 and 𝑏, we integrate our probability density function with respect to 𝑥 between the limits 𝑎 and 𝑏. In this question, we do not have an expression for the function 𝑓 of 𝑥. We’ll therefore just use the graph to calculate the probability.

We need to calculate the probability that 𝑋 is greater than or equal to four and less than or equal to five. This is the area that lies between the two 𝑋-values of four and five. The area we’ll need to calculate is in the shape of a trapezoid or trapezium. And we know that the area of a trapezoid is equal to 𝑝 plus 𝑞 divided by two multiplied by ℎ, where 𝑝 and 𝑞 are the lengths of the parallel sides and ℎ is the perpendicular distance between them.

The length of one of our parallel sides is equal to one-quarter. The portion of our graph that slopes downwards has a constant slope. This means that as five is halfway between four and six on the 𝑥-axis, the height on the 𝑦-axis will be halfway between zero and one-quarter. This is equal to one-eighth. The area of the trapezoid is therefore equal to one-quarter plus one-eighth divided by two all multiplied by one. One-quarter plus one-eighth is equal to three-eighths. Therefore, we need to divide three-eighths by two. We don’t need to keep writing the one as multiplying any value by one gives us the same answer. Three-eighths divided by two is equal to three sixteenths. We can therefore conclude that the probability that 𝑋 is greater than or equal to four and less than or equal to five is three sixteenths.

In our next question, we’ll use integration to calculate the probability that 𝑋 is greater than some value.

Let 𝑋 be a continuous random variable with the probability density function 𝑓 of 𝑥 which is equal to one over 63 for 𝑥 greater than or equal to nine and less than or equal to 72 and equal to zero otherwise. Find the probability that 𝑋 is greater than 64.

We know that when dealing with a continuous random variable with probability density function 𝑓 of 𝑥, then the probability that 𝑋 is greater than or equal to 𝑎 and less than or equal to 𝑏 is equal to the integral of 𝑓 of 𝑥 with respect to 𝑥 between the limits 𝑎 and 𝑏. It is important to note that as our random variable is continuous, it doesn’t matter whether the inequality sign is greater than or greater than or equal to. We can see from the question that 𝑓 of 𝑥 only has a nonzero value when 𝑥 lies between nine and 72 inclusive.

This means that the probability that 𝑋 is greater than 64 is the same as the probability that 𝑋 is greater than or equal to 64 and less than or equal to 72. We can calculate this by integrating the function one over 63 with respect to 𝑥 between these two limits. Integrating any constant with respect to 𝑥 gives us this constant multiplied by 𝑥. Therefore, we have 𝑥 over 63 between the limits 64 and 72. Substituting in our limits gives us 72 over 63 minus 64 over 63. As the denominators are the same, we simply subtract the numerators, giving us eight over 63. The probability that 𝑋 is greater than 64 is therefore equal to eight over 63.

We could also have answered this question graphically. We are told that the probability density function 𝑓 of 𝑥 is equal to one over 63 when 𝑥 is between nine and 72 and is equal to zero otherwise. The graph of the function will therefore be as shown. As we need to calculate the probability that 𝑋 is greater than 64, we need to calculate the area of the rectangle shown. This rectangle has a width of eight and a height of one over 63. Multiplying these gives us an answer of eight over 63. This confirms that our answer for the probability that 𝑋 is greater than 64 is correct.

In our next example, we need to calculate the value of an unknown in our probability density function.

Let 𝑋 be a continuous random variable with probability density function 𝑓 of 𝑥 equal to 𝑎𝑥 where 𝑥 is greater than or equal to one and less than or equal to five and equal to zero otherwise. Determine the value of 𝑎.

We know that for any continuous random variable with probability density function 𝑓 of 𝑥, the integral of 𝑓 of 𝑥 with respect to 𝑥 for all values between negative ∞ and ∞ is equal to one. In this question, 𝑓 of 𝑥 is equal to zero when 𝑥 is greater than five and less than one. This means that we are only interested in the area where 𝑥 is greater than or equal to one and less than or equal to five. The integral of 𝑎𝑥 with respect to 𝑥 between one and five is equal to one. As 𝑎 is a constant, integrating 𝑎𝑥 with respect to 𝑥 gives us 𝑎𝑥 squared over two.

We need to substitute our limits of five and one into this expression. When 𝑥 is equal to five, 𝑎𝑥 squared over two is equal to 25𝑎 over two. When 𝑥 is equal to one, we have one 𝑎 over two or 𝑎 over two. 25𝑎 over two minus 𝑎 over two is equal to one. Simplifying the left-hand side gives us 24𝑎 over two, which in turn simplifies to 12𝑎. We can then divide both sides of this equation by 12 such that 𝑎 is equal to one twelfth. If the probability density function 𝑓 of 𝑥 is equal to 𝑎𝑥 when 𝑥 is greater than or equal to one and less than or equal to five and equal to zero otherwise, then 𝑎 is equal to one twelfth.

We will now look at a couple of slightly more difficult examples.

Let 𝑋 be a continuous random variable with the probability density function 𝑓 of 𝑥 which is equal to 𝑥 over eight when 𝑥 is greater than two and less than three, equal to one over 48 when 𝑥 is greater than three and less than 36, and equal to zero otherwise. Find the probability that 𝑋 is greater than or equal to 11 and less than or equal to 24.

We know that when dealing with a continuous random variable with probability density function 𝑓 of 𝑥, the probability that 𝑥 is greater than or equal to 𝑎 and less than or equal to 𝑏 is equal to the definite integral of 𝑓 of 𝑥 with respect to 𝑥 between our limits 𝑎 and 𝑏. In this question, our values of 𝑎 and 𝑏 are 11 and 24, respectively. These two values lie between three and 36. This means that we will use the second part of our function 𝑓 of 𝑥. The probability that 𝑋 is greater than or equal to 11 and less than or equal to 24 is equal to the integral of one over 48 with respect to 𝑥 between 11 and 24.

Integrating one over 48 with respect to 𝑥 gives us 𝑥 over 48. We then need to substitute the limits 24 and 11 into this expression. This gives us 24 over 48 minus 11 over 48. As the denominators are the same, we simply subtract the numerators. The probability that 𝑋 is greater than or equal to 11 and less than or equal to 24 is 13 over 48.

In our final example, we will once again need to calculate the value of a constant in a probability density function.

Let 𝑋 be a continuous random variable with probability density function 𝑓 of 𝑥 equal to four 𝑥 plus 𝑘 divided by 21 if 𝑥 is greater than or equal to three and less than or equal to four and equal to zero otherwise. Find the value of 𝑘.

We know that when dealing with a continuous random variable, the integral of the probability density function 𝑓 of 𝑥 with respect to 𝑥 between the values negative ∞ and ∞ is equal to one. In this question, we know that 𝑓 of 𝑥 is equal to zero when 𝑥 is greater than four and less than three. This means that the integral of 𝑓 of 𝑥 with respect to 𝑥 between three and four must be equal to one. We need to integrate four 𝑥 plus 𝑘 over 21 with respect to 𝑥 between the limits three and four. To make our integration easier, we can factor out one over 21. Integrating four 𝑥 with respect to 𝑥 gives us four 𝑥 squared over two. This simplifies to two 𝑥 squared. Integrating the constant 𝑘 with respect to 𝑥 gives us 𝑘𝑥. We get the expression one over 21 multiplied by two 𝑥 squared plus 𝑘𝑥 between the limits four and three.

Substituting 𝑥 equals four into the brackets gives us two multiplied by four squared plus 𝑘 multiplied by four. This simplifies to 32 plus four 𝑘. Substituting 𝑥 equals three gives us two multiplied by three squared plus 𝑘 multiplied by three. This simplifies to 18 plus three 𝑘. Our expression simplifies to one over 21 multiplied by 32 plus four 𝑘 minus 18 plus three 𝑘. By collecting like terms, this, in turn, simplifies to one over 21 multiplied by 14 plus 𝑘. This is the integral of our probability density function 𝑓 of 𝑥 between the limits three and four.

We now have an equation, one over 21 multiplied by 14 plus 𝑘 is equal to one. Multiplying both sides of this equation by 21, we have 14 plus 𝑘 is equal to 21. Finally, we subtract 14 from both sides, giving us 𝑘 is equal to seven. If the probability density function 𝑓 of 𝑥 is equal to four 𝑥 plus 𝑘 divided by 21 when 𝑥 is greater than or equal to three and less than or equal to four, then 𝑘 is equal to seven.

We will know summarize the key points from this video. We saw in this video that a continuous random variable can take infinitely many values. A probability density function allows us to calculate the probability of the random variable falling between a range of values. For any probability density function 𝑓 of 𝑥, 𝑓 of 𝑥 must be greater than or equal to zero. And the integral of 𝑓 of 𝑥 with respect to 𝑥 between all values between negative ∞ and ∞ must equal one.

When trying to calculate the probability that 𝑥 lies between two values such that 𝑥 is greater than or equal to 𝑎 and less than or equal to 𝑏, we integrate our probability density function 𝑓 of 𝑥 with respect to 𝑥 between the limits 𝑎 and 𝑏.

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