### 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 π of π₯, we have that π of π₯ must be greater
than or equal to zero. Also, the integral of π of π₯ with
respect to π₯ between the limits negative β and β is equal to one.

As integration calculates the area
under a graph and above the π₯-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 π₯ lies
between two values. As weβre dealing with continuous
functions, the probability that π is greater than π and less than π will be the
same as the probability that π is greater than or equal to π and less than or
equal to π. Either way, we can calculate the
probability by integrating our probability density function π of π₯ with respect to
π₯ between the limits π and π.

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

Let π be a continuous random
variable with the probability density function π of π₯ represented by the following
graph. Find the probability that π 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 π₯-axis is equal to one. We also know that to calculate the
probability of our random variable taking a value between π and π, we integrate
our probability density function with respect to π₯ between the limits π and
π. In this question, we do not have an
expression for the function π of π₯. Weβll therefore just use the graph
to calculate the probability.

We need to calculate the
probability that π is greater than or equal to four and less than or equal to
five. This is the area that lies between
the two π-values of four and five. The area weβll need to calculate is
in the shape of a trapezoid or trapezium. And we know that the area of a
trapezoid is equal to π plus π divided by two multiplied by β, where π and π are
the lengths of the parallel sides and β 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 π₯-axis, the height on the π¦-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 donβt 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 π is greater than or equal to four and less than or equal to five
is three sixteenths.

In our next question, weβll use
integration to calculate the probability that π is greater than some value.

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

We know that when dealing with a
continuous random variable with probability density function π of π₯, then the
probability that π is greater than or equal to π and less than or equal to π is
equal to the integral of π of π₯ with respect to π₯ between the limits π and
π. It is important to note that as our
random variable is continuous, it doesnβt matter whether the inequality sign is
greater than or greater than or equal to. We can see from the question that
π of π₯ only has a nonzero value when π₯ lies between nine and 72 inclusive.

This means that the probability
that π is greater than 64 is the same as the probability that π 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 π₯ between these two
limits. Integrating any constant with
respect to π₯ gives us this constant multiplied by π₯. Therefore, we have π₯ 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 π 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 π of π₯ is equal to one over 63 when π₯ 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 π 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 π 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 π be a continuous random
variable with probability density function π of π₯ equal to ππ₯ where π₯ is
greater than or equal to one and less than or equal to five and equal to zero
otherwise. Determine the value of π.

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

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

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

Let π be a continuous random
variable with the probability density function π of π₯ which is equal to π₯ over
eight when π₯ is greater than two and less than three, equal to one over 48 when π₯
is greater than three and less than 36, and equal to zero otherwise. Find the probability that π 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 π of π₯, the
probability that π₯ is greater than or equal to π and less than or equal to π is
equal to the definite integral of π of π₯ with respect to π₯ between our limits π
and π. In this question, our values of π
and π 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 π of π₯. The probability that π 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 π₯ between 11 and 24.

Integrating one over 48 with
respect to π₯ gives us π₯ 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 π 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 π be a continuous random
variable with probability density function π of π₯ equal to four π₯ plus π divided
by 21 if π₯ is greater than or equal to three and less than or equal to four and
equal to zero otherwise. Find the value of π.

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

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

We now have an equation, one over
21 multiplied by 14 plus π is equal to one. Multiplying both sides of this
equation by 21, we have 14 plus π is equal to 21. Finally, we subtract 14 from both
sides, giving us π is equal to seven. If the probability density function
π of π₯ is equal to four π₯ plus π divided by 21 when π₯ is greater than or equal
to three and less than or equal to four, then π 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 π of π₯, π of π₯ must be greater than or equal to zero. And the integral of π of π₯ with
respect to π₯ between all values between negative β and β must equal one.

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