### Video Transcript

In this video, weβll learn how to
describe the probability density function of a continuous random variable and use it
to find the probability of an event. Weβll begin by reminding ourselves
of the properties of discrete random variables and then define those of continuous
random variables. Weβll then use the properties of
continuous random variables in some examples to find an unknown constant and
probabilities for events given a probability density function.

We recall that a discrete random
variable is a random variable that can take only discrete values. There may be a finite number of
these possible values or a countably infinite number of possible values. We can find the probability that π
is a particular value, and this is defined by a probability distribution function π
of π₯. And the sum of the probabilities is
equal to one. Suppose, for example, that π is
the number thrown on one toss of a fair die. Then the possible π₯-values are
one, two, three, four, five, and six. Then the probability that π is
equal to one of these values is one over six. And thatβs the value of the
probability distribution function π of π₯ for π equal to π₯π. The sum of all of the π of π₯π is
equal to one.

For a discrete random variable
then, the variable values are discrete. The variable values for a
continuous random variable, on the other hand, can take any value within a given
range. And this means that between any two
of our values of π₯, thereβs an infinite number of other values of π₯. A continuous random variable π is
characterized by a probability density function π of π₯. That is where 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 for a continuous random
variable, the function is never negative and the total area under the graph is equal
to one. In our first example, weβll see how
we can use this definition to find an unknown constant in a given function.

Let π be a continuous random
variable with probability density function π of π₯ is equal to ππ₯ 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, weβre given that
π of π₯ is a probability density function of a continuous random variable. And weβre asked to find the
constant π. To do this, letβs 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 letβs begin by examining the
first condition; thatβs 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 ππ₯. 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 letβs 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 ππ₯
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, thatβs 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 thatβs the area. Dividing both sides by 12, we can
solve for π. And so the value of π is equal to
one over 12.

In this example, we found an
unknown constant in a probability density function. Letβs turn our attention now to
calculating probabilities for continuous random variables.

When calculating probabilities for
a continuous random variable π, we consider the probability that π lies within a
particular interval. And if we call this interval πΌ,
then the probability that π₯ one is less than or equal to π is less than or equal
to π₯ two is equal to the area under the curve π of π₯ on the interval with
boundaries π₯ one and π₯ two. So the probability is the area
under the probability density function π of π₯.

Now we recall that the total area
under the probability density curve is equal to one. And in the sample space, the sum of
all the probabilities must be one. Unlike for a discrete random
variable, however, for a continuous random variable, we cannot specify the
probability that π is a particular value. And thatβs because the area under
the curve for a specific value of π doesnβt really exist. It corresponds to the area carved
out by an infinitesimally thin line. So instead, we calculate the
probability that π is between two values. That is the area under the curve
between two values of π₯.

Now, when the graph of a
probability density function forms a simple geometric shape, such as a triangle, a
trapezoid, or a rectangle, we can use geometric formulae for the area to find the
probabilities of events. Letβs look at an example.

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 weβll 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 thatβs 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.

In this example, we were given the
graph of a probability density function. In our next example, we will find
the probability of an event when the probability density function is given in its
algebraic form.

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

In this example, we need to find
the probability of an event for a continuous random variable when the event is π is
greater than 64. Weβre given a probability density
function, so letβs begin by graphing this function. The function takes the value one
over 63 when π₯ is between nine and 72 and a zero otherwise.

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 π of π₯ over the interval representing the
event. In our case then, we need to find
the area under this graph over the interval 64 to β. However, since we know that this
function is equal to zero for π₯ greater than 72, we need only find the area under
the curve for π between 64 and 72. That is the area of the highlighted
area on the graph, which is a rectangle. And this area gives us the
probability of the given event.

We see that the base of the
rectangle has length 72 minus 64; that is eight units. And the height of the rectangle is
one over 63. And we know, of course, that the
area of a rectangle is the base times the height, which in our case is eight
multiplied by one over 63. And thatβs eight over 63 squared
units. Hence, the probability that π is
greater than 64 is eight over 63. And we note that this is a
reasonable answer for a probability, since eight over 63 lies between zero and
one.

Sometimes the graph of a
probability density function is a piecewise-defined function consisting of numerous
subfunctions. In such cases, to find the
probability of an event, we only need to draw the portion of the graph relevant to
the given event. Letβs see how this works in our
final example.

Let π be a continuous random
variable with the probability density function π of π₯ is equal to π₯ over eight
for π₯ between two and three, π of π₯ is one over 48 for π₯ between three and 36,
and π of π₯ is zero otherwise. Find the probability that π is
between 11 and 24.

In this example, we need to find
the probability of an event for a continuous random variable when the event is π is
between 11 and 24. Now 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 π of π₯ over the interval representing the event,
that is, with boundaries π₯ one and π₯ two.

In our case then, this means that
we need to find the area under the graph over the interval with boundaries 11 and
24. And since we do only need to find
the area over this interval, we donβt need to find the graph of the function outside
of this region. In particular, the smallest
possible π₯ in our region is 11 and the largest is 24. And both of these values lie within
the range of the second subfunction for π of π₯. That is the function defined
between three and 36.

And so we need to only draw the
graph for this subfunction. For values of π₯ between three and
36 then, π of π₯ is the constant function one over 48. Now highlighting the region under
the curve between π₯ is 11 and 24, we see that our area is that of a rectangle. And the area of this rectangle is
the probability of the given event.

We see that the base of our
rectangle has length 24 minus 11; that is 13 units. And the height of the rectangle is
one over 48 units. And we know that the area of a
rectangle is its base times its height. And so the area of our rectangle is
13 multiplied by one over 48; that is 13 over 48 units squared. And so we find that, with the
probability density function as shown, the probability that π is between 11 and 24
is 13 over 48. And we know that this is a
reasonable answer for a probability, since 13 over 48 lies between zero and one.

Letβs complete this video by
recapping a few of the important concepts weβve covered. We know that a continuous random
variable takes values in a continuum. A 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 π¦ is equal to π of π₯ is equal to
one. And finally, the probability of the
event that π lies between π₯ one and π₯ two is the area under the curve π¦ is equal
to π of π₯ on the interval πΌ with boundaries π₯ one and π₯ two.