Question Video: Using Probability Density Functions of Continuous Random Variables to Find Probabilities

Probability Density Function using Integration Method, Lecture | Sabaq.pk |
Probability Density Function using Integration Method, Lecture | Sabaq.pk |

Video Transcript

Let 𝑋 be a continuous random
variable with probability density function 𝑓 of 𝑥 equals 𝑎 over 62 𝑥 for 𝑥
greater than or equal to 30 but less than or equal to 32 and zero otherwise. Find the probability that 𝑋 is
greater than or equal to 30.5 and less than or equal to 31.5.

We recall first that for continuous
random variable 𝑋, the probability that 𝑥 lies in a given interval is equal to the
area under the graph of its probability density function 𝑓 of 𝑥 between the
endpoints of that interval. In this case, the probability
density function is a straight line because 𝑓 of 𝑥 is a linear function of 𝑥. And we’re looking for the
probability that 𝑋 lies between 30.5 and 31.5. We also recall that we can find the
area under any curve, and indeed the area under a straight line, using
integration.

In general, the probability that 𝑋
lies in the interval from 𝑎 to 𝑏 is found by evaluating the definite integral from
𝑎 to 𝑏 of the probability density function 𝑓 of 𝑥 with respect to 𝑥. In this case then, we need to
evaluate the integral from 30.5 to 31.5 of 𝑎 over 62 𝑥 with respect to 𝑥. Now, this isn’t a difficult
integral to evaluate. But the problem is we don’t know
the value of this constant 𝑎. We’ll need to work this out
first. And to do so, we need to recall a
probability of probability density functions, which is that for a probability
density function 𝑓 of 𝑥, the integral from negative ∞ to positive ∞ of 𝑓 of 𝑥
with respect to 𝑥 must be equal to one, which is another way of saying that the
area under the entire curve is equal to one.

The probability density function in
this question is only nonzero on the interval from 30 to 32, so we know that the
integral from 30 to 32 of 𝑎 over 62 𝑥 with respect to 𝑥 must be equal to one. Evaluating this integral will give
an equation that we can solve in order to determine the value of 𝑎. We recall then that to integrate a
power of 𝑥 where the power is not equal to negative one, we increase the power, or
exponent, by one and then divide by the new power. So the integral of 𝑥 with respect
to 𝑥 is 𝑥 squared over two. And we multiply by the constant 𝑎
over 62. This simplifies to give 𝑎𝑥
squared over 124. And evaluating this between the
limits of 32 and 30 gives the answer one.

Substituting in the limits of 30
and 32 gives 32 squared over 124 multiplied by 𝑎 minus 30 squared over 124
multiplied by 𝑎 is equal to one. We can factor by 𝑎 and also
evaluate 32 squared and 30 squared to give 𝑎 multiplied by 1024 over 124 minus 900
over 124 is equal to one. That simplifies to 𝑎 multiplied by
124 over 124 is equal to one. But of course, 124 over 124 is just
one. So we have 𝑎 multiplied by one
equals one. And this tells us that the value of
𝑎 is one. So having found the value of 𝑎,
we’re now able to substitute this into the integral we wrote down earlier to enable
us to find the probability that 𝑋 is between 30.5 and 31.5.

We have then the integral from 30.5
to 31.5 of one over 62 𝑥 with respect to 𝑥. Integrating as before, this gives
𝑥 squared over 124 evaluated between 30.5 and 31.5. Substituting in the limits, we have
31.5 squared over 124 minus 30.5 squared also over 124. This evaluates to 62 over 124,
which simplifies to one-half. So by recalling that the integral
from negative ∞ to positive ∞ of any probability density function 𝑓 of 𝑥 with
respect to 𝑥 must be equal to one, we were able to determine the value of this
unknown constant 𝑎 and then use integration to determine the probability that 𝑋 is
in the given interval. We found the probability that 𝑋 is
greater than or equal to 30.5 but less than or equal to 31.5 to be one-half.

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