### 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.