### Video Transcript

If 𝑋 is a continuous random variable with probability density function 𝑓 of 𝑥 equals 𝑥 over eight if 𝑥 is greater than or equal to zero or less than four and equal to zero otherwise, find part one, the probability that 𝑋 is greater than two, and part two, the probability that 𝑋 is greater than one and less than three.

When we are given the probability density function of a continuous random variable, we can find the probability between two limits by integrating. The probability of 𝑥 is equal to the integral of 𝑓 of 𝑥 with respect to 𝑥 between the two limits 𝑎 and 𝑏. The first part of our question wants us to calculate the probability that 𝑋 is greater than two. This suggests that there is no upper limit. However, as the value of 𝑓 of 𝑥 has an upper limit of four, we can integrate between two and four. The probability that 𝑋 is greater than two is the same as the probability is greater than two and less than four in this case.

We, therefore, need to integrate the function 𝑥 over eight between four and two. Integrating 𝑥 gives us 𝑥 squared over two. Therefore, integrating 𝑥 over eight gives us 𝑥 squared over 16, as eight multiplied by two is equal to 16. Our next step is to substitute in our limits four and two and subtract the answers. Substituting in 𝑥 equals four gives us 16 over 16, as four squared equals 16.

Substituting in 𝑥 equals two gives us four over 16, as two squared is equal to four. Subtracting these two answers gives us 12 over 16 or twelve sixteenths. This fraction can be simplified by dividing the numerator and denominator by four. 12 divided by four is equal to three. And 16 divided by four is equal to four. This means that twelve sixteenths is equal to three-quarters. The probability that 𝑋 is greater than two is equal to three-quarters or 0.75.

For the second part of the question, we need to repeat the same process. This time we need to integrate 𝑥 over eight between the limits three and one. We already know that 𝑥 over eight integrates to 𝑥 squared over 16. Substituting in our two limits this time gives us nine over 16 minus one over 16. Three squared is equal to nine. And one squared is equal to one. Once again as the denominators are the same, we can just subtract the numerators. This gives us an answer of eight over 16.

The highest common factor of eight and 16 is eight. So we can divide the numerator and denominator by eight. Eight divided by eight is equal to one. And 16 divided by eight is equal to two. This means that the fraction can be simplified to one-half. The probability that 𝑋 is greater than one and less than three for this continuous random variable is one-half.