I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 8.5 : Probability
1. Let,
\[f\left( x \right) = \left\{ {\begin{array}{ll}{\displaystyle \frac{3}{{37760}}{x^2}\left( {20 - x} \right)}&{{\mbox{if }}2 \le x \le 18}\\0&{{\mbox{otherwise}}}\end{array}} \right.\]- Show that \(f\left( x \right)\) is a probability density function.
- Find \(P\left( {X \le 7} \right)\).
- Find \(P\left( {X \ge 7} \right)\).
- Find \(P\left( {3 \le X \le 14} \right)\).
- Determine the mean value of \(X\).
Okay, to show that this function is a probability density function we can first notice that in the range \(2 \le x \le 18\) the function is positive and will be zero everywhere else and so the first condition is satisfied.
The main thing that we need to do here is show that \(\int_{{ - \infty }}^{\infty }{{f\left( x \right)\,dx}} = 1\) .
\[\begin{align*}\int_{{ - \infty }}^{\infty }{{f\left( x \right)\,dx}} & = \int_{2}^{{18}}{{\frac{3}{{37760}}{x^2}\left( {20 - x} \right)\,dx}}\\ & = \frac{3}{{37760}}\int_{2}^{{18}}{{20{x^2} - {x^3}\,dx}} = \left. {\frac{3}{{37760}}\left( {\frac{{20}}{3}{x^3} - \frac{1}{4}{x^4}} \right)} \right|_2^{18} = 1\end{align*}\]The integral is one and so this is in fact a probability density function.
b Find \(P\left( {X \le 7} \right)\). Show Solution
First note that because of our limits on \(x\) for which the function is not zero this is equivalent to \(P\left( {2 \le X \le 7} \right)\). Here is the work for this problem.
\[P\left( {X \le 7} \right) = P\left( {2 \le X \le 7} \right) = \int_{2}^{7}{{\frac{3}{{37760}}{x^2}\left( {20 - x} \right)\,dx}} = \left. {\frac{3}{{37760}}\left( {\frac{{20}}{3}{x^3} - \frac{1}{4}{x^4}} \right)} \right|_2^7 = \require{bbox} \bbox[2pt,border:1px solid black]{{0.130065}}\]Note that we made use of the fact that we’ve already done the indefinite integral itself in the first part. All we needed to do was change limits from that part to match the limits for this part.
c Find \(P\left( {X \ge 7} \right)\). Show Solution
First note that because of our limits on \(x\) for which the function is not zero this is equivalent to \(P\left( {7 \le X \le 18} \right)\). Here is the work for this problem.
\[P\left( {X \ge 7} \right) = P\left( {7 \le X \le 18} \right) = \int_{7}^{{18}}{{\frac{3}{{37760}}{x^2}\left( {20 - x} \right)\,dx}} = \left. {\frac{3}{{37760}}\left( {\frac{{20}}{3}{x^3} - \frac{1}{4}{x^4}} \right)} \right|_7^{18} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.869935}}\]Note that we made use of the fact that we’ve already done the indefinite integral itself in the first part. All we needed to do was change limits from that part to match the limits for this part.
d Find \(P\left( {3 \le X \le 14} \right)\). Show Solution
Not much to do here other than compute the integral.
\[P\left( {3 \le X \le 14} \right) = \int_{3}^{{14}}{{\frac{3}{{37760}}{x^2}\left( {20 - x} \right)\,dx}} = \left. {\frac{3}{{37760}}\left( {\frac{{20}}{3}{x^3} - \frac{1}{4}{x^4}} \right)} \right|_3^{14} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.677668}}\]Note that we made use of the fact that we’ve already done the indefinite integral itself in the first part. All we needed to do was change limits from that part to match the limits for this part.
e Determine the mean value of \(X\). Show Solution
For this part all we need to do is compute the following integral.
\[\begin{align*}\mu & = \int_{{ - \infty }}^{\infty }{{x\,f\left( x \right)\,dx}} = \int_{2}^{{18}}{{x\left[ {\frac{3}{{37760}}{x^2}\left( {20 - x} \right)} \right]\,dx}}\\ & = \frac{3}{{37760}}\int_{2}^{{18}}{{20{x^3} - {x^4}\,dx}} = \left. {\frac{3}{{37760}}\left( {5{x^4} - \frac{1}{5}{x^5}} \right)} \right|_2^{18} = 11.6705\end{align*}\]The mean value of \(X\) is then 11.6705.