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### Section 8.5 : Probability

1. Let,

$f\left( x \right) = \left\{ {\begin{array}{ll}{\displaystyle \frac{3}{{37760}}{x^2}\left( {20 - x} \right)}&{{\mbox{if }}2 \le x \le 18}\\0&{{\mbox{otherwise}}}\end{array}} \right.$
1. Show that $$f\left( x \right)$$ is a probability density function.
2. Find $$P\left( {X \le 7} \right)$$.
3. Find $$P\left( {X \ge 7} \right)$$.
4. Find $$P\left( {3 \le X \le 14} \right)$$.
5. Determine the mean value of $$X$$.
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a Show that $$f\left( x \right)$$ is a probability density function. Show Solution

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.