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### Section 2-5 : Probability

In this last application of integrals that we’ll be looking at we’re going to look at probability. Before actually getting into the applications we need to get a couple of definitions out of the way.

Suppose that we wanted to look at the age of a person, the height of a person, the amount of time spent waiting in line, or maybe the lifetime of a battery. Each of these quantities have values that will range over an interval of integers. Because of this these are called continuous random variables. Continuous random variables are often represented by $$X$$.

Every continuous random variable, $$X$$, has a probability density function, $$f\left( x \right)$$. Probability density functions satisfy the following conditions.

1. $$f\left( x \right) \ge 0$$ for all $$x$$.

2. $$\displaystyle \int_{{\, - \infty }}^{{\,\infty }}{{f\left( x \right)\,dx}} = 1$$

Probability density functions can be used to determine the probability that a continuous random variable lies between two values, say $$a$$ and $$b$$. This probability is denoted by $$P\left( {a \le X \le b} \right)$$ and is given by,

$P\left( {a \le X \le b} \right) = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}$

Let’s take a look at an example of this.

Example 1 Let $$f\left( x \right) = \frac{{{x^3}}}{{5000}}\left( {10 - x} \right)$$ for $$0 \le x \le 10$$ and $$f\left( x \right) = 0$$ for all other values of $$x$$. Answer each of the following questions about this function.
1. Show that $$f\left( x \right)$$ is a probability density function.
2. Find $$P\left( {1 \le X \le 4} \right)$$
3. Find $$P\left( {x \ge 6} \right)$$
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a Show that $$f\left( x \right)$$ is a probability density function. Show Solution

First note that in the range $$0 \le x \le 10$$ is clearly positive and outside of this range we’ve defined it to be zero.

So, to show this is a probability density function we’ll need to show that $$\int_{{\, - \infty }}^{{\,\infty }}{{f\left( x \right)\,dx}} = 1$$.

\begin{align*}\int_{{\, - \infty }}^{{\,\infty }}{{f\left( x \right)\,dx}} & = \int_{{\,0}}^{{\,10}}{{\frac{{{x^3}}}{{5000}}\left( {10 - x} \right)\,dx}}\\ & = \left. {\left( {\frac{{{x^4}}}{{2000}} - \frac{{{x^5}}}{{25000}}} \right)} \right|_0^{10}\\ & = 1\end{align*}

Note the change in limits on the integral. The function is only non-zero in these ranges and so the integral can be reduced down to only the interval where the function is not zero.

b Find $$P\left( {1 \le X \le 4} \right)$$ Show Solution

In this case we need to evaluate the following integral.

\begin{align*}P\left( {1 \le X \le 4} \right) & = \int_{{\,1}}^{{\,4}}{{\frac{{{x^3}}}{{5000}}\left( {10 - x} \right)\,dx}}\\ & = \left. {\left( {\frac{{{x^4}}}{{2000}} - \frac{{{x^5}}}{{25000}}} \right)} \right|_1^4\\ & = 0.08658\end{align*}

So the probability of $$X$$ being between 1 and 4 is 8.658%.

c Find $$P\left( {x \ge 6} \right)$$ Show Solution

Note that in this case $$P\left( {x \ge 6} \right)$$ is equivalent to $$P\left( {6 \le X \le 10} \right)$$ since 10 is the largest value that $$X$$ can be. So the probability that $$X$$ is greater than or equal to 6 is,

\begin{align*}P\left( {X \ge 6} \right) & = \int_{{\,6}}^{{\,10}}{{\frac{{{x^3}}}{{5000}}\left( {10 - x} \right)\,dx}}\\ & = \left. {\left( {\frac{{{x^4}}}{{2000}} - \frac{{{x^5}}}{{25000}}} \right)} \right|_6^{10}\\ & = 0.66304\end{align*}

This probability is then 66.304%.

Probability density functions can also be used to determine the mean of a continuous random variable. The mean is given by,

$\mu = \int_{{\, - \infty }}^{\infty }{{xf\left( x \right)\,dx}}$

Let’s work one more example.

Example 2 It has been determined that the probability density function for the wait in line at a counter is given by, $f\left( t \right) = \left\{ {\begin{array}{ll}0&{{\mbox{if }}t < 0}\\{0.1{{\bf{e}}^{ - \,\frac{t}{{10}}}}}&{{\mbox{if }}t \ge 0}\end{array}} \right.$
1. Verify that this is in fact a probability density function.
2. Determine the probability that a person will wait in line for at least 6 minutes.
3. Determine the mean wait in line.
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a Verify that this is in fact a probability density function. Show Solution

This function is clearly positive or zero and so there’s not much to do here other than compute the integral.

\begin{align*}\int_{{\, - \infty }}^{{\,\infty }}{{f\left( t \right)\,dt}} & = \int_{{\,0}}^{{\,\infty }}{{0.1{{\bf{e}}^{ - \,\frac{t}{{10}}}}\,dt}}\\ & = \mathop {\lim }\limits_{u \to \infty } \int_{{\,0}}^{{\,u}}{{0.1{{\bf{e}}^{ - \,\frac{t}{{10}}}}\,dt}}\\ & = \mathop {\lim }\limits_{u \to \infty } \left. {\left( { - {{\bf{e}}^{ - \,\frac{t}{{10}}}}} \right)} \right|_0^u\\ & = \mathop {\lim }\limits_{u \to \infty } \left( {1 - {{\bf{e}}^{ - \,\frac{u}{{10}}}}} \right) = 1\end{align*}

So it is a probability density function.

b Determine the probability that a person will wait in line for at least 6 minutes. Show Solution

The probability that we’re looking for here is $$P\left( {X \ge 6} \right)$$.

\begin{align*}P\left( {X \ge 6} \right) & = \int_{{\,6}}^{{\,\infty }}{{0.1{{\bf{e}}^{ - \,\frac{t}{{10}}}}\,dt}}\\ & = \mathop {\lim }\limits_{u \to \infty } \int_{{\,6}}^{{\,u}}{{0.1{{\bf{e}}^{ - \,\frac{t}{{10}}}}\,dt}}\\ & = \mathop {\lim }\limits_{u \to \infty } \left. {\left( { - {{\bf{e}}^{ - \,\frac{t}{{10}}}}} \right)} \right|_6^u\\ & = \mathop {\lim }\limits_{u \to \infty } \left( {{{\bf{e}}^{ - \,\frac{6}{{10}}}} - {{\bf{e}}^{ - \,\frac{u}{{10}}}}} \right) = {{\bf{e}}^{ - \,\frac{3}{5}}} = 0.548812\end{align*}

So the probability that a person will wait in line for more than 6 minutes is 54.8811%.

c Determine the mean wait in line. Show Solution

Here’s the mean wait time.

\begin{align*}\mu & = \int_{{\, - \infty }}^{\infty }{{t\,f\left( t \right)\,dt}}\\ & = \int_{{\,0}}^{{\,\infty }}{{0.1t\,{{\bf{e}}^{ - \,\frac{t}{{10}}}}\,dt}}\\ & = \mathop {\lim }\limits_{u \to \infty } \int_{{\,0}}^{{\,u}}{{0.1t\,{{\bf{e}}^{ - \,\frac{t}{{10}}}}\,dt}}\hspace{0.25in}\hspace{0.25in}{\mbox{integrating by parts}}....\\ & = \mathop {\lim }\limits_{u \to \infty } \left. {\left( { - \left( {t + 10} \right){{\bf{e}}^{ - \,\frac{t}{{10}}}}} \right)} \right|_0^u\\ & = \mathop {\lim }\limits_{u \to \infty } \left( {10 - \left( {u + 10} \right){{\bf{e}}^{ - \,\frac{u}{{10}}}}} \right) = 10\end{align*}

So, it looks like the average wait time is 10 minutes.