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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.