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Paul

January 27, 2020

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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.

- \(f\left( x \right) \ge 0\) for all \(x\).
- \(\displaystyle \int_{{\, - \infty }}^{{\,\infty }}{{f\left( x \right)\,dx}} = 1\)

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,

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

- Show that \(f\left( x \right)\) is a probability density function.
- Find \(P\left( {1 \le X \le 4} \right)\)
- Find \(P\left( {x \ge 6} \right)\)

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,

Let’s work one more example.

- Verify that this is in fact a probability density function.
- Determine the probability that a person will wait in line for at least 6 minutes.
- Determine the mean wait in line.

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.