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### Section 8.5 : Probability

2. For a brand of light bulb the probability density function of the life span of the light bulb is given by the function below, where t is in months.

$f\left( t \right) = \left\{ {\begin{array}{ll}{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}}&{{\mbox{if }}t \ge 0}\\0&{{\mbox{if }}t < 0}\end{array}} \right.$
1. Verify that $$f\left( t \right)$$ is a probability density function.
2. What is the probability that a light bulb will have a life span less than 8 months?
3. What is the probability that a light bulb will have a life span more than 20 months?
4. What is the probability that a light bulb will have a life span between 14 and 30 months?
5. Determine the mean value of the life span of the light bulbs.
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a Verify that $$f\left( t \right)$$ is a probability density function. Show Solution

Okay, to show that this function is a probability density function we can first notice that the exponential portion is always positive regardless of the value of $$t$$ we plug in and the remainder of the function is always zero 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_{0}^{\infty }{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \mathop {\lim }\limits_{n \to \infty } \int_{0}^{n}{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}}\\ & = \mathop {\lim }\limits_{n \to \infty } \left. {\left( { - {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right)} \right|_0^n = \mathop {\lim }\limits_{n \to \infty } \left( { - {{\bf{e}}^{ - \,\,\frac{n}{{25}}}} + 1} \right) = 0 + 1 = 1\end{align*}

The integral is one and so this is in fact a probability density function.

For this integral do not forget to properly deal with the infinite limit! If you don’t recall how to deal with these kinds of integrals go back to the Improper Integral section and do a quick review!

b What is the probability that a light bulb will have a life span less than 8 months? Show Solution

What this problem is really asking us to compute is $$P\left( {X \le 8} \right)$$. Also, because of our limits on $$t$$ for which the function is not zero this is equivalent to $$P\left( {0 \le X \le 8} \right)$$. Here is the work for this problem.

$P\left( {X \le 8} \right) = P\left( {0 \le X \le 8} \right) = \int_{0}^{8}{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \left. { - {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right|_0^8 = \require{bbox} \bbox[2pt,border:1px solid black]{{0.273851}}$

c What is the probability that a light bulb will have a life span more than 20 months? Show Solution

What this problem is really asking us to compute is $$P\left( {X \ge 20} \right)$$. Here is the work for this problem.

\begin{align*}P\left( {X \ge 20} \right) & = \int_{{20}}^{\infty }{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \mathop {\lim }\limits_{n \to \infty } \int_{{20}}^{n}{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}}\\ & = \mathop {\lim }\limits_{n \to \infty } \left. {\left( { - {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right)} \right|_{20}^n = \mathop {\lim }\limits_{n \to \infty } \left( { - {{\bf{e}}^{ - \,\,\frac{n}{{25}}}} + {{\bf{e}}^{ - \,\,\frac{{20}}{{25}}}}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{0.449329}}\end{align*}

For this integral do not forget to properly deal with the infinite limit! If you don’t recall how to deal with these kinds of integrals go back to the Improper Integral section and do a quick review!

d What is the probability that a light bulb will have a life span between 14 and 30 months? Show Solution

What this problem is really asking us to compute is $$P\left( {14 \le X \le 30} \right)$$. Here is the work for this problem.

$P\left( {14 \le X \le 30} \right) = \int_{{14}}^{{30}}{{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \left. { - {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right|_{14}^{30} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.270015}}$

e Determine the mean value of the life span of the light bulbs. Show Solution

For this part all we need to do is compute the following integral.

\begin{align*}\mu & = \int_{{ - \infty }}^{\infty }{{t\,f\left( t \right)\,dt}} = \int_{0}^{\infty }{{0.04t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = \mathop {\lim }\limits_{n \to \infty } \int_{0}^{n}{{0.04t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}}\\ & = \mathop {\lim }\limits_{n \to \infty } \left. {\left[ { - t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}} - 25{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}} \right]} \right|_0^n = \mathop {\lim }\limits_{n \to \infty } \left[ { - n{{\bf{e}}^{ - \,\,\frac{n}{{25}}}} - 25{{\bf{e}}^{ - \,\,\frac{n}{{25}}}} - \left( { - 25} \right)} \right]\\ & = \mathop {\lim }\limits_{n \to \infty } \left[ { - \frac{n}{{{{\bf{e}}^{\frac{n}{{25}}}}}} - 25{{\bf{e}}^{ - \,\,\frac{n}{{25}}}} + 25} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ { - \frac{1}{{\frac{1}{{25}}{{\bf{e}}^{\frac{n}{{25}}}}}}} \right] - 25\left( 0 \right) + 25 = \require{bbox} \bbox[2pt,border:1px solid black]{{25}}\end{align*}

The mean value of the life span of the light bulbs is then 25 months.

We had to use integration by parts to do the integral. Here is that work if you need to see it.

\begin{align*} & \int{{0.04t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}u = 0.04t\,\,\,\,\,\,du = 0.04dt\,\,\,\,\,\,\,\,\,\,\,\,dv = {{\bf{e}}^{ - \,\,\frac{t}{{25}}}}dt\,\,\,\,\,\,v = - 25{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\\ & \int{{0.04t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = - t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}} + \int{{{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\,dt}} = - t{{\bf{e}}^{ - \,\,\frac{t}{{25}}}} - 25{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}\end{align*}

Also, for the limit of the first term we used L’Hospital’s Rule to do the limit.