Q. 8.10 - A First Course in Probability

Question

Civil engineers believe that W, the amount of weight (in units of $1000$pounds) that a certain span of a bridge can withstand without structural damage resulting, is normally distributed with a mean of $400$and standard deviation of$40$. Suppose that the weight (again, in units of $1000$pounds) of a car is a random variable with a mean of $3$and standard deviation$.3$. Approximately how many cars would have to be on the bridge span for the probability of structural damage to exceed$.1$?

Step-by-Step Solution

Answer

The smallest $n$for which the probability of structural damage exceeds $.1$is$n=\mathbf{117}$.

Step 1 Given Information.

Civil engineers believe that W, the amount of weight (in units of $1000$pounds) that a certain span of a bridge can withstand without structural damage resulting, is normally distributed with a mean of $400$and standard deviation of$40$. Suppose that the weight (again, in units of $1000$pounds) of a car is a random variable with a mean of $3$and a standard deviation of$.3$.

Step 2 Explanation.

Let $W$represent the amount of weight (in units of $1000$pounds) that a certain span of a bridge can withstand without structural damage resulting. It is given that this random variable is normally distributed with mean ${\mu }_W=400$and standard deviation${\sigma }_W=40$.

Additionally, let $C_i$represents the weight (in units of $1000$pounds) of $i$th car, and let $C_W$be the total weight of $n$cars:

$C_W=C_1+C_2+\cdots +C_n.$

Since the weights of cars are independent random variables with mean ${\mu }_C=3$and standard deviation ${\sigma }_C=.3$the mean and the variance of the random variable $C_W$are:

$E\left[C_W\right]=n{\mu }_C=3n,Var\left(C_W\right)=n{\sigma }_C^2=.09n$

At first, notice that the probability of structural damage corresponds to the following probabilities:

$P\left\{C_W\ge W\right\}=P\left\{C_W-W\ge 0\right\}.$

Therefore, let's consider the random variable$C_W-W$. Because of the independence of $C_i$and also obviously of the independence between $C_W$and$W$, we have:

$E\left[C_W-W\right]=E\left[C_W\right]-E\left[W\right]=3n-400$

and

$Var\left(C_W-W\right)=Var\left(C_W\right)+Var\left(W\right)=.09n+{40}^2=.09n+1600.$

Step 3 Explanation.

The question is: how many cars would have to be on the bridge span for the probability of structural damage to e$P\left\{C_W-W\ge 0\right\}>.1?$xceed .l? In other words, how large needs $n$to be so that

To approximate the probability$P\left\{C_W-W\ge 0\right\}$. we use the central limit theorem and in that case, we get:

$.1<P\left\{C_W-W\ge 0\right\}=P\left\{\frac{\left(C_W-W\right)-E\left[C_W-W\right]}{\sqrt{Var\left(C_W-W\right)}}\ge \frac{0-E\left[C_W-W\right]}{\sqrt{Var\left(C_W-W\right)}}\right\}=$

$$\begin{gathered} 1-P\left\{\frac{\left(C_W-W\right)-E\left[C_W-W\right]}{\sqrt{Var\left(C_W-W\right)}}<\frac{0-E\left[C_W-W\right]}{\sqrt{Var\left(C_W-W\right)}}\right\}= \\ 1-P\left\{\frac{\left(C_W-W\right)-(3n-400)}{\sqrt{.09n+1600}}<\frac{400-3n}{\sqrt{.09n+1600}}\right\}\approx \\ 1-\Phi \left(\frac{400-3n}{\sqrt{.09n+1600}}\right)\Rightarrow \Phi \left(\frac{400-3n}{\sqrt{.09n+1600}}\right)<.9 \\ \stackrel{\text{ Table 5.1 (textbook, Chapter 5) }}{\Rightarrow }\frac{400-3n}{\sqrt{.09n+1600}}<1.28 \\ \Uparrow \\ \underset{¯}{9n^2-2400.147456+157378.56<0} \\ \Uparrow \\ n\in (116.211,150.4721) \end{gathered}$$

But, since$n\in \mathrm{\mathbb{N}}$, the smallest $n$for which the probability of structural damage exceeds $.1$is$n=\mathbf{117}$.