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civil engineers belive 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 mean 400 and standard deviation 40. suppose that the weight (again,in units of 1000 pounds) of a car is a random variable with mean 3 and standard deviation .3.How many cars would have to be on the bridge span for the probability of structural damage to exceed .1?

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After 30 questions posted on the site, you still omit every indication of what you know, what you tried, why this failed, and so on? –  Did Jul 7 '12 at 9:52
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Nitpicking: weight cannot be negative, hence the Gaussian assumption is unrealistic. –  Aditya Jul 7 '12 at 21:45

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up vote 3 down vote accepted

Make the obvious independence assumptions, and proceed along the lines described by Michael Chernick. Since everything has normal distribution, there is no need to make a normal approximation.

Let $X_i$ be the weight of the $i$-th car, and let $S=\sum X_i$. Then $S$ has normal distribution with mean $3n$ and variance $(0.3)^2 n$. Let $Y=S-W$. Then $Y$ has normal distribution with mean $3n-400$, and variance $v_n=(0.3)^2 n+(40)^2$.

We want to find smallest $n$ such that $\Pr(Y \gt 0) \gt 0.1$. To do that, we solve the equation $\Pr(Y \gt 0) = 0.1$. We have $$\Pr(Y\gt 0)=\Pr\left(\frac{Y-(3n-400)}{\sqrt{v_n}}\gt \frac{400-3n}{\sqrt{v_n}}\right)=\Pr\left(Z \gt \frac{400-3n}{\sqrt{v_n}}\right),$$ where $Z$ is standard normal.

But the point $c$ such that $\Pr(Z \gt c)=0.1$ is approximately given by $c=1.28$. So we need to solve the equation $$\frac{400-3n}{\sqrt{(0.3)^2 n+(40)^2}}=1.28.\tag{$1$}$$ Multiply through by the denominator, and square both sides. We get a quadratic equation in $n$, with somewhat messy coefficients. Solve as usual. We want the smaller root. (The larger root is extraneous, introduced when we squared both sides.)

Or else solve Equation $(1)$ directly by using the "Solve" button on a calculator, or any one of the many programs that do such things.

Our root is roughly $116.2$. So the number of cars such that the probability of damage is $\gt 0.1$ is $117$.

Remark: If we think about it, we can see that there is no need to solve a quadratic. For it is clear that the number of cars must be less than $400/3$. So the contribution of the cars to the variance of $Y$ is at most $(400/3)(0.09)$, which is $12$. The contribution of the bridge strength variance to the variance of $Y$ is $1600$, whose bulk makes $12$ negligible. Thus we can assume that $Y$ has standard deviation $40$, and for all practical purposes we want to solve the linear equation $$400-3n=(1.28)(40).$$

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I think the problem is intended to bring in the central limit theorem. The n car are iid random variables with mean 3. So the random variable, Sn, that is the sum of the weight of n cars has mean 3n and variance (0.3)$^2$n So you would apply the central limit theorem and say since Sn is approximately normal with mean 3n and variance 0.30$^2$n. Then you need to solve for the value of n that has probability 0.1 in the region below 400. You want P(Sn>W)=0.1. To get that you need to find where the densities cross as function of n (of course assuming the mean for Sn is below 400 the mean for W). Then find n so that the part of the distribution of Sn to the right of the cross point is 0.1.

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