Suppose bridges can tolerate mass without structural damage following $N(200000,20000^2)$ and that car weights follow $N(1300,130^2)$. Unit is kg for both random variables. I'm trying to find out the amount of cars needed on the bridge for causing structural damage with probability greater than $0.1$.

I calculated $\frac{X-200000}{20000}<1.645\implies X<232 900$, where $1.645$ is the $0.95$ decile (I don't know if decile is the correct term here) of normal distribution and so if the cars weigh more than $232900kg$ or less than $200000kg-32900kg=167100kg$, the bridge will suffer structural damage with a probability of $0.1$. This seems flawed, because cars weighing less means that one car, two cars etc. might cause structural damage with a $5\%$ probability.

Then I tried to calculate amount of cars, the probability for previously calculated car weights set to $1$. In other words, how many cars are needed to guarantee, that their total weight exeeds $232900kg$. But I can't seem to cross the bridge from a single (standardised) random variable to the sum. If the sum of $n$ car weights follows $(200000\cdot n,20000^2\cdot\sqrt{n})$, then how to standardise it and solve for $n$? I found that $Z_n=\frac{S_n-\mu\cdot n}{\sqrt{n}\cdot\sigma}$, where $S_n$ is the sum of $n$ car weights. But from that I got $n$ next to zero.

My approach might be completely wrong so any help and corrections are appreciated. This is a new kind of problem for me.

edit: corrected an error

  • 1
    $\begingroup$ How do you arrive at your conclusion "so if the cars weigh more than ... or less than"? The latter part of it is inconsistent with your calculation. $\endgroup$
    – whuber
    Commented Apr 8, 2022 at 14:38
  • 1
    $\begingroup$ I suppose normal distribution is symmetric. So if the other direction is 32900kg, then also the other is? ... oh, now I see the mistake. I didn't calculate anything with the lower part, though. I was just trying to find the number of cars to outweigh the upper part. $\endgroup$
    – jvkloc
    Commented Apr 8, 2022 at 14:49
  • $\begingroup$ Should the first equation be$$\dfrac{X-200000}{20000}$$ $\endgroup$
    – Xi'an
    Commented Apr 8, 2022 at 16:11
  • $\begingroup$ Corrected that one, too. $\endgroup$
    – jvkloc
    Commented Apr 8, 2022 at 19:39

1 Answer 1


A complete solution requires some familiar basic steps.

  1. Describe the problem in statistical language.

  2. The solution will require finding a quantile of a random variable. Because this variable turns out to have a Normal distribution, at a minimum we will need to compute its expectation and variance (its parameters).

  3. Now that we know the parameters of this distribution, we can compute any probabilities we wish. They will depend on the number of cars, giving us an equation relating the number of cars and the failure probability.

  4. Solve the equation.

Before we go any further, let's do some reality checks. If we were being so crude as to ignore all uncertainty, we would estimate the bridge could hold about $200000/1300 \approx 154$ cars. That number of cars intuitively ought to correspond to a failure probability near $0.5.$ Thus, a probability of $0.1$ will be reached with fewer cars. The insight afforded by these probability models concerns how many fewer. That is, at what point should we seriously start worrying about the bridge? After 10 cars? 100 cars? 153 cars?

I read the question as follows:

Suppose the total loading a bridge can tolerate is a random variable $X$ with a Normal$(\mu,\sigma^2)$ distribution. Suppose further that there is a line of cars whose weights are random variables $Y_i,$ $i=1,2,3,\ldots,$ with a common Normal($\nu,\tau^2)$ distribution. Assume $X$ and all the $Y_i$ are independent. Drive the cars one by one onto the bridge so that after car $n$ is placed, the total weight is $S_n = Y_1 + Y_2 + \cdots + Y_n.$ What is the smallest $n$ for which the chance of bridge failure (computed before driving any cars onto the bridge!) exceeds $p = 0.1$?

To obtain an answer, we ought to find the chance of bridge failure in terms of $n.$ This is readily obtained by expressing the event in terms of the random variables as

$$\Pr(\text{failure with } n \text{ cars}) = \Pr(S_n \gt X) = \Pr(X - S_n \lt 0),$$

thus reducing the calculations to a single random variable $X-S_n.$

The expectation is

$$E[X - S_n] = E[X] - E[Y_1] - E[Y_2] - \cdots - E[Y_n] = \mu - n\nu$$

and, because the variables are independent, their variances add:

$$\operatorname{Var}(X - S_n) = \operatorname{Var}(X) + \operatorname{Var}(Y_1) + \cdots + \operatorname{Var}(Y_n) = \sigma^2 + n\tau^2.$$

Moreover, because all these distributions are Normal, $X-S_n$ has a Normal distribution -- and we now know its parameters. Thus,

$$X - S_n\ \sim\ \operatorname{Normal}(\mu - n\nu, \sigma^2 + n\tau^2).$$

Let $\Phi$ be the standard normal distribution function (CDF). Upon standardizing $X-S_n$ and recalling that normal distributions are continuous, we obtain

$$\Pr(X - S_n) \lt 0 = \Pr(X - S_n \le 0) = \Phi\left(\frac{\mu - n\nu}{\sqrt{\sigma^2 + n\tau^2}}\right).$$

A systematic search, perhaps aided by a plot of this probability, will determine the solution.


Just plot the horizontal line at level $p,$ find where it crosses the plot, and read down to find $n$ (here written $n^{*}(p)$).

If you would like a formula for $n,$ you will need to invert $\Phi$ using its quantile function $\Phi^{-1}:$

$$\Phi^{-1}(p) = \frac{\mu - n\nu}{\sqrt{\sigma^2 + n\tau^2}}.$$

Squaring both sides and clearing the denominator gives a quadratic equation for $n,$

$$0 = (\nu^2)n^2 - \left(2\mu\nu + \Phi^{-1}(p)^2\tau^2\right) n + \mu^2 - \Phi^{-1}(p)^2 \sigma^2.$$

Solve it any way you please, such as with the Quadratic Formula.

Such a quadratic equation has two solutions, equally spaced around $\mu/\nu = 200000/1300 = 153.85.$ They will correspond to $p$ and $1-p$ because $\Phi^{-1}(p)^2 = \Phi^{-1}(1-p)^2.$ Pick the smaller solution when $p \lt 1-p$ and otherwise pick the larger solution. You will get a fractional value $n^*.$ Thus, the bridge has a chance $p$ of tolerating any whole number of (random) cars less than or equal to $n^*,$ but putting any more cars than that on the bridge will make the risk of failure exceed $p.$

For the data in the question, the higher value of $n^{*}$ is $173.6.$ This means the chance of bridge failure will first exceed $1-0.1 = 0.9$ when $n=174$ cars are driven onto the bridge. I leave it to you to find the answer to the question, which is the lower value of $n^{*}.$

  • $\begingroup$ Thank you very much, this looks like a wonderfully educational answer. I'll accept it after I have had the time to go through it. $\endgroup$
    – jvkloc
    Commented Apr 9, 2022 at 7:30
  • $\begingroup$ Can I please explore the r implementation? thanks. $\endgroup$
    – Maximilian
    Commented Jun 23, 2023 at 20:17
  • 1
    $\begingroup$ @Maximilian I did not save it, but curve(1 - pnorm((mu - n * nu) / sqrt(sigma^2 + n * tau^2)), 100, 200, xname = "n") ought to get you started. $\endgroup$
    – whuber
    Commented Jun 23, 2023 at 20:53
  • $\begingroup$ @Whuber: many thanks, can you also help with tau parameter? Not quite sure how to obtain this. Also not getting the right outcomes, from the optim(). $\endgroup$
    – Maximilian
    Commented Jun 26, 2023 at 15:14
  • $\begingroup$ @Max I'm sorry, but I don't understand what you mean by "help with." $\tau$ is one of the numbers stated in the problem. $\endgroup$
    – whuber
    Commented Jun 26, 2023 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.