**(Cambridge) 15B Question 6**

So we know that , and we are looking for a multiple, say \(n\), of W such that \(\Pr(nW<680)=0.99\). Where \(nW\sim \mathcal N (\mu=82n, \sigma^2 = 81n^2)\). My trouble is that, on my CAS (TI Nspire CX), I can't figure out a way to numerically solve using a NormCDF function for the requisite value of \(n\). The Cambridge solutions use a guess and check method and, unless I am mistaken, their values of sigma for their guesses don't seem accurate.

I was considering using the formula for the normal distribution and using a numerical approximation to solve, but I'm not sure if that would work. Anyone have any suggestions?

Hey Tau,

In this question, we are

**NOT** concerned with the distribution of \(nW\). This does

**NOT** represent the distribution of the combined weight of \(n\) randomly selected persons, and actually represents the distribution of \(n\) times the weight of a

**single** randomly selected person.

First, we define a new random variable \(U\) so that \(U=W_1+\dots+W_n\), where \(W_i\), \(i=1,...,n\), are independent identically distributed (i.i.d) copies of \(W\). Then, we have \begin{align*}\text{E}(U)&=\text{E}(W_1+\dots+W_n)\\ &=\text{E}(W_1)+\dots+\text{E}(W_n)\\ &=n\text{E}(W),\end{align*}\begin{align*}\text{Var}(U)&=\text{Var}(W_1+\dots+W_n)\\ &=\text{Var}(W_1)+\dots+\text{Var}(W_n)\\&=n\text{Var}(W)\end{align*} Notice that for the variance obtained is \(n\text{Var}(W)\) instead of \(n^2\text{Var}(W)\).

Since the sum of i.i.d normal variables is still a normal variable, \[U\sim\mathcal{N}(\mu=82n,\ \sigma^2=81n).\] Thus, where \(Z\sim \mathcal{N}(0,\ 1)\), we solve the following: \begin{align*}&\Pr(U<680)\geq 0.99\\ \implies &\Pr\left(Z<\frac{680-82n}{9\sqrt{n}}\right)\geq 0.99\\ \implies &\frac{680-82n}{9\sqrt{n}}\geq 2.32635...\\ \implies &n\leq 7.58928... \end{align*}That is, maximum is \(n=7\) people.