(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.