Normal Distribution: Comprehensive normal — HCI 2025 H2 Math Prelim Paper 2
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Question
In this question you should state the parameters of any distributions you use.
At a burger shop, the wait time, \(W\) (in minutes), is defined to be the time from when a customer places an order at the counter until the food is collected. It was proposed that \(W\) is modelled by \(\mathrm{N}(2,1.5^{2})\).
(a) Give a reason why this model is not suitable.
A new model for \(W\) is given by \(\mathrm{N}(5,1^{2})\).
(b) Find the range of values of \(k\) such that at least \(90\%\) of customers experience a wait time longer than \(k\) minutes.
A customer bought burgers from the shop on three independent occasions, with wait times denoted by \(W_{1}\), \(W_{2}\) and \(W_{3}\) respectively. Let \(\bar{W}=\dfrac{W_{1}+W_{2}+W_{3}}{3}\).
(c) Find the values of \(\operatorname{Var}(\bar{W}-W_{1})\) and \(\operatorname{Var}(\bar{W})+\operatorname{Var}(W_{1})\) and hence, determine whether \(\operatorname{Var}(\bar{W}-W_{1})=\operatorname{Var}(\bar{W})+\operatorname{Var}(W_{1})\).
(d) Find the probability that the mean wait time is within one minute of the wait time on the first occasion.
On the \(4^{\text{th}}\) occasion, the customer went to the restroom immediately after the order was placed. The time spent in the restroom, in \(T\) minutes, is modelled by \(T\sim\mathrm{N}(7,1.5^{2})\). Assume that the time taken to walk to the restroom is negligible and \(W\) and \(T\) follow independent normal distributions.
(e) The customer is in a rush and will not wait for more than \(3\) minutes after leaving the restroom. Given that the burger is not ready for collection after the customer leaves the restroom, find the probability that the customer will leave the shop without collecting the burger.
To reduce the wait time for customers, the shop later installs self-order kiosks. The wait time from when a customer places an order at the kiosk until the food is collected is modelled as being reduced by \(20\%\) compared to that at the counter.
(f) \(8\) customers who ordered at the counter and \(8\) customers who ordered at the kiosk are randomly selected and their wait times are recorded. The wait time of each customer is independent of the others. Find the probability that exactly \(2\) of these \(16\) customers have a wait time of at least \(5\) minutes.
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(a) According to the model, \(\mathrm{P}(W<0)=0.0912\), which means \(9.12\%\) of the orders will have a negative wait time, which is not realistic.
(b) \(\mathrm{P}(W>k)\ge 0.9\)
\(0<k\le 3.71\)OR \[ \mathrm{P}\!\left(Z>\frac{k-5}{1}\right)\ge 0.9 \quad\Longleftrightarrow\quad 0<\frac{k-5}{1}\le -1.28155 \quad\Longleftrightarrow\quad 0<k\le 3.71. \]
(c) \[\begin{aligned} \mathrm{Var}(\bar W-W_{1}) &= \mathrm{Var}\!\left(\frac{W_{1}+W_{2}+W_{3}}{3}-W_{1}\right) \\ &= \mathrm{Var}\!\left(\frac{-2W_{1}+W_{2}+W_{3}}{3}\right) \\ &= \frac{1}{9}(4+1+1)=\frac{2}{3}. \end{aligned}\] \[ \mathrm{Var}(\bar W)+\mathrm{Var}(W_{1})=\frac{1}{3}+1=\frac{4}{3}. \] Hence \(\mathrm{Var}(\bar W-W_{1})\neq\mathrm{Var}(\bar W)+\mathrm{Var}(W_{1})\).
(d) \(\dfrac{-2W_{1}+W_{2}+W_{3}}{3}\sim\mathrm{N}\!\left(0,\,\dfrac{2}{3}\right)\).
Required probability \[\begin{aligned} &= \mathrm{P}(|\bar W-W_{1}|\le 1) \\ &= \mathrm{P}(-1\le\bar W-W_{1}\le 1) \\ &= 0.779\ (3\,\text{s.f.}). \end{aligned}\]
(e) \(W-T\sim\mathrm{N}(5-7,\,1^{2}+1.5^{2})=\mathrm{N}(-2,\,3.25).\) \[\begin{aligned} \mathrm{P}(W-T>3\mid W>T) &= \frac{\mathrm{P}(W-T>3\text{ and }W>T)}{\mathrm{P}(W>T)} \\ &= \frac{\mathrm{P}(W-T>3)}{\mathrm{P}(W-T>0)} \\ &= 0.0208\ (3\,\text{s.f.}). \end{aligned}\]
(f) Let \(K\) be the wait time of a customer who placed order at the kiosk. \[ K=0.8W\sim\mathrm{N}\!\bigl(0.8\times 5,\ 0.8^{2}\times 1^{2}\bigr) \] \[ \mathrm{P}(K\ge 5)=0.10565. \]
Let \(X\) and \(Y\) be the number of customers who placed order at the counter and kiosk respectively, out of \(8\), with at least \(5\) minutes waiting time. \[ X\sim\mathrm{B}(8,0.5)\qquad Y\sim\mathrm{B}(8,0.10565) \]
Required probability \[\begin{aligned} &= \mathrm{P}(X=2\cap Y=0)+\mathrm{P}(X=1\cap Y=1)+\mathrm{P}(X=0\cap Y=2) \\ &= \mathrm{P}(X=2)\times\mathrm{P}(Y=0)+\mathrm{P}(X=1)\times\mathrm{P}(Y=1)+\mathrm{P}(X=0)\times\mathrm{P}(Y=2) \\ &= 0.0575. \end{aligned}\]