| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single sum threshold probability |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal distributions with clear setup and standard calculations. While it requires understanding independence assumptions and combining multiple normal variables, the question provides all parameters explicitly and follows a routine template. The multi-part structure and context make it slightly above average, but the mathematical techniques are standard S3 fare with no novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(P\) = time to serve a customer at a standard checkout. \(Q = P_1+P_2+P_3\), \([Q\sim] \mathbf{N}(720, 1200)\) | B1 | For N(720,1200) or \(N\!\left(12,\frac{1}{3}\right)\). Maybe awarded if used in standardisation |
| \(P(Q<660) = P\!\left(Z < \pm\frac{660-720}{\sqrt{1200}}\right) [= P(Z < -1.732\ldots)]\) | M1 | For standardising using 660, their mean \(\neq 240\) or 4 and their standard deviation \(\neq 20\) or \(\frac{1}{3}\). If no distribution given the mean and sd must be correct in the standardisation. Allow \(\pm\) stand |
| \(= 0.0418\) (Calculator gives 0.04163…) awrt 0.041/0.042 | A1 | awrt 0.041 or awrt 0.042 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assume the time taken to serve customers is independent | B1 | Correct assumption. Must have context of customers or time and independence (allow random) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R\) = time to serve a customer at an express checkout. \(S=(P_1+P_2+P_3)-(R_1+\ldots+R_7)\), \([S\sim]\mathbf{N}(20, 1648)\) | M1 A1 | M1 for \(N(\pm20,\ldots)\) or \(N\!\left(\frac{1}{3},\ldots\right)\); A1 for \(N(\pm20, 1648)\) or \(N\!\left(\frac{1}{3},\frac{103}{225}\right)\) maybe awarded if used in standardisation |
| \(P(S>0) = P\!\left(Z > \pm\frac{0-20}{\sqrt{1648}}\right) [= P(Z > -0.492\ldots)]\) | M1 | For standardising using 0 and mean of \(\pm20\) or \(\pm\frac{1}{3}\) and their standard deviation. The 0 may be implied by having just the mean on the numerator. Allow \(\pm\) stand |
| \(= 0.6879\) (Calculator gives 0.6888…) awrt 0.688/0.689 | A1 | awrt 0.688 to 0.689 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $P$ = time to serve a customer at a standard checkout. $Q = P_1+P_2+P_3$, $[Q\sim] \mathbf{N}(720, 1200)$ | B1 | For N(720,1200) or $N\!\left(12,\frac{1}{3}\right)$. Maybe awarded if used in standardisation |
| $P(Q<660) = P\!\left(Z < \pm\frac{660-720}{\sqrt{1200}}\right) [= P(Z < -1.732\ldots)]$ | M1 | For standardising using 660, their mean $\neq 240$ or 4 and their standard deviation $\neq 20$ or $\frac{1}{3}$. If no distribution given the mean and sd must be correct in the standardisation. Allow $\pm$ stand |
| $= 0.0418$ (Calculator gives 0.04163…) **awrt 0.041/0.042** | A1 | awrt 0.041 or awrt 0.042 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume the time taken to serve customers is independent | B1 | Correct assumption. Must have context of customers or time and independence (allow random) |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R$ = time to serve a customer at an express checkout. $S=(P_1+P_2+P_3)-(R_1+\ldots+R_7)$, $[S\sim]\mathbf{N}(20, 1648)$ | M1 A1 | M1 for $N(\pm20,\ldots)$ or $N\!\left(\frac{1}{3},\ldots\right)$; A1 for $N(\pm20, 1648)$ or $N\!\left(\frac{1}{3},\frac{103}{225}\right)$ maybe awarded if used in standardisation |
| $P(S>0) = P\!\left(Z > \pm\frac{0-20}{\sqrt{1648}}\right) [= P(Z > -0.492\ldots)]$ | M1 | For standardising using 0 and mean of $\pm20$ or $\pm\frac{1}{3}$ and their standard deviation. The 0 may be implied by having just the mean on the numerator. Allow $\pm$ stand |
| $= 0.6879$ (Calculator gives 0.6888…) **awrt 0.688/0.689** | A1 | awrt 0.688 to 0.689 |7 At a particular supermarket, the times taken to serve each customer in a queue at a standard checkout may be modelled by a normal distribution with mean 240 seconds and standard deviation 20 seconds.
There is a queue of 3 customers at a standard checkout.\\
Making a reasonable assumption about the times taken to serve these customers,
\begin{enumerate}[label=(\alph*)]
\item find the probability that the total time taken to serve the 3 customers will be less than 11 minutes.
\item State the assumption you have made in part (a)
In the supermarket there is also an express checkout, which is reserved for customers buying 10 or fewer items. The time taken to serve a customer at this express checkout may be modelled by a normal distribution with mean 100 seconds and standard deviation 8 seconds.
On a particular day Jiang has 8 items to pay for and has to choose whether to join a queue of 3 customers waiting at a standard checkout or a queue of 7 customers waiting at the express checkout.
Using a similar assumption to that made in part (a),
\item find the probability that the total time taken to serve the 3 customers at the standard checkout will exceed the total time taken to serve the 7 customers at the express checkout.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2023 Q7 [8]}}