| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Estimate from summary statistics |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard normal distribution techniques: calculating mean/variance from summary statistics, reading a histogram, justifying normality, and applying normal probability calculations. All parts are routine textbook exercises requiring no novel insight, though part (f) requires understanding linear transformations of normal distributions. Slightly easier than average due to the step-by-step scaffolding. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Standing charge per day in pence | 7.8 |
| Charge per litre in pence | 0.18 |
| Answer | Marks |
|---|---|
| \(260\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(31\) | B1 | mark the final answer |
| Answer | Marks | Guidance |
|---|---|---|
| any 2 distinct reasons e.g. (approximately) symmetrical (about the mean) | B1 | ignore extra comments unless they contradict an otherwise correct answer |
| e.g. approximately bell-shaped / unimodal; e.g. data is continuous | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [variance is] \(\mathbf{awrt}\ 62.2\) or [sd is] \(7.89\) seen | M1 | NB \(62.15567...\) may be implied by sd \(= 7.88\) or \(7.89\); NB \(0.263047...\) from \(\sqrt{62.2}\) or \(0.263133...\) from \(7.89\); NB \(0.262871...\) from \(7.88\), \(0.262973...\) from unrounded sd |
| \(0.26287 - 0.263134\) or \(0.26\) | A1 | allow B2 for correct answer unsupported |
| Answer | Marks | Guidance |
|---|---|---|
| \(B(28, p)\) used, where \(p\) is value calculated in (d) | M1 | |
| \(0.888 \leq p < 0.896\) | A1 | may be given to 2 sf; allow B2 for correct answer unsupported |
| Answer | Marks | Guidance |
|---|---|---|
| \(7.8 + 0.18 \times 260\) | M1 | |
| \(0.18^2 \times 62.2\) | M1 | or \(0.18 \times \sqrt{62.2}\) |
| \(N(54.6,\ 2.0138 - 2.02)\) | A1 | allow e.g. \(1.42^2\) for variance |
## Question 18:
### Part (a):
$260$ | B1 |
### Part (b):
$31$ | B1 | mark the final answer
### Part (c):
any 2 distinct reasons e.g. (approximately) symmetrical (about the mean) | B1 | ignore extra comments unless they contradict an otherwise correct answer
e.g. approximately bell-shaped / unimodal; e.g. data is continuous | B1 |
### Part (d):
[variance is] $\mathbf{awrt}\ 62.2$ **or** [sd is] $7.89$ seen | M1 | NB $62.15567...$ may be implied by sd $= 7.88$ or $7.89$; NB $0.263047...$ from $\sqrt{62.2}$ or $0.263133...$ from $7.89$; NB $0.262871...$ from $7.88$, $0.262973...$ from unrounded sd
$0.26287 - 0.263134$ or $0.26$ | A1 | allow B2 for correct answer unsupported
### Part (e):
$B(28, p)$ used, where $p$ is value calculated in **(d)** | M1 |
$0.888 \leq p < 0.896$ | A1 | may be given to 2 sf; allow B2 for correct answer unsupported
### Part (f):
$7.8 + 0.18 \times 260$ | M1 |
$0.18^2 \times 62.2$ | M1 | or $0.18 \times \sqrt{62.2}$
$N(54.6,\ 2.0138 - 2.02)$ | A1 | allow e.g. $1.42^2$ for variance
Based on the visible content in these pages, the appendix page (page 26) shows a table for **Exemplar responses for Q2(b)** but the table cells are **empty** — no content has been filled in. The response and mark columns exist but contain no data.
The final page is a contact/copyright page with no mark scheme content.
Therefore, **no mark scheme content can be extracted** from these particular pages as the exemplar response table is blank.
To get the full mark scheme content, you would need the earlier pages of the H640/02 June 2023 mark scheme document.18 Riley is investigating the daily water consumption, in litres, of his household.\\
He records the amount used for a random sample of 120 days from the previous twelve-month period.
The daily water consumption, in litres, is denoted by $x$.
Summary statistics for Riley's sample are given below.\\
$\sum \mathrm { x } = 31164.7 \sum \mathrm { x } ^ { 2 } = 8101050.91 \mathrm { n } = 120$
\begin{enumerate}[label=(\alph*)]
\item Calculate the sample mean giving your answer correct to $\mathbf { 3 }$ significant figures.
Riley displays the data in a histogram.\\
\includegraphics[max width=\textwidth, alt={}, center]{11788aaf-98fb-4a78-8a40-a40743b1fe15-13_832_1383_934_242}
\item Find the number of days on which between 255 and 260 litres were used.
\item Give two reasons why a Normal distribution may be an appropriate model for the daily consumption of water.
Riley uses the sample mean and the sample variance, both correct to $\mathbf { 3 }$ significant figures, as parameters of a Normal distribution to model the daily consumption of water.
\item Use Riley's model to calculate the probability that on a randomly chosen day the household uses less than 255 litres of water.
\item Calculate the probability that the household uses less than 255 litres of water on at least 5 days out of a random sample of 28 days.
The company which supplies the water makes charges relating to water consumption which are shown in the table below.
\begin{center}
\begin{tabular}{ | l | l | }
\hline
Standing charge per day in pence & 7.8 \\
\hline
Charge per litre in pence & 0.18 \\
\hline
\end{tabular}
\end{center}
\item Adapt Riley's model for daily water consumption to model the daily charges for water consumption.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2023 Q18 [11]}}