| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find percentile from probability |
| Difficulty | Standard +0.3 This is a straightforward normal distribution question requiring standard techniques: finding μ from symmetry, using inverse normal tables to find σ, applying linear transformations, and calculating a probability. All steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| 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 | Marks | Guidance |
| \([\mu=]19\) | B1 (AO1.1) | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.96=\frac{21.548-19}{\sigma}\) | M1 (AO3.1a) | Or \(-1.96=\frac{16.452-19}{\sigma}\); may be implied by final answer; NB 1.959963985… rounded to 3 or more sf; M0 if \(z=2\) |
| \([\sigma=]\) awrt \(1.3\) | A1 (AO1.1) | |
| \([\sigma^2=]\) awrt \(1.69\) | A1 (AO1.1) | Allow B3 for awrt 1.69 unsupported |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\mu=]4\times \textit{their } 19+5\) | M1 (AO2.1) | |
| \([\sigma^2=]4^2\times \textit{their } 1.69\) or \(\sigma=4\times \textit{their } 1.3\) | M1 (AO1.1) | NB 27.04 |
| \([Y\sim]\text{N}(81, 5.2^2)\) oe | A1 (AO1.1) | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.04175\) or \(0.0417\) or \(0.042\) BC | B1 (AO1.1) | NB 0.0417462427103 |
| [1] |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\mu=]19$ | B1 (AO1.1) | |
| **[1]** | | |
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.96=\frac{21.548-19}{\sigma}$ | M1 (AO3.1a) | Or $-1.96=\frac{16.452-19}{\sigma}$; may be implied by final answer; NB 1.959963985… rounded to 3 or more sf; M0 if $z=2$ |
| $[\sigma=]$ awrt $1.3$ | A1 (AO1.1) | |
| $[\sigma^2=]$ awrt $1.69$ | A1 (AO1.1) | Allow B3 for awrt 1.69 unsupported |
| **[3]** | | |
## Question 10(iii) A:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\mu=]4\times \textit{their } 19+5$ | M1 (AO2.1) | |
| $[\sigma^2=]4^2\times \textit{their } 1.69$ or $\sigma=4\times \textit{their } 1.3$ | M1 (AO1.1) | NB 27.04 |
| $[Y\sim]\text{N}(81, 5.2^2)$ oe | A1 (AO1.1) | |
| **[3]** | | |
## Question 10(iii) B:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.04175$ or $0.0417$ or $0.042$ BC | B1 (AO1.1) | NB 0.0417462427103 |
| **[1]** | | |
---10 The screenshot in Fig. 10 shows the probability distribution for the continuous random variable $X$, where $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d8ff9511-aff7-45ea-ba55-e6667e8ba760-06_515_1009_338_529}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
The area of each of the unshaded regions under the curve is 0.025 . The lower boundary of the shaded region is at 16.452 and the upper boundary of the shaded region is at 21.548 .
\begin{enumerate}[label=(\roman*)]
\item Calculate the value of $\mu$.
\item Calculate the value of $\sigma ^ { 2 }$.
\item $Y$ is the random variable given by $Y = 4 X + 5$.\\
(A) Write down the distribution of $Y$.\\
(B) Find $\mathrm { P } ( \mathrm { Y } > 90 )$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2018 Q10 [8]}}