| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Estimate from summary statistics |
| Difficulty | Moderate -0.3 This is a straightforward statistics question requiring standard calculations (mean, variance from summary statistics) and basic Normal distribution applications (probability calculations). The multi-part structure and need to interpret continuity correction for discrete marks adds slight complexity, but all techniques are routine A-level content with no novel problem-solving required. |
| Spec | 2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (a) | mean 112.4 isw or 112 isw |
| variance 8.8 or √8.82 cao isw | B1 | |
| B1 | 1.1 | |
| 1.1 | B0 for 8.757 explicitly rounded to 8.8 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (b) | N(their 112.4, their 8.8) |
| N(a, b) | M1 | |
| A1 | 3.3 | |
| 1.1 | allow M1 for 8.8² or √8.8 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (c) | P(mark < 104.5) or P(mark < 105) found |
| Answer | Marks |
|---|---|
| oe | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.4 |
| Answer | Marks |
|---|---|
| 3.5a | may see N(−∞,104.5,112.4,√8.8) |
| Answer | Marks |
|---|---|
| M1 | FT their distribution |
| Answer | Marks | Guidance |
|---|---|---|
| compares their mark with 105 | M1 | |
| 104.7 or 104.3 is close to 105 so good fit | A1 | |
| 9 | (d) | P(mark between 114.5 and 115.5) found |
| Answer | Marks |
|---|---|
| or 16.5 to 16.534 so allow 16 or 17 | M1 |
| A1 | 3.4 |
| 3.5a | NB awrt 0.0915 or awrt 0.0807 implies M1 |
Question 9:
9 | (a) | mean 112.4 isw or 112 isw
variance 8.8 or √8.82 cao isw | B1
B1 | 1.1
1.1 | B0 for 8.757 explicitly rounded to 8.8
[2]
9 | (b) | N(their 112.4, their 8.8)
N(a, b) | M1
A1 | 3.3
1.1 | allow M1 for 8.8² or √8.8
a = 112.4 or 112 and b = 8.8 or 2.972
[2]
9 | (c) | P(mark < 104.5) or P(mark < 105) found
from their distribution in part (b)
205 × their non-zero 0.00387
0.79 to 0.794 or 1.17 to 1.175 so consistent
oe | M1
M1
A1 | 3.4
3.1a
3.5a | may see N(−∞,104.5,112.4,√8.8)
NB 0.00387 or 0.0063(06) implies M1
NB 0.00573 or 0.00914 implies M1
NB 0.00379(69..) or 0.00619(81…) may imply M1 FT use of
variance = 8.757
NB 0.200(199…) and 0.184(665…) may imply M1 FT use of
sd = 8.8
if probability is correctly found to be 0 eg from use of
8.8
N(112.4, ) allow M1 only – no further marks available
205
1
or compare (≈ 0.00488) with their non-zero 0.00387
205
or probabilities similar so consistent oe
[3]
M1 | FT their distribution
Alternatively
1
InvNorm( ,112.4,√8.8) or
205
1
InvNorm( ,112,√8.8) used to find their
205
mark
compares their mark with 105 | M1
104.7 or 104.3 is close to 105 so good fit | A1
9 | (d) | P(mark between 114.5 and 115.5) found
18.75 to 18.77 so allow 18 or 19
or 16.5 to 16.534 so allow 16 or 17 | M1
A1 | 3.4
3.5a | NB awrt 0.0915 or awrt 0.0807 implies M1
unsupported answers score M0
[2]
M1
FT their distributionAt the beginning of the academic year, all the pupils in year 12 at a college take part in an assessment. Summary statistics for the marks obtained by the 2021 cohort are given below.
$n = 205 \quad \sum x = 23042 \quad \sum x^2 = 2591716$
Marks may only be whole numbers, but the Head of Mathematics believes that the distribution of marks may be modelled by a Normal distribution.
\begin{enumerate}[label=(\alph*)]
\item Calculate
\begin{itemize}
\item The mean mark
\item The variance of the marks
\end{itemize}
[2]
\item Use your answers to part (a) to write down a possible Normal model for the distribution of marks. [2]
\end{enumerate}
One candidate in the cohort scored less than 105.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Determine whether the model found in part (b) is consistent with this information. [3]
\item Use the model to calculate an estimate of the number of candidates who scored 115 marks. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2022 Q9 [9]}}