| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Outliers from box plot or summary statistics |
| Difficulty | Moderate -0.8 This is a straightforward A-level statistics question requiring standard box plot interpretation (identifying outliers using IQR rule), reading medians, and comparing basic summary statistics. All parts involve routine recall and application of well-practiced techniques with no problem-solving insight required. The pre-release material context adds minimal difficulty since the statistical methods are standard. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers2.02i Select/critique data presentation |
| mean | standard deviation | |
| men | 82.69 kg | 19.98 kg |
| women | 72.5 kg | 19.95 kg |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| not all the data were available | B1 | LDS advantage; must refer to data not being available or reference to #N/A |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(57.7 - 1.5 \times (82.05 - 57.7)\) or \(82.05 + 1.5 \times (82.05 - 57.7)\) seen | M1 | |
| outliers \(< 21.175\) or outliers \(> 118.575\) | A1 | given correct to 1 dp or better; both regions needed; allow non-strict inequalities |
| (hence all outliers in interval) \((118.575, 132.2]\) (since no outliers in lower tail) | A1 | allow eg between 118.6 and 132.2; allow strict or non-strict inequalities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| should not be removed since no reason to eg doubt that it's genuine data / eg suspect it's been misrecorded / eg doubt since from (US) government | B1 | LDS advantage |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| a typical man is heavier than a typical woman, [since \(79.9 > 69.5\)] | B1 | allow eg an average man is heavier than an average woman; do not allow eg men are heavier than women on average |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| mean weight for men is greater than mean weight for women, so distribution for men is located further along the number line than the distribution for women (by about 10 kg) | B1 | allow mean weight for men greater than mean weight for women, so men are heavier than women (by about 10 kg); must refer to mean or average |
| standard deviations (or variances) are approximately equal, so similar dispersion about the mean / variation in weights for men and women | B1 | must refer to standard deviation or variance |
# Question 14:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| not all the data were available | B1 | LDS advantage; must refer to data not being available or reference to #N/A |
**[1]**
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $57.7 - 1.5 \times (82.05 - 57.7)$ or $82.05 + 1.5 \times (82.05 - 57.7)$ seen | M1 | |
| outliers $< 21.175$ or outliers $> 118.575$ | A1 | given correct to 1 dp or better; both regions needed; allow non-strict inequalities |
| (hence all outliers in interval) $(118.575, 132.2]$ (since no outliers in lower tail) | A1 | allow eg between 118.6 and 132.2; allow strict or non-strict inequalities |
**[3]**
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| should **not** be removed since no reason to eg doubt that it's genuine data / eg suspect it's been misrecorded / eg doubt since from (US) government | B1 | LDS advantage |
**[1]**
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **a typical** man is heavier than **a typical** woman, [since $79.9 > 69.5$] | B1 | allow eg **an average** man is heavier than **an average** woman; do not allow eg men are heavier than women **on average** |
**[1]**
## Part (e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **mean weight for men** is greater than **mean weight for women**, so **distribution for men** is located further along the number line than the **distribution for women** (by about 10 kg) | B1 | allow **mean weight for men** greater than **mean weight for women**, so **men are heavier than women** (by about 10 kg); must refer to mean or average |
| **standard deviations** (or variances) are approximately equal, so similar dispersion about the mean / variation in weights for men and women | B1 | must refer to standard deviation or variance |
**[2]**
---14 The pre-release material contains medical data for 103 women and 97 men.\\
The boxplot represents the weights in kg of 101 of the women from the pre-release material.\\
\includegraphics[max width=\textwidth, alt={}, center]{8e48bbd3-2166-49e7-8906-833261f331ca-09_421_1232_735_244}
\begin{enumerate}[label=(\alph*)]
\item Use your knowledge of the pre-release material to give a reason why the weights of all 103 women were not included in the diagram.
\item Determine the range of values in which any outliers lie.
\item Use your knowledge of the pre-release material to explain whether these outliers should be removed from any further analysis of the data.
\item The median weight of men in the sample was found to be 79.9 kg .
Explain what may be inferred by comparing the median weight of men with the median weight of women.
Further analysis of the weights of both men and women is carried out. The table shows some of the results.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& mean & standard deviation \\
\hline
men & 82.69 kg & 19.98 kg \\
\hline
women & 72.5 kg & 19.95 kg \\
\hline
\end{tabular}
\end{center}
\item Use the information in the table to make two inferences about the distribution of the weights of men compared with the distribution of the weights of women.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2024 Q14 [8]}}