OCR MEI Paper 2 2021 November — Question 10 9 marks

Exam BoardOCR MEI
ModulePaper 2 (Paper 2)
Year2021
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeDescribe or suggest sampling method
DifficultyModerate -0.8 This is a straightforward statistics question testing basic concepts: understanding stratified sampling rationale, describing distribution shape from a stem-and-leaf diagram, explaining limitations of the mode, and calculating median/IQR. All parts require standard recall and routine application of A-level statistics techniques with no novel problem-solving or complex reasoning required.
Spec2.01c Sampling techniques: simple random, opportunity, etc2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers

10 Ben has an interest in birdwatching. For many years he has identified, at the start of the year, 32 days on which he will spend an hour counting the number of birds he sees in his garden. He divides the year into four using the Meteorological Office definition of seasons. Each year he uses stratified sampling to identify the 32 days on which he will count the birds in his garden, drawn equally from the four seasons. Ben's data for 2019 are shown in the stem and leaf diagram in Fig. 10.1. \begin{table}[h]
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60
\captionsetup{labelformat=empty} \caption{Fig. 10.1}
\end{table}
  1. Suggest a reason why Ben chose to use stratified sampling instead of simple random sampling.
  2. Describe the shape of the distribution.
  3. Explain why the mode is not a useful measure of central tendency in this case.
  4. For Ben's sample, determine
    Ben found a boxplot for the sample of size 32 he collected using stratified sampling in 2015. The boxplot is shown in Fig. 10.2. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{c9d14a4d-a1c8-42ad-9c0b-42cef6b3612f-06_483_1163_1982_242} \captionsetup{labelformat=empty} \caption{Fig. 10.2}
    \end{figure} In 2016 Ben replaced his hedge with a garden fence.
    Ben now believes that
    Jane says she can tell that the data for 2015 is definitely uniformly distributed by looking at the boxplot.
  5. Explain why Jane is wrong.

Question 10(a):
AnswerMarks Guidance
AnswerMarks Guidance
To account for seasonal variation in birds oeB1
or to account for different weather conditions which affects what birds are seen oe
[1]
Question 10(b):
AnswerMarks Guidance
AnswerMarks Guidance
positive skew caoB1
[1]
Question 10(c):
AnswerMarks Guidance
AnswerMarks Guidance
e.g. in lower quartile (or lower tail); much lower than median (or middle) oe; frequency of 3 and others have frequency of 2; distribution is skewed (so median more appropriate)B1 NB mode is 9
[1]
Question 10(d):
AnswerMarks Guidance
AnswerMarks Guidance
median is \(19.5\)B1
LQ is between \(8^{\text{th}}\) and \(9^{\text{th}}\) value or UQ is between \(24^{\text{th}}\) and \(25^{\text{th}}\) value soiM1 allow for sight of 11 or 30
\(\text{IQR} = 30 - 11 = 19\)A1
[3]
Question 10(e):
AnswerMarks Guidance
AnswerMarks Guidance
\(2015\): \(32 \leq \text{median} \leq 34\), much bigger than (their \(19.5\)) in \(2019\) which supports Ben's first statement oeB1 if B0B0 allow SC1 for true – median smaller in 2019 oe and false – IQR smaller in 2019 oe
\(2015\): \(31 \leq \text{IQR} \leq 34\) was (much) bigger than (their \(19\)) in \(2019\) which does not support Ben's second statement oeB1
[2]
Question 10:
Part (f):
AnswerMarks Guidance
Other distributions are symmetrical (so could be e.g. Normal)B1 AO2.4
## Question 10(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| To account for seasonal variation in **birds oe** | B1 | |
| or to account for different weather conditions which affects what **birds** are seen **oe** | | |
| | **[1]** | |

---

## Question 10(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| positive skew **cao** | B1 | |
| | **[1]** | |

---

## Question 10(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. in lower quartile (or lower tail); much lower than median (or middle) **oe**; frequency of 3 and others have frequency of 2; distribution is skewed (so median more appropriate) | B1 | NB mode is 9 |
| | **[1]** | |

---

## Question 10(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| median is $19.5$ | B1 | |
| LQ is between $8^{\text{th}}$ and $9^{\text{th}}$ value or UQ is between $24^{\text{th}}$ and $25^{\text{th}}$ value **soi** | M1 | allow for sight of 11 or 30 |
| $\text{IQR} = 30 - 11 = 19$ | A1 | |
| | **[3]** | |

---

## Question 10(e):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2015$: $32 \leq \text{median} \leq 34$, much bigger than (their $19.5$) in $2019$ which supports Ben's first statement **oe** | B1 | if **B0B0** allow **SC1** for true – median smaller in 2019 **oe and** false – IQR smaller in 2019 **oe** |
| $2015$: $31 \leq \text{IQR} \leq 34$ was (much) bigger than (their $19$) in $2019$ which does **not** support Ben's second statement **oe** | B1 | |
| | **[2]** | |

# Question 10:

## Part (f):
Other distributions are symmetrical (so could be e.g. Normal) | **B1** | AO2.4 | Can't tell how values distributed in each quartile

---
10 Ben has an interest in birdwatching.

For many years he has identified, at the start of the year, 32 days on which he will spend an hour counting the number of birds he sees in his garden.

He divides the year into four using the Meteorological Office definition of seasons. Each year he uses stratified sampling to identify the 32 days on which he will count the birds in his garden, drawn equally from the four seasons.

Ben's data for 2019 are shown in the stem and leaf diagram in Fig. 10.1.

\begin{table}[h]
\begin{center}
\begin{tabular}{ l | l l l l l l l l l l l l }
0 & 3 & 5 & 9 & 9 & 9 &  &  &  &  &  &  \\
1 & 0 & 0 & 1 & 1 & 2 & 4 & 5 & 6 & 7 & 8 & 9 \\
2 & 0 & 1 & 4 & 6 & 7 & 8 & 9 &  &  &  &  \\
3 & 0 & 0 & 2 & 3 &  &  &  &  &  &  &  \\
4 & 0 & 3 & 6 &  &  &  &  &  &  &  &  \\
5 & 1 &  &  &  &  &  &  &  &  &  &  \\
6 & 0 &  &  &  &  &  &  &  &  &  &  \\
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 10.1}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Suggest a reason why Ben chose to use stratified sampling instead of simple random sampling.
\item Describe the shape of the distribution.
\item Explain why the mode is not a useful measure of central tendency in this case.
\item For Ben's sample, determine

\begin{itemize}
  \item the median,
  \item the interquartile range.
\end{itemize}

Ben found a boxplot for the sample of size 32 he collected using stratified sampling in 2015.

The boxplot is shown in Fig. 10.2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{c9d14a4d-a1c8-42ad-9c0b-42cef6b3612f-06_483_1163_1982_242}
\captionsetup{labelformat=empty}
\caption{Fig. 10.2}
\end{center}
\end{figure}

In 2016 Ben replaced his hedge with a garden fence.\\
Ben now believes that

\begin{itemize}
  \item he sees fewer birds in his garden,
  \item the number of birds he sees in his garden is more variable.
\item With reference to Fig. 10.2 and your answer to part (d), comment on whether there is any evidence to support Ben's belief.
\end{itemize}

Jane says she can tell that the data for 2015 is definitely uniformly distributed by looking at the boxplot.
\item Explain why Jane is wrong.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Paper 2 2021 Q10 [9]}}