| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Histogram from continuous grouped data |
| Difficulty | Moderate -0.8 This is a straightforward grouped data question requiring standard calculations of mean from a frequency table with unequal class widths. Part (a) is routine midpoint calculation, part (b) tests basic understanding of grouped data limitations, and part (c) requires simple comparison. All techniques are standard AS-level statistics with no problem-solving insight needed. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| \(0 \leqslant t < 1\) | \(1 \leqslant t < 2\) | \(2 \leqslant t < 3\) | \(3 \leqslant t < 4\) | \(4 \leqslant t < 6\) | \(6 \leqslant t < 8\) | ||
| 28 | 36 | 42 | 31 | 24 | 12 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.8\) to \(2.81\) BC | B1 (AO 1.1) | NB \(2.8063583815029\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| eg the data is already grouped; eg we do not have the original raw data; eg we are using the mid-point of the intervals; eg we are assuming the data are uniformly distributed across each interval | B1 [1] | Any one valid reason |
| Answer | Marks | Guidance |
|---|---|---|
| eg using upper class limit in each case gives mean is 3.4…so it is possible that mean is more than 3 | B1 [1] | Simply stating eg the mean could be 3.4 is insufficient; or eg need 520 or more and using the upper limits gives 590 |
## Question 3:
### Part (a)
$2.8$ to $2.81$ BC | B1 (AO 1.1) | NB $2.8063583815029\ldots$
**[1 mark]**
## Question 3:
### Part (b):
eg the data is already grouped; eg we do not have the original raw data; eg we are using the mid-point of the intervals; eg we are assuming the data are uniformly distributed across each interval | **B1 [1]** | Any one valid reason
### Part (c):
eg using upper class limit in each case gives mean is 3.4…so it is possible that mean is more than 3 | **B1 [1]** | Simply stating eg the mean could be 3.4 is insufficient; or eg need 520 or more and using the upper limits gives 590
---3 Fig. 3 shows the time Lorraine spent in hours, $t$, answering e-mails during the working day. The data were collected over a number of months.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Time in hours, \\
$t$ \\
\end{tabular} & $0 \leqslant t < 1$ & $1 \leqslant t < 2$ & $2 \leqslant t < 3$ & $3 \leqslant t < 4$ & $4 \leqslant t < 6$ & $6 \leqslant t < 8$ \\
\hline
\begin{tabular}{ l }
Number of \\
days \\
\end{tabular} & 28 & 36 & 42 & 31 & 24 & 12 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Calculate an estimate of the mean time per day that Lorraine spent answering e-mails over this period.
\item Explain why your answer to part (a) is an estimate.
When Lorraine accepted her job, she was told that the mean time per day spent answering e-mails would not be more than 3 hours.
\item Determine whether, according to the data in Fig. 3, it is possible that the mean time per day Lorraine spends answering e-mails is in fact more than 3 hours.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2019 Q3 [3]}}