| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate mean and standard deviation from histogram |
| Difficulty | Moderate -0.8 This is a routine S1 histogram question requiring standard techniques: reading frequency density to find frequencies, calculating median from cumulative frequency, and computing mean from grouped data. All methods are textbook procedures with no problem-solving insight required, making it easier than average but not trivial due to the multi-step calculations involved. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Total area of bars \(= 400\) small squares | B1 | Correct attempt to calculate whole area (400 small squares). Accept \(160+120+60+40+20\) or \(80+60+30+20+10=200\) or frequencies: \(16+12+6+4+2\) or cm² \(6.4+4.8+2.4+1.6+0.8=16\) or key: 10 small squares = 1 person |
| Area required \(= 40\times4+20\times6+6\times10 = 340\) small squares | B1 | Correct attempt to calculate required area. Accept \(160+120+60\) or frequencies: \(16+12+6\) or cm² \(6.4+4.8+2.4=13.6\) |
| No. of staff \(= \text{"340"}\times\dfrac{40}{\text{"400"}} = 34\) | M1, A1 | M1 for correct expression using their 400 and their 340. A1 for 34. An answer of 34 will usually score 4/4 unless incorrect working seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Median is \((2+)\dfrac{4}{12}\times3=3\) or \((5-)\dfrac{8}{12}\times3=3\) | M1A1 | M1 for \(\dfrac{20-16}{12}\times(5-2)\) or \(\dfrac{20.5-16}{12}\times(5-2)\) or similar. A1 for 3 (if using \(n+1\) accept 3.125 or awrt 3.13) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mean \(= \dfrac{\sum fx}{40} = \dfrac{1\times16+3.5\times12+7.5\times6+15\times4+25\times2}{40} = \dfrac{213}{40} = 5.325\) | M1, A1 | M1 for attempt at \(\dfrac{\sum fx}{40}\) where at least 3 correct products of \(\sum fx\) are seen, or \(\sum fx =\) awrt 200 (1 sf). A1 for 5.325 or exact equivalent e.g. \(\dfrac{213}{40}\), accept 5.33. Accept 5h 19 mins or 5h 20 mins. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (Positive) skew but not negative, or there are outliers (which affect mean) | B1 | For a reason e.g. data is skewed. Allow mention of "extreme values" or "outliers". Do not allow "negative skew" or "anomalies". |
| Median | dB1 | Dependent on mentioning skew for choosing median. Allow B0B1 for "Choose median since the data has negative skew". |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Total area of bars $= 400$ small squares | B1 | Correct attempt to calculate whole area (400 small squares). Accept $160+120+60+40+20$ or $80+60+30+20+10=200$ or frequencies: $16+12+6+4+2$ or cm² $6.4+4.8+2.4+1.6+0.8=16$ or key: 10 small squares = 1 person |
| Area required $= 40\times4+20\times6+6\times10 = 340$ small squares | B1 | Correct attempt to calculate required area. Accept $160+120+60$ or frequencies: $16+12+6$ or cm² $6.4+4.8+2.4=13.6$ |
| No. of staff $= \text{"340"}\times\dfrac{40}{\text{"400"}} = 34$ | M1, A1 | M1 for correct expression using their 400 and their 340. A1 for 34. An answer of 34 will usually score 4/4 unless incorrect working seen. |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Median is $(2+)\dfrac{4}{12}\times3=3$ **or** $(5-)\dfrac{8}{12}\times3=3$ | M1A1 | M1 for $\dfrac{20-16}{12}\times(5-2)$ or $\dfrac{20.5-16}{12}\times(5-2)$ or similar. A1 for 3 (if using $n+1$ accept 3.125 or awrt 3.13) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean $= \dfrac{\sum fx}{40} = \dfrac{1\times16+3.5\times12+7.5\times6+15\times4+25\times2}{40} = \dfrac{213}{40} = 5.325$ | M1, A1 | M1 for attempt at $\dfrac{\sum fx}{40}$ where at least 3 correct products of $\sum fx$ are seen, or $\sum fx =$ awrt 200 (1 sf). A1 for 5.325 or exact equivalent e.g. $\dfrac{213}{40}$, accept 5.33. Accept 5h 19 mins or 5h 20 mins. |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (Positive) skew but not negative, **or** there are outliers (which affect mean) | B1 | For a reason e.g. data is skewed. Allow mention of "extreme values" or "outliers". Do not allow "negative skew" or "anomalies". |
| Median | dB1 | Dependent on mentioning skew for choosing median. Allow B0B1 for "Choose median since the data has negative skew". |8. A manager records the number of hours of overtime claimed by 40 staff in a month.
The histogram in Figure 1 represents the results.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a839a89a-17f0-473b-ac10-bcec3dbe97f7-26_1107_1513_406_210}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Calculate the number of staff who have claimed less than 10 hours of overtime in the month.
\item Estimate the median number of hours of overtime claimed by these 40 staff in the month.
\item Estimate the mean number of hours of overtime claimed by these 40 staff in the month.
The manager wants to compare these data with overtime data he collected earlier to find out if the overtime claimed by staff has decreased.
\item State, giving a reason, whether the manager should use the median or the mean to compare the overtime claimed by staff.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2014 Q8 [10]}}