| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Find median and quartiles from stem-and-leaf diagram |
| Difficulty | Easy -1.2 This is a straightforward S1 question testing basic statistical concepts: reading values from a box plot (IQR, range), identifying skewness from a diagram, updating a box plot with changed data points, and interpreting a correlation coefficient. All parts require standard recall and routine application of definitions with minimal problem-solving or multi-step reasoning. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| 1 | 1 | 5 | 8 | 9 | |||
| 2 | 0 | 2 | 5 | 8 | 9 | ||
| 3 | 3 | 5 | 5 | 7 | 8 | 9 | \(\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([IQR = 47 - 33 =] \mathbf{14}\) | B1 | For 14 |
| \([\text{Range} = 54 - 11 =] \mathbf{43}\) | B1 | For 43 |
| e.g. \(Q_2 - Q_1 (-9) > (5 =) Q_3 - Q_2\) Therefore negative (skew) | M1, A1 | For a suitable reason or calculation (allow longer whisker on left etc). For negative skew (dep on M1 seen) "left skew" etc is A0 [Condone incorrect "9" or "5"] |
| \(25 \to 37 \Rightarrow \text{new } Q_1 = 35\) (may be on box plot) \([54 \to 60\) (implies upper whisker now at 60) but no change to \(Q_2]\) New IQR = 12 so need to re-calculate for outliers Outliers now [ > 47 + 18 = 65 or] < 35 – 18 = 17 | B1, M1, A1 | For new lower quartile at 35 (stated or on box plot). For finding the new IQR (< 14) and attempting to re-calculate for outliers. For at least the correct lower limit of 17 seen. |
| Box Plot Box and two whiskers with median still at 42 Lower quartile at their 35 (≠ 33) and upper quartile unchanged at 47 Two outliers at 11 and 15 Lower whisker at 18 (or 17) and upper whisker at 60 | M1, A1ft, A1, A1 | For drawing a box with only two whiskers and median at 42 (all points + 0.5 square). For lower quartile of "35" (changed from 33) and upper quartile unchanged at 47. For only two outliers at 11 and 15 (no overlap with whisker). For lower whisker ending at 18 (or 17) and upper whisker ending at 60. Correct box plot scores all except 1st M1A1 (i.e. 5/7) this M1A1 requires some working |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[IQR = 47 - 33 =] \mathbf{14}$ | B1 | For 14 |
| $[\text{Range} = 54 - 11 =] \mathbf{43}$ | B1 | For 43 |
| e.g. $Q_2 - Q_1 (-9) > (5 =) Q_3 - Q_2$ Therefore **negative (skew)** | M1, A1 | For a suitable reason or calculation (allow longer whisker on left etc). For negative skew (dep on M1 seen) "left skew" etc is A0 [Condone incorrect "9" or "5"] |
| $25 \to 37 \Rightarrow \text{new } Q_1 = 35$ (may be on box plot) $[54 \to 60$ (implies upper whisker now at 60) but no change to $Q_2]$ New IQR = 12 so need to re-calculate for outliers Outliers now [ > 47 + 18 = 65 or] < 35 – 18 = 17 | B1, M1, A1 | For new lower quartile at 35 (stated or on box plot). For finding the new IQR (< 14) and attempting to re-calculate for outliers. For at least the correct lower limit of 17 seen. |
| **Box Plot** Box and two whiskers with median still at 42 Lower quartile at their 35 (≠ 33) and upper quartile unchanged at 47 Two outliers at 11 and 15 Lower whisker at 18 (or 17) and upper whisker at 60 | M1, A1ft, A1, A1 | For drawing a box with only two whiskers and median at 42 (all points + 0.5 square). For lower quartile of "35" (changed from 33) and upper quartile unchanged at 47. For only two outliers at 11 and 15 (no overlap with whisker). For lower whisker ending at 18 (or 17) **and** upper whisker ending at 60. **Correct box plot scores all except 1st M1A1 (i.e. 5/7) this M1A1 requires some working** |
**Notes:**
**(a)(i)** 1st B1 for 14
2nd B1 for 43
**(b)** M1 for a suitable reason or calculation (allow longer whisker on left etc)
A1 for negative skew (dep on M1 seen) "left skew" etc is A0 [Condone incorrect "9" or "5"]
**(c)** B1 for new lower quartile at 35 (stated or on box plot)
1st M1 for finding the new IQR (< 14) and attempting to re-calculate for outliers
1st A1 for at least the correct lower limit of 17 seen
2nd M1 for drawing a box with only two whiskers and median at 42 (all points + 0.5 square)
2nd A1ft for lower quartile of "35" (changed from 33) and upper quartile unchanged at 47
3rd A1 for only two outliers at 11 and 15 (no overlap with whisker)
4th A1 for lower whisker ending at 18 (or 17) **and** upper whisker ending at 60
**Correct box plot scores all except 1st M1A1 (i.e. 5/7) this M1A1 requires some working**
**(d)** M1 for comment that pmcc is "small" so little correlation (just saying < 0 is not enough)
Allow e.g. "not significant" or "not relevant" or $-0.5 < r < 0.5$ or "not close to – 1 " but "no correlation" is M0
A1 for suggesting the complaint is not supported e.g. "little evidence to support claim"
Dep on M1 seen NB M1A0 is possible
---2. Chi wanted to summarise the scores of the 39 competitors in a village quiz. He started to produce the following stem and leaf diagram
Key: 2|5 is a score of 25
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Score}
\begin{tabular}{ l | l l l l l l l }
1 & 1 & 5 & 8 & 9 & & & \\
2 & 0 & 2 & 5 & 8 & 9 & & \\
3 & 3 & 5 & 5 & 7 & 8 & 9 & $\ldots$ \\
\end{tabular}
\end{center}
\end{table}
He did not complete the stem and leaf diagram but instead produced the following box plot.\\
\includegraphics[max width=\textwidth, alt={}, center]{9ac7647f-b291-4a64-9518-fa6438a0cc7d-04_357_1237_772_356}
Chi defined an outlier as a value that is
$$\text { greater than } Q _ { 3 } + 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right)$$
or\\
less than $Q _ { 1 } - 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right)$
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the interquartile range
\item the range.
\end{enumerate}\item Describe, giving a reason, the skewness of the distribution of scores.
Albert and Beth asked for their scores to be checked.\\
Albert's score was changed from 25 to 37\\
Beth's score was changed from 54 to 60
\item On the grid on page 5, draw an updated box plot.
Show clearly any calculations that you used.
Some of the competitors complained that the questions were biased towards the younger generation. The product moment correlation coefficient between the age of the competitors and their score in the quiz is - 0.187
\item State, giving a reason, whether or not the complaint is supported by this statistic.
\includegraphics[max width=\textwidth, alt={}, center]{9ac7647f-b291-4a64-9518-fa6438a0cc7d-05_360_1242_2238_351}
Turn over for a spare grid if you need to redraw your box plot.
\includegraphics[max width=\textwidth, alt={}, center]{9ac7647f-b291-4a64-9518-fa6438a0cc7d-07_367_1246_2261_351}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2019 Q2 [13]}}