| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Identify outliers using IQR rule |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing standard procedures: finding quartiles from ordered data, applying the IQR outlier rule (formula given), drawing a box plot, and calculating correlation coefficient using given summary statistics. All parts are routine recall and application of formulas with no problem-solving insight required. Easier than average A-level due to being early statistics content with explicit formulas provided. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers5.08a Pearson correlation: calculate pmcc |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_2 = 53\), \(Q_1 = 35\), \(Q_3 = 60\) | B1, B1, B1 (3) | 1st B1 for median, 2nd B1 for lower quartile, 3rd B1 for upper quartile |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_3 - Q_1 = 25 \Rightarrow Q_1 - 1.5 \times 25 = -2.5\) (no outlier) | M1 | M1 for attempt to find one limit |
| \(Q_3 + 1.5 \times 25 = 97.5\) (so 110 is an outlier) | A1 (2) | A1 for both limits found and correct. No explicit comment about outliers needed |
| Answer | Marks | Guidance |
|---|---|---|
| [Box plot with whiskers, outlier marked with *] | M1, A1ft, A1ft (3) | M1 for box and two whiskers; 1st A1ft for correct position of box, median and quartiles (follow through values); 2nd A1ft for 17 and 77 or "their" 97.5 and *. If 110 is not an outlier score A0. Must label outlier. Accuracy should be within correct square so 97 or 98 will do for 97.5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum y = 461\), \(\sum y^2 = 24219 \therefore S_{yy} = 24219 - \frac{461^2}{10} = 2966.9\) | B1, B1, B1cso (3) | 1st B1 for \(\sum y\); 2nd B1 for \(\sum y^2\) or at least three correct terms of \(\sum(y-\bar{y})^2\); 3rd B1 for complete correct expression leading to 2966.9 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{-18.3}{\sqrt{3463.6 \times 2966.9}} = \frac{-18.3}{3205.64...} = -0.0057\) AWRT \(-0.006\) or \(-6 \times 10^{-3}\) | M1, A1 (2) | M1 for attempt at correct expression for \(r\); can ft their \(S_{yy}\) for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r\) suggests correlation is close to zero so parent's claim is not justified | B1 (1) | B1 for comment rejecting parent's claim on basis of weak or zero correlation. "Negative correlation so comment is true" scores B0. Weak negative or weak positive correlation OK as basis for rejection |
## Question 4:
**(a)**
$Q_2 = 53$, $Q_1 = 35$, $Q_3 = 60$ | B1, B1, B1 (3) | 1st B1 for median, 2nd B1 for lower quartile, 3rd B1 for upper quartile
**(b)**
$Q_3 - Q_1 = 25 \Rightarrow Q_1 - 1.5 \times 25 = -2.5$ (no outlier) | M1 | M1 for attempt to find one limit
$Q_3 + 1.5 \times 25 = 97.5$ (so 110 is an outlier) | A1 (2) | A1 for both limits found and correct. No explicit comment about outliers needed
**(c)**
[Box plot with whiskers, outlier marked with *] | M1, A1ft, A1ft (3) | M1 for box and two whiskers; 1st A1ft for correct position of box, median and quartiles (follow through values); 2nd A1ft for 17 and 77 or "their" 97.5 and *. If 110 is not an outlier score A0. Must label outlier. Accuracy should be within correct square so 97 or 98 will do for 97.5
**(d)**
$\sum y = 461$, $\sum y^2 = 24219 \therefore S_{yy} = 24219 - \frac{461^2}{10} = 2966.9$ | B1, B1, B1cso (3) | 1st B1 for $\sum y$; 2nd B1 for $\sum y^2$ or at least three correct terms of $\sum(y-\bar{y})^2$; 3rd B1 for complete correct expression leading to 2966.9
**(e)**
$r = \frac{-18.3}{\sqrt{3463.6 \times 2966.9}} = \frac{-18.3}{3205.64...} = -0.0057$ AWRT $-0.006$ or $-6 \times 10^{-3}$ | M1, A1 (2) | M1 for attempt at correct expression for $r$; can ft their $S_{yy}$ for M1
**(f)**
$r$ suggests correlation is close to zero so parent's claim is not justified | B1 (1) | B1 for comment rejecting parent's claim on basis of weak or zero correlation. "Negative correlation so comment is true" scores B0. Weak negative or weak positive correlation OK as basis for rejection
---4. In a study of how students use their mobile telephones, the phone usage of a random sample of 11 students was examined for a particular week.
The total length of calls, $y$ minutes, for the 11 students were
$$17,23,35,36,51,53,54,55,60,77,110$$
\begin{enumerate}[label=(\alph*)]
\item Find the median and quartiles for these data.
A value that is greater than $Q _ { 3 } + 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right)$ or smaller than $Q _ { 1 } - 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right)$ is defined as an outlier.
\item Show that 110 is the only outlier.
\item Using the graph paper on page 15 draw a box plot for these data indicating clearly the position of the outlier.
The value of 110 is omitted.
\item Show that $S _ { y y }$ for the remaining 10 students is 2966.9
These 10 students were each asked how many text messages, $x$, they sent in the same week.
The values of $S _ { x x }$ and $S _ { x y }$ for these 10 students are $S _ { x x } = 3463.6$ and $S _ { x y } = - 18.3$.
\item Calculate the product moment correlation coefficient between the number of text messages sent and the total length of calls for these 10 students.
A parent believes that a student who sends a large number of text messages will spend fewer minutes on calls.
\item Comment on this belief in the light of your calculation in part (e).
\includegraphics[max width=\textwidth, alt={}, center]{d5d000c7-de42-461a-ba05-6c8b2c333780-09_611_1593_297_178}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2009 Q4 [14]}}