| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from raw data |
| Difficulty | Easy -1.3 This is a routine S1 statistics question requiring standard calculations (mean, standard deviation, quartiles) from given data with helpful summations provided. All techniques are straightforward textbook procedures with no problem-solving insight needed, making it easier than average A-level questions. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Patient | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) | \(I\) | \(J\) | \(K\) | \(L\) | ||
| 160 | 390 | 169 | 175 | 125 | 420 | 171 | 250 | 210 | 258 | 186 | 243 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| mean \(= \frac{2757}{12} = 229.75\) | M1, A1 | AWRT 230; M1 for using \(\frac{\sum x}{n}\) with credible numerator and \(n=12\) |
| sd \(= \sqrt{\frac{724961}{12} - (229.75)^2} = 87.34045\) | M1, A1 | AWRT 87.3; accept \(s =\) AWRT 91.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Q_2 = \frac{1}{2}(186 + 210) = 198\) | B1 | 1st B1 for median = 198 only |
| \(Q_1 = \frac{1}{2}(169 + 171) = 170\) | B1 | 2nd B1 for lower quartile |
| \(Q_3 = \frac{1}{2}(250 + 258) = 254\) | B1 | 3rd B1 for upper quartile |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Q_3 + 1.5(Q_3 - Q_1) = 254 + 1.5(254 - 170) = 380\) | M1, A1 | Accept AWRT 370–392 |
| Patients \(F\) (420) and \(B\) (390) are outliers | B1ft, B1ft | ft their limit in range (258, 420); if more points given score B0 for second B |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{Q_1 - 2Q_2 + Q_3}{Q_3 - Q_1} = \frac{170 - 2 \times 198 + 254}{254 - 170} = 0.\dot{3}\) | M1, A1 | AWRT 0.33; M1 for \(\geq 2\) correct substitutions |
| Positive skew | A1ft | Follow through their value/sign; "positive correlation" scores A0 |
## Question 2:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| mean $= \frac{2757}{12} = 229.75$ | M1, A1 | AWRT 230; M1 for using $\frac{\sum x}{n}$ with credible numerator and $n=12$ |
| sd $= \sqrt{\frac{724961}{12} - (229.75)^2} = 87.34045$ | M1, A1 | AWRT 87.3; accept $s =$ AWRT 91.2 |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Q_2 = \frac{1}{2}(186 + 210) = 198$ | B1 | 1st B1 for median = 198 only |
| $Q_1 = \frac{1}{2}(169 + 171) = 170$ | B1 | 2nd B1 for lower quartile |
| $Q_3 = \frac{1}{2}(250 + 258) = 254$ | B1 | 3rd B1 for upper quartile |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Q_3 + 1.5(Q_3 - Q_1) = 254 + 1.5(254 - 170) = 380$ | M1, A1 | Accept AWRT 370–392 |
| Patients $F$ (420) and $B$ (390) are outliers | B1ft, B1ft | ft their limit in range (258, 420); if more points given score B0 for second B |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{Q_1 - 2Q_2 + Q_3}{Q_3 - Q_1} = \frac{170 - 2 \times 198 + 254}{254 - 170} = 0.\dot{3}$ | M1, A1 | AWRT 0.33; M1 for $\geq 2$ correct substitutions |
| Positive skew | A1ft | Follow through their value/sign; "positive correlation" scores A0 |
---2. Cotinine is a chemical that is made by the body from nicotine which is found in cigarette smoke. A doctor tested the blood of 12 patients, who claimed to smoke a packet of cigarettes a day, for cotinine. The results, in appropriate units, are shown below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Patient & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ & $I$ & $J$ & $K$ & $L$ \\
\hline
\begin{tabular}{ c }
Cotinine \\
level, $X$ \\
\end{tabular} & 160 & 390 & 169 & 175 & 125 & 420 & 171 & 250 & 210 & 258 & 186 & 243 \\
\hline
\end{tabular}
\end{center}
$$\text { [You may use } \sum x ^ { 2 } = 724 \text { 961] }$$
\begin{enumerate}[label=(\alph*)]
\item Find the mean and standard deviation of the level of cotinine in a patient's blood.
\item Find the median, upper and lower quartiles of these data.
A doctor suspects that some of his patients have been smoking more than a packet of cigarettes per day. He decides to use $\mathrm { Q } _ { 3 } + 1.5 \left( \mathrm { Q } _ { 3 } - \mathrm { Q } _ { 1 } \right)$ to determine if any of the cotinine results are far enough away from the upper quartile to be outliers.
\item Identify which patient(s) may have been smoking more than a packet of cigarettes a day. Show your working clearly.
Research suggests that cotinine levels in the blood form a skewed distribution.\\
One measure of skewness is found using $\frac { \left( Q _ { 1 } - 2 Q _ { 2 } + Q _ { 3 } \right) } { \left( Q _ { 3 } - Q _ { 1 } \right) }$.
\item Evaluate this measure and describe the skewness of these data.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2008 Q2 [14]}}