| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Coding to simplify calculation |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question testing standard procedures: finding midpoints, calculating mean/standard deviation from grouped frequency data, linear interpolation for median, and applying coding transformations. All techniques are straightforward textbook applications with no problem-solving insight required, making it easier than average but not trivial due to the multiple parts and calculations involved. |
| Spec | 2.02g Calculate mean and standard deviation2.02h Recognize outliers5.02c Linear coding: effects on mean and variance |
| Coded Time (v) | Frequency ( \(\boldsymbol { f }\) ) | Coded Time Midpoint (m) |
| \(0 \leqslant v < 5\) | 20 | 2.5 |
| \(5 \leqslant v < 10\) | 24 | \(a\) |
| \(10 \leqslant v < 15\) | 16 | 12.5 |
| \(15 \leqslant v < 20\) | 14 | 17.5 |
| \(20 \leqslant v < 30\) | 6 | \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Mark |
| (a) | 7.5 and 25 | B1 |
| (b) | Mean \(= 10.3125\) awrt 10.3 | B1 |
| (c) | \(\sigma = \sqrt{\frac{120125}{80} - 10.3125^2}\) | M1 |
| (c) | \(= 6.6188...\) \((s = 6.6605...)\) awrt 6.62 | A1 |
| (d) | Median \(= \{5\}+\frac{20}{24}\times5\) or \(\{10\}-\frac{4}{24}\times5\) | M1 |
| (d) | \(= 9.16\overline{6}\) awrt 9.17 | A1 |
| (e) | Mean \(>\) median \(\therefore\) positive skew | M1A1 |
| (f) | \(t = 10v+5\); Mean \(= 10\times10.3125+5\) | M1 |
| (f) | \(= 108.125\) awrt 108 | A1 |
| (f) | \(\sigma = 10\times6.6188\) | M1 |
| (f) | \(= 66.188...\) (66.605 from \(s\)) awrt 66.2 | A1 |
## Question 4:
| Part | Answer/Working | Mark | Guidance |
|---|---|---|---|
| (a) | 7.5 and 25 | B1 | Both values correct (may be seen in table) |
| (b) | Mean $= 10.3125$ awrt **10.3** | B1 | Do not allow improper fractions |
| (c) | $\sigma = \sqrt{\frac{120125}{80} - 10.3125^2}$ | M1 | Correct expression including square root (allow ft from their mean) |
| (c) | $= 6.6188...$ $(s = 6.6605...)$ awrt **6.62** | A1 | Allow $s =$ awrt 6.66 |
| (d) | Median $= \{5\}+\frac{20}{24}\times5$ or $\{10\}-\frac{4}{24}\times5$ | M1 | Correct fraction: $\frac{20}{24}\times5$ or if using $n+1$ for $\frac{20.5}{24}\times5$; also $-\frac{4}{24}\times5$ |
| (d) | $= 9.16\overline{6}$ awrt **9.17** | A1 | If using $n+1$ accept awrt 9.27 |
| (e) | Mean $>$ median $\therefore$ positive skew | M1A1 | Correct comparison of their (b) and (d); allow $Q_3-Q_2 > Q_2-Q_1$ only if $Q_1=5$ and $Q_3=15$ both seen |
| (f) | $t = 10v+5$; Mean $= 10\times10.3125+5$ | M1 | For $10\times\text{'their mean'}+5$ |
| (f) | $= 108.125$ awrt **108** | A1 | Use of decoded data must be fully correct i.e. $8650/80$ |
| (f) | $\sigma = 10\times6.6188$ | M1 | For $10\times\text{'their sd'}$ |
| (f) | $= 66.188...$ (66.605 from $s$) awrt **66.2** | A1 | Use of decoded data must be fully correct i.e. $\sqrt{\frac{1285750}{80}-\left(\frac{8650}{80}\right)^2}$ |\begin{enumerate}
\item A researcher recorded the time, $t$ minutes, spent using a mobile phone during a particular afternoon, for each child in a club.
\end{enumerate}
The researcher coded the data using $v = \frac { t - 5 } { 10 }$ and the results are summarised in the table below.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Coded Time (v) & Frequency ( $\boldsymbol { f }$ ) & Coded Time Midpoint (m) \\
\hline
$0 \leqslant v < 5$ & 20 & 2.5 \\
\hline
$5 \leqslant v < 10$ & 24 & $a$ \\
\hline
$10 \leqslant v < 15$ & 16 & 12.5 \\
\hline
$15 \leqslant v < 20$ & 14 & 17.5 \\
\hline
$20 \leqslant v < 30$ & 6 & $b$ \\
\hline
\end{tabular}
\end{center}
$$\text { (You may use } \sum f m = 825 \text { and } \sum f m ^ { 2 } = 12012.5 \text { ) }$$
(a) Write down the value of $a$ and the value of $b$.\\
(b) Calculate an estimate of the mean of $v$.\\
(c) Calculate an estimate of the standard deviation of $v$.\\
(d) Use linear interpolation to estimate the median of $v$.\\
(e) Hence describe the skewness of the distribution. Give a reason for your answer.\\
(f) Calculate estimates of the mean and the standard deviation of the time spent using a mobile phone during the afternoon by the children in this club.
$\_\_\_\_$ VAYV SIHI NI JIIIM ION OC\\
VJYV SIHI NI JIIIM ION OC\\
VJYV SIHI NI JLIYM ION OC
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel S1 2018 Q4 [12]}}