| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | March |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Coding to simplify calculation |
| Difficulty | Easy -1.2 This is a straightforward application of coding (linear transformation) to simplify calculations - a standard S1 technique. Students need to subtract 1760 from each value, calculate mean and standard deviation of the coded data, then reverse the transformation. The arithmetic is routine and the method is a textbook exercise requiring only recall of transformation properties (mean shifts by the constant, standard deviation unchanged). |
| Spec | 2.02g Calculate mean and standard deviation |
| 1761.6 | 1758.5 | 1762.3 | 1761.4 | 1759.4 | 1759.1 |
| 1762.5 | 1761.9 | 1762.4 | 1761.9 | 1762.8 | 1761.0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.6, -1.5, 2.3, 1.4, -0.6, -0.9, 2.5, 1.9, 2.4, 1.9, 2.8, 1.0\) | M1 | Subtracting 1760, allow max 2 slips |
| Mean \(= 1.23\) | A1 | |
| \(sd = 1.39\) | A1 | |
| Mean of \(x = 1761.23\), sd of \(x = 1.39\) | A1\(\checkmark\) | ft their coded mean and sd. *SR B1 correct mean and sd without use of coded process* |
# Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.6, -1.5, 2.3, 1.4, -0.6, -0.9, 2.5, 1.9, 2.4, 1.9, 2.8, 1.0$ | M1 | Subtracting 1760, allow max 2 slips |
| Mean $= 1.23$ | A1 | |
| $sd = 1.39$ | A1 | |
| Mean of $x = 1761.23$, sd of $x = 1.39$ | A1$\checkmark$ | ft their coded mean and sd. *SR B1 correct mean and sd without use of coded process* |
---1 Twelve values of $x$ are shown below.
\begin{center}
\begin{tabular}{ l l l l l l }
1761.6 & 1758.5 & 1762.3 & 1761.4 & 1759.4 & 1759.1 \\
1762.5 & 1761.9 & 1762.4 & 1761.9 & 1762.8 & 1761.0 \\
\end{tabular}
\end{center}
Find the mean and standard deviation of $( x - 1760 )$. Hence find the mean and standard deviation of $x$. [4]\\
\hfill \mbox{\textit{CAIE S1 2017 Q1 [4]}}