| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Adding data values |
| Difficulty | Moderate -0.8 This is a straightforward application of formulas for updating mean and standard deviation when adding a data value. It requires recall of the definitions (sum = n × mean, variance formula) and careful arithmetic, but involves no problem-solving insight or conceptual difficulty beyond basic S1 content. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Sigma x = 14 \times 31.2 = 436.8\) | M1 | |
| \(\text{new } \Sigma x = 436.8 + 42 = 478.8\) | M1 | |
| \(\text{new mean} = \frac{478.8}{15} = 31.9 \text{ years}\) | A1 | |
| \(\Sigma x^2 = 14(7.4^2 + 31.2^2) = 14394.8\) | M1 | |
| \(\text{new } \Sigma x^2 = 14394.8 + 42^2 = 16158.8\) | M1 | |
| \(\text{new std. dev.} = \sqrt{\frac{16158.8}{15} - 31.9^2} = 7.6 \text{ years}\) | M1 A1 | (7) |
$\Sigma x = 14 \times 31.2 = 436.8$ | M1 |
$\text{new } \Sigma x = 436.8 + 42 = 478.8$ | M1 |
$\text{new mean} = \frac{478.8}{15} = 31.9 \text{ years}$ | A1 |
$\Sigma x^2 = 14(7.4^2 + 31.2^2) = 14394.8$ | M1 |
$\text{new } \Sigma x^2 = 14394.8 + 42^2 = 16158.8$ | M1 |
$\text{new std. dev.} = \sqrt{\frac{16158.8}{15} - 31.9^2} = 7.6 \text{ years}$ | M1 A1 | (7)
---\begin{enumerate}
\item An adult evening class has 14 students. The ages of these students have a mean of 31.2 years and a standard deviation of 7.4 years.
\end{enumerate}
A new student who is exactly 42 years old joins the class. Calculate the mean and standard deviation of the 15 students now in the group.\\
\hfill \mbox{\textit{Edexcel S1 Q1 [7]}}