| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two or more different variables |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for linear combinations of independent normal random variables. Part (a) requires summing three normals and finding a probability between two values. Part (b) requires finding P(B > C) which is equivalent to P(B - C > 0). Both parts use bookwork formulas with no conceptual challenges—students simply need to recall that sums/differences of independent normals are normal, calculate the combined mean and variance, then use normal tables. This is slightly easier than average as it's purely procedural with clear signposting. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Task | Mean | Standard deviation |
| \(A\) | 225 | 38 |
| \(B\) | 165 | 23 |
| \(C\) | 185 | 27 |
The three tasks most frequently carried out in a garage are $A$, $B$ and $C$. For each of the tasks the times, in minutes, taken by the garage mechanics are assumed to be normally distributed with means and standard deviations given in the following table.
\begin{center}
\begin{tabular}{|c|c|c|}
\hline
Task & Mean & Standard deviation \\
\hline
$A$ & 225 & 38 \\
$B$ & 165 & 23 \\
$C$ & 185 & 27 \\
\hline
\end{tabular}
\end{center}
Assuming that the times for the three tasks are independent, calculate the probability that
\begin{enumerate}[label=(\alph*)]
\item the total time taken by a single randomly chosen mechanic to carry out all three tasks lies between 533 and 655 minutes, [5]
\item a randomly chosen mechanic takes longer to carry out task $B$ than task $C$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q3 [10]}}