| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Different variables, one observation each |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal distributions with clear setup. Part (a) requires forming R - B ~ N(2.3, 25) and finding P(R - B > 3). Part (b) extends to 4R - 5B with variance calculation. Both are standard S3 techniques with no conceptual surprises, though part (b) requires careful arithmetic with the variance of the sum. Part (c) is a simple contextual comment. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks |
|---|---|
| ∴ \(A \sim N(14.5 - 12.2, 16.0 + 9.0) = N(2.3, 25)\) | M1, A1 |
| \(P(A > 3) = P(Z > \frac{3-2.3}{5})\) | M1 |
| \(= P(Z > 0.14) = 1 - 0.5557 = 0.4443\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| ∴ \(C \sim N(4×14.5 - 5×12.2, 4×16 + 5×9) = N(-3, 109)\) | M1, A2 | |
| \(P(C > 0) = P(Z > \frac{0+3}{\sqrt{109}})\) | M1 | |
| \(= P(Z > 0.29) = 1 - 0.6141 = 0.3859\) | M1, A1 | |
| (c) e.g. likely to use smaller blocks higher up the tower | B1 | (12) |
**(a)** let A = amount side length of red cube is longer than blue cube
∴ $A \sim N(14.5 - 12.2, 16.0 + 9.0) = N(2.3, 25)$ | M1, A1 |
$P(A > 3) = P(Z > \frac{3-2.3}{5})$ | M1 |
$= P(Z > 0.14) = 1 - 0.5557 = 0.4443$ | M1, A1 |
**(b)** let C = amount red tower is taller than blue tower
∴ $C \sim N(4×14.5 - 5×12.2, 4×16 + 5×9) = N(-3, 109)$ | M1, A2 |
$P(C > 0) = P(Z > \frac{0+3}{\sqrt{109}})$ | M1 |
$= P(Z > 0.29) = 1 - 0.6141 = 0.3859$ | M1, A1 |
**(c)** e.g. likely to use smaller blocks higher up the tower | B1 | (12) |
---5. A child is playing with a set of red and blue wooden cubes. The side length of the red cubes is normally distributed with a mean of 14.5 cm and a variance of $16.0 \mathrm {~cm} ^ { 2 }$. The side length of the blue cubes is normally distributed with a mean of 12.2 cm and a variance of $9.0 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen red cube will have a side length of more than 3 cm greater than a randomly chosen blue cube.
The child makes two towers, one from 4 red cubes and one from 5 blue cubes. Assuming that the cubes for each colour of tower were chosen at random,
\item find the probability that the red tower is taller than the blue tower.
\item Explain why the assumption that the cubes for each tower were chosen at random is unlikely to be realistic.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q5 [12]}}