| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | All components random including container |
| Difficulty | Standard +0.8 This is a Further Maths Statistics question requiring understanding of linear combinations of normal variables with multiple components. Part (a) is straightforward (sum of independent normals), but part (b) requires forming the distribution of L+12E-2(S+6E)=L-2S, which demands careful algebraic manipulation and conceptual understanding of how to handle the comparison of two composite random variables. The multi-step nature and the non-trivial setup in part (b) place this above average difficulty. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(W = E_1 + E_2 + ... + E_6 + S \sim \text{N}(6\times60 + 24, 6\times3^2 + 1.8^2)\) | M1 | Setting up normal model with correct mean |
| \([W \sim \text{N}(384, 57.24)]\) | M1 | Use of correct variance |
| \(P(W < 387) = 0.6541...\) awrt 0.654 | A1 | awrt 0.654 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X = E_1 + E_2 + ... + E_{12} + L \sim \text{N}(12\times60 + 40, 12\times3^2 + 2.1^2)\) | M1 | Setting up normal model for large box of eggs, sight of \(12\times60+40\) |
| \([X \sim \text{N}(760, 112.41)]\) | ||
| \(2W \sim \text{N}(\text{"384"}\times2, 2^2\times\text{"57.24"})[= \text{N}(768, 228.96)]\) | M1 | Sight or use of 2 times their mean from (a) and 4 times their variance from (a) |
| \(X - 2W \sim \text{N}(760-768, 112.41+228.96) \rightarrow \text{N}(-8, 341.37)\) | M1A1 | Setting up distribution for the difference with mean of their "760" – "768". Correct mean and variance |
| \(P(X - 2W > 0) = 0.3325...\) awrt 0.333 | A1 | awrt 0.333 |
# Question 4:
## Part 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $W = E_1 + E_2 + ... + E_6 + S \sim \text{N}(6\times60 + 24, 6\times3^2 + 1.8^2)$ | M1 | Setting up normal model with correct mean |
| $[W \sim \text{N}(384, 57.24)]$ | M1 | Use of correct variance |
| $P(W < 387) = 0.6541...$ awrt **0.654** | A1 | awrt **0.654** |
## Part 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X = E_1 + E_2 + ... + E_{12} + L \sim \text{N}(12\times60 + 40, 12\times3^2 + 2.1^2)$ | M1 | Setting up normal model for large box of eggs, sight of $12\times60+40$ |
| $[X \sim \text{N}(760, 112.41)]$ | | |
| $2W \sim \text{N}(\text{"384"}\times2, 2^2\times\text{"57.24"})[= \text{N}(768, 228.96)]$ | M1 | Sight or use of 2 times their mean from (a) and 4 times their variance from (a) |
| $X - 2W \sim \text{N}(760-768, 112.41+228.96) \rightarrow \text{N}(-8, 341.37)$ | M1A1 | Setting up distribution for the difference with mean of their "760" – "768". Correct mean and variance |
| $P(X - 2W > 0) = 0.3325...$ awrt **0.333** | A1 | awrt **0.333** |
---\begin{enumerate}
\item The weights of eggs, $E$ grams, follow a normal distribution, $\mathrm { N } \left( 60,3 ^ { 2 } \right)$
\end{enumerate}
The weights of empty small boxes, $S$ grams, follow a normal distribution, $\mathrm { N } \left( 24,1.8 ^ { 2 } \right)$\\
The weights of empty large boxes, $L$ grams, follow a normal distribution, $\mathrm { N } \left( 40,2.1 ^ { 2 } \right)$\\
Small boxes of eggs contain 6 randomly selected eggs.\\
Large boxes of eggs contain 12 randomly selected eggs.\\
(a) Find the probability that the total weight of a randomly selected small box of 6 eggs weighs less than 387 grams.\\
(b) Find the probability that a randomly selected large box of 12 eggs weighs more than twice a randomly selected small box of 6 eggs.
\hfill \mbox{\textit{Edexcel FS2 2023 Q4 [8]}}