| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Difficulty | Challenging +1.2 This question requires understanding of linear combinations of normal variables and independence, which are core Further Maths topics. Part (a) involves forming L - 4S and finding its distribution (requiring careful variance calculation with the coefficient), then a probability calculation. Parts (b) and (c) are more routine applications. The multi-step nature and need to construct appropriate linear combinations elevates this above average difficulty, but it follows standard Further Stats patterns without requiring novel insight. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities2.04e Normal distribution: as model N(mu, sigma^2)5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(X = L - 4S\) then \(E(X) = 19.6 - 4 \times 4.8\) | M1 | Selecting and using an appropriate model i.e. \(\pm(L - 4S)\). May be implied by 0.4 |
| \(= 0.4\) | A1 | 0.4 o.e |
| \(\text{Var}(X) = \text{Var}(L) + 4^2 \text{Var}(S) = 0.6^2 + 16 \times 0.3^2\) | M1 | For realising the need to use \(\text{Var}(L) + 4^2\text{Var}(S)\). Allow use of 0.6 for \(\text{Var}(L)\) instead of \(0.6^2\) and/or 0.3 for \(\text{Var}(S)\) instead of \(0.3^2\) may be implied by 1.8 |
| \(= 1.8\) | A1 | 1.8 only |
| \(\text{P}(X > 0) = [\text{P}(Z > \frac{0-0.4}{\sqrt{1.8}} = -0.298...-)]\) | M1 | For realising \(\text{P}(X > 0)\) is required and an attempt to find it e.g. \(\frac{0-0.4}{\sqrt{\text{"their Var}(X)"}}\) but do not allow a negative \(\text{Var}(X)\) |
| \(= 0.617202...\) awrt 0.617 | A1 | awrt 0.617 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = S_1 + S_2 + S_3 + S_4\) (May be implied by 0.36) | M1 | Selecting and using an appropriate model i.e. \(s_1 + s_2 + s_3 + s_4\); may be implied by 0.36 |
| \(T \sim N(19.2, 0.36)\) | B1 | 19.2 only |
| \(E(T) = 19.2\) \(\text{Var}(T) = 0.36\) or \(0.6^2\) | A1 | 0.36 |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(Y = L - T\) \(E(Y) = E(L) - E(T) = [0.4]\) | M1 | Setting up and using the model \(Y = L - T\). May be implied by \(E(Y) = E(L) - E(T)\) |
| \(\text{Var}(Y) = \text{Var}(L) + \text{Var}(T) = [0.72]\) | M1 | Using \(\text{Var}(Y) = \text{Var}(L) + \text{Var}(T)\) |
| Require \(\text{P}(-0.2 < Y < 0.2)\) | M1 | Dealing with the modulus and realising they need to find \(\text{P}(-0.2 < Y < 0.2)\) |
| \(= 0.16708...\) awrt 0.167 | A1 | awrt 0.167 |
**(a)**
| Let $X = L - 4S$ then $E(X) = 19.6 - 4 \times 4.8$ | M1 | Selecting and using an appropriate model i.e. $\pm(L - 4S)$. May be implied by 0.4 |
| $= 0.4$ | A1 | 0.4 o.e |
| $\text{Var}(X) = \text{Var}(L) + 4^2 \text{Var}(S) = 0.6^2 + 16 \times 0.3^2$ | M1 | For realising the need to use $\text{Var}(L) + 4^2\text{Var}(S)$. Allow use of 0.6 for $\text{Var}(L)$ instead of $0.6^2$ and/or 0.3 for $\text{Var}(S)$ instead of $0.3^2$ may be implied by 1.8 |
| $= 1.8$ | A1 | 1.8 only |
| $\text{P}(X > 0) = [\text{P}(Z > \frac{0-0.4}{\sqrt{1.8}} = -0.298...-)]$ | M1 | For realising $\text{P}(X > 0)$ is required and an attempt to find it e.g. $\frac{0-0.4}{\sqrt{\text{"their Var}(X)"}}$ but do not allow a negative $\text{Var}(X)$ |
| $= 0.617202...$ awrt **0.617** | A1 | awrt 0.617 |
**(b)**
| $T = S_1 + S_2 + S_3 + S_4$ (May be implied by 0.36) | M1 | Selecting and using an appropriate model i.e. $s_1 + s_2 + s_3 + s_4$; may be implied by 0.36 |
| $T \sim N(19.2, 0.36)$ | B1 | 19.2 only |
| $E(T) = 19.2$ $\text{Var}(T) = 0.36$ **or** $0.6^2$ | A1 | 0.36 |
**(c)**
| Let $Y = L - T$ $E(Y) = E(L) - E(T) = [0.4]$ | M1 | Setting up and using the model $Y = L - T$. May be implied by $E(Y) = E(L) - E(T)$ |
| $\text{Var}(Y) = \text{Var}(L) + \text{Var}(T) = [0.72]$ | M1 | Using $\text{Var}(Y) = \text{Var}(L) + \text{Var}(T)$ |
| Require $\text{P}(-0.2 < Y < 0.2)$ | M1 | Dealing with the modulus and realising they need to find $\text{P}(-0.2 < Y < 0.2)$ |
| $= 0.16708...$ awrt **0.167** | A1 | awrt 0.167 |
---\begin{enumerate}
\item Scaffolding poles come in two sizes, long and short. The length $L$ of a long pole has the normal distribution $\mathrm { N } \left( 19.6,0.6 ^ { 2 } \right)$. The length $S$ of a short pole has the normal distribution N(4.8, 0.32). The random variables $L$ and $S$ are independent.
\end{enumerate}
A long pole and a short pole are selected at random.\\
(a) Find the probability that the length of the long pole is more than 4 times the length of the short pole. Show your working clearly.
Four short poles are selected at random and placed end to end in a row. The random variable $T$ represents the length of the row.\\
(b) Find the distribution of $T$.\\
(c) Find $\mathrm { P } ( | L - T | < 0.2 )$
\hfill \mbox{\textit{Edexcel FS2 Q5 [13]}}