| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sum versus sum comparison |
| Difficulty | Challenging +1.2 This question tests linear combinations of normal random variables with multiple parts requiring standard techniques: finding distributions of sums and differences (part a), working with a scaled comparison (part b), and applying conditional probability with cost calculation (part c). While part (c) requires careful setup of the conditional expectation, all parts follow established Further Stats methods without requiring novel insight. The multi-step nature and Further Maths context place it above average difficulty, but it remains a straightforward application question. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X = L_1+L_2+L_3 \sim N\!\left(594,\sqrt{75}^2\right)\) | M1 | For an attempt at \(X\) or \(Y\) — expression or implied by one correct distribution |
| A1 | For a correct distribution for \(X\) or implied by \(E(D)=2\) | |
| \(Y = S_1+\ldots+S_8 \sim N\!\left(592,\sqrt{72}^2\right)\) | A1 | For a correct distribution for \(Y\) or implied by \(\text{Var}(D)=147\) |
| \(P(X>Y) = P(D>0)\) where \(D \sim N\!\left(2,\sqrt{147}^2\right)\) | M1 | For a correct strategy — attempt \(X-Y\) and \(P(D>0)\) statement |
| A1ft | For a correct distribution for \(D\) ft their \(X\) and \(Y\) | |
| \(= 0.56551\ldots\) awrt 0.566 | A1 | For awrt \(0.566\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(W = L - \frac{8}{3}S \Rightarrow W \sim N\!\left(\frac{2}{3},\ 25+\frac{64}{9}\times 9\right) = N\!\left(\frac{2}{3},\sqrt{89}^2\right)\) | M1, M1, A1 | 1st M1 for attempt at correct model; 2nd M1 for correct expression for variance of form \(L-kS\) or \(kL-S\); A1 for fully correct distribution |
| \(P(W>0) = 0.528168\ldots\) awrt 0.528 | M1A1 | For a correct probability statement using their distribution; for awrt \(0.528\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F = L_1+\ldots+L_5 \sim N\!\left(990,\sqrt{125}^2\right)\) | M1 | For a correct start to solve the problem, attempt at \(F\) and correct mean |
| A1 | For a correct distribution | |
| \(P(F < 1000) = 0.814455\ldots\) (o.e.) | A1 | For using this model to find \(P(F<1000)=\) awrt \(0.814\) or \(P(F>1000)=\) awrt \(0.186\) |
| \(E(\text{cost of Rosa's plan}) = 430\times\text{"0.814"}+400\times(1-\text{"0.814"})\) | M1 | For a correct strategy to solve the problem i.e. attempt at expected cost ft their prob |
| \(= £424.43\) | A1 | For awrt £424 |
| Buying 14 small panels costs \(14\times 30 = £420\), so Rosa's plan is likely to be more expensive | A1 | For a correct conclusion with comparison with £420 and reject Rosa's plan |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X = L_1+L_2+L_3 \sim N\!\left(594,\sqrt{75}^2\right)$ | M1 | For an attempt at $X$ or $Y$ — expression or implied by one correct distribution |
| | A1 | For a correct distribution for $X$ or implied by $E(D)=2$ |
| $Y = S_1+\ldots+S_8 \sim N\!\left(592,\sqrt{72}^2\right)$ | A1 | For a correct distribution for $Y$ or implied by $\text{Var}(D)=147$ |
| $P(X>Y) = P(D>0)$ where $D \sim N\!\left(2,\sqrt{147}^2\right)$ | M1 | For a correct strategy — attempt $X-Y$ and $P(D>0)$ statement |
| | A1ft | For a correct distribution for $D$ ft their $X$ and $Y$ |
| $= 0.56551\ldots$ awrt **0.566** | A1 | For awrt $0.566$ |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $W = L - \frac{8}{3}S \Rightarrow W \sim N\!\left(\frac{2}{3},\ 25+\frac{64}{9}\times 9\right) = N\!\left(\frac{2}{3},\sqrt{89}^2\right)$ | M1, M1, A1 | 1st M1 for attempt at correct model; 2nd M1 for correct expression for variance of form $L-kS$ or $kL-S$; A1 for fully correct distribution |
| $P(W>0) = 0.528168\ldots$ awrt **0.528** | M1A1 | For a correct probability statement using their distribution; for awrt $0.528$ |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = L_1+\ldots+L_5 \sim N\!\left(990,\sqrt{125}^2\right)$ | M1 | For a correct start to solve the problem, attempt at $F$ and correct mean |
| | A1 | For a correct distribution |
| $P(F < 1000) = 0.814455\ldots$ (o.e.) | A1 | For using this model to find $P(F<1000)=$ awrt $0.814$ or $P(F>1000)=$ awrt $0.186$ |
| $E(\text{cost of Rosa's plan}) = 430\times\text{"0.814"}+400\times(1-\text{"0.814"})$ | M1 | For a correct strategy to solve the problem i.e. attempt at expected cost ft their prob |
| $= £424.43$ | A1 | For awrt £424 |
| Buying 14 small panels costs $14\times 30 = £420$, so Rosa's plan is likely to be more expensive | A1 | For a correct conclusion with comparison with £420 and reject Rosa's plan |7 Fence panels come in two sizes, large and small. The lengths of the large panels are normally distributed with mean 198 cm and standard deviation 5 cm . The lengths of the small panels are normally distributed with mean 74 cm and standard deviation 3 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the total length of a random sample of 3 large panels is greater than the total length of a random sample of 8 small panels.
One large panel and one small panel are selected at random.
\item Find the probability that the length of the large panel is more than $\frac { 8 } { 3 }$ times the length of the small panel.
Rosa needs 1000 cm of fencing. The large panels cost $\pounds 80$ each and the small panels cost $\pounds 30$ each. Rosa's plan is to buy 5 large panels and measure the total length. If the total length is less than 1000 cm she will then buy one small panel as well.
\item Calculate whether or not the expected cost of Rosa's plan is cheaper than simply buying 14 small panels.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 2020 Q7 [17]}}