| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2022 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sum or total of normal variables |
| Difficulty | Standard +0.8 This is a multi-part question requiring understanding of linear combinations of normal variables, including finding distributions of sums and differences (2B vs C), weighted combinations (cost calculations), and working backwards from probability constraints. While the individual techniques are standard S3 content, the question requires careful setup of multiple different linear combinations and inverse normal calculations, making it moderately challenging but within typical Further Maths Statistics scope. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Let \(X\) represent \(B_1 + B_2 - C_1\) | ||
| \(X \sim \text{N}(0.268, 0.015633)\) awrt \(0.0156\) | M1 A1 | M1 for setting up normal distribution with mean \(0.268\); A1 for correct variance \((= 0.015633)\) or standard deviation \((= 0.125\ldots)\) |
| \(P(X < 0) = P\left(Z < \dfrac{0 - 0.268}{\sqrt{"0.015633"}}(= -2.14)\right)\) | M1 | for standardising with \(0\), \(0.268\) and their standard deviation |
| \((= 1 - 0.9838) = 0.0162\) | A1 | awrt \(0.0162\) (allow awrt \(0.0160\) from calculator) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Let \(Y\) represent \(2.5B_1 + 3C_1 + 3C_2\) | ||
| \(Y \sim \text{N}(6.918, 0.071478)\) awrt \(6.92\), \(0.0715\) | M1 A1 | M1 for setting up normal distribution with mean awrt \(6.92\); A1 for correct variance \((= 0.071478)\) or standard deviation \((= 0.267\ldots)\) |
| \(P(Y > 7) = P\left(Z > \dfrac{7 - \text{"6.918"}}{\sqrt{\text{"0.071478"}}}(= 0.31)\right)\) | M1 | for standardising with \(7\), \(0.071478\) and their standard deviation |
| \((= 1 - 0.6217) = 0.3783\) (Calculator gives \(0.3795\ldots\)) \(\quad 0.378 - 0.380\) | A1 | for answer between \(0.378 - 3.80\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Mean \(= 2.94w\) | B1 | for \(2.94w\) |
| Standard deviation \(= 0.084\sqrt{5}\, w \quad (= 0.188w)\) | B1 | for \(0.084\sqrt{5}\,w\) or awrt \(0.188w\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\dfrac{6 - 2.94w}{0.084\sqrt{5}\, w} = {-1.2816}\) | M1; B1 | M1 for standardising using their mean and standard deviation \(= z\) where \(1 < \ |
| \(-1.2816 \times 0.084\sqrt{5}\, w + 2.94w \ldots 6\) | dM1 | dependent on M1, for solving their inequality |
| \(w \ldots 2.22\ldots \quad\) So \(w = 2.23\) | A1 | awrt (£)\(2.23\) |
## Question 7:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Let $X$ represent $B_1 + B_2 - C_1$ | | |
| $X \sim \text{N}(0.268, 0.015633)$ awrt $0.0156$ | M1 A1 | M1 for setting up normal distribution with mean $0.268$; A1 for correct variance $(= 0.015633)$ or standard deviation $(= 0.125\ldots)$ |
| $P(X < 0) = P\left(Z < \dfrac{0 - 0.268}{\sqrt{"0.015633"}}(= -2.14)\right)$ | M1 | for standardising with $0$, $0.268$ and their standard deviation |
| $(= 1 - 0.9838) = 0.0162$ | A1 | awrt $0.0162$ (allow awrt $0.0160$ from calculator) |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Let $Y$ represent $2.5B_1 + 3C_1 + 3C_2$ | | |
| $Y \sim \text{N}(6.918, 0.071478)$ awrt $6.92$, $0.0715$ | M1 A1 | M1 for setting up normal distribution with mean awrt $6.92$; A1 for correct variance $(= 0.071478)$ or standard deviation $(= 0.267\ldots)$ |
| $P(Y > 7) = P\left(Z > \dfrac{7 - \text{"6.918"}}{\sqrt{\text{"0.071478"}}}(= 0.31)\right)$ | M1 | for standardising with $7$, $0.071478$ and their standard deviation |
| $(= 1 - 0.6217) = 0.3783$ (Calculator gives $0.3795\ldots$) $\quad 0.378 - 0.380$ | A1 | for answer between $0.378 - 3.80$ |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Mean $= 2.94w$ | B1 | for $2.94w$ |
| Standard deviation $= 0.084\sqrt{5}\, w \quad (= 0.188w)$ | B1 | for $0.084\sqrt{5}\,w$ or awrt $0.188w$ |
### Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\dfrac{6 - 2.94w}{0.084\sqrt{5}\, w} = {-1.2816}$ | M1; B1 | M1 for standardising using their mean and standard deviation $= z$ where $1 < \|z\| < 1.5$; B1 for $-1.28$ |
| $-1.2816 \times 0.084\sqrt{5}\, w + 2.94w \ldots 6$ | dM1 | dependent on M1, for solving their inequality |
| $w \ldots 2.22\ldots \quad$ So $w = 2.23$ | A1 | awrt (£)$2.23$ |\begin{enumerate}
\item A market stall sells vegetables. Two of the vegetables sold are broccoli heads and cabbages.
\end{enumerate}
The weights of these broccoli heads, $B$ kilograms, follow a normal distribution
$$B \sim \mathrm {~N} \left( 0.588,0.084 ^ { 2 } \right)$$
The weights of these cabbages, $C$ kilograms, follow a normal distribution
$$C \sim \mathrm {~N} \left( 0.908,0.039 ^ { 2 } \right)$$
(a) Find the probability that the total weight of two randomly chosen broccoli heads is less than the weight of a randomly chosen cabbage.
Broccoli heads cost $\pounds 2.50$ per kg and cabbages cost $\pounds 3.00$ per kg.
Jaymini buys 1 broccoli head and 2 cabbages, chosen randomly.\\
(b) Find the probability that she pays more than £7
The market stall offers a discount for buying 5 or more broccoli heads. The price with the discount is $\pounds w$ per kg.
Let $\pounds D$ be the price with the discount of 5 broccoli heads.\\
(c) Find, in terms of $w$, the mean and standard deviation of $D$
Given that $\mathrm { P } ( D < 6 ) < 0.1$\\
(d) find the smallest possible value of $w$, giving your answer to 2 decimal places.
\hfill \mbox{\textit{Edexcel S3 2022 Q7 [14]}}