| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2005 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Same variable, two observations |
| Difficulty | Standard +0.3 This is a standard S3 question on linear combinations of normal variables requiring routine application of formulas (variance of differences/sums) and normal probability calculations. While it has multiple parts and requires careful bookkeeping of variances, all techniques are textbook exercises with no novel insight needed—slightly easier than average A-level difficulty. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Guidance |
| (a) Let \(W = C_1 - C_2\) is \(W \sim N(0, 16)\) Normal | M1, B1; M | |
| 0; 16 | ||
| \(P(\ | W\ | > 6) = 2 P(W > 6)\) |
| \(= 2 \times P\left(Z > \frac{6-0}{\sqrt{16}}\right)\) Standardising, their \(\sigma\) | M1 | |
| \(= 2 \times P(Z > 1.5)\) | ||
| \(= 2 \times (1 - 0.9332) = 0.1336\) | A1 (6) | |
| (b) Let \(W_e\) ~ \(C - L\) \(\therefore W \sim N(5, 25)\) | B1; B1 | 5; 25 |
| \(P(W > 0) = P\left(Z > \frac{±5}{\sqrt{25}}\right)\) | M1; A1 | |
| \(= P(Z < 1)\) | M1 (\(\rho > 0.5\)) | |
| \(= 0.6443\) | A1 (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Guidance |
| (c) Let \(W = C_1 + \cdots + C_{24} + B\) \(\therefore E(W) = 24 \times 360 + 100 = 8500\) | B1 | |
| \(\text{Var}(W) = 24 \times 6 + 2^2 = 196\) | B1 | |
| \(P(8510 < W < 8520) = P\left(\frac{8510-8500}{\sqrt{196}} \leq Z \leq \frac{8520-8500}{\sqrt{196}}\right)\) | M1 | |
| \(= P(0.71... \leq Z \leq 1.43....)\) AWET | A√ A1√ | |
| \(= 0.9236 - 0.7611 = 0.1625\) | 0.161 - 0.143 | A1 (6) |
| (d) All random variables are independent. | B1 (1) |
| Content | Marks | Guidance |
|---------|-------|----------|
| **(a)** Let $W = C_1 - C_2$ is $W \sim N(0, 16)$ **Normal** | M1, B1; M | |
| | | 0; 16 |
| $P(\|W\| > 6) = 2 P(W > 6)$ | M1 | |
| $= 2 \times P\left(Z > \frac{6-0}{\sqrt{16}}\right)$ **Standardising, their $\sigma$** | M1 | |
| $= 2 \times P(Z > 1.5)$ | | |
| $= 2 \times (1 - 0.9332) = 0.1336$ | A1 (6) | |
| **(b)** Let $W_e$ ~ $C - L$ $\therefore W \sim N(5, 25)$ | B1; B1 | 5; 25 |
| $P(W > 0) = P\left(Z > \frac{±5}{\sqrt{25}}\right)$ | M1; A1 | |
| $= P(Z < 1)$ | M1 ($\rho > 0.5$) | |
| $= 0.6443$ | A1 (6) | |
---
# Question 7(c)
| Content | Marks | Guidance |
|---------|-------|----------|
| **(c)** Let $W = C_1 + \cdots + C_{24} + B$ $\therefore E(W) = 24 \times 360 + 100 = 8500$ | B1 | |
| $\text{Var}(W) = 24 \times 6 + 2^2 = 196$ | B1 | |
| $P(8510 < W < 8520) = P\left(\frac{8510-8500}{\sqrt{196}} \leq Z \leq \frac{8520-8500}{\sqrt{196}}\right)$ | M1 | |
| $= P(0.71... \leq Z \leq 1.43....)$ **AWET** | A√ A1√ | |
| $= 0.9236 - 0.7611 = 0.1625$ | **0.161 - 0.143** | A1 (6) |
| **(d)** All random variables are independent. | B1 (1) | |A manufacturer produces two flavours of soft drink, cola and lemonade. The weights, $C$ and $L$, in grams, of randomly selected cola and lemonade cans are such that $C \sim \text{N}(350, 8)$ and $L \sim \text{N}(345, 17)$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the weights of two randomly selected cans of cola will differ by more than 6 g. [6]
\end{enumerate}
One can of each flavour is selected at random.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the probability that the can of cola weighs more than the can of lemonade. [6]
\end{enumerate}
Cans are delivered to shops in boxes of 24 cans. The weights of empty boxes are normally distributed with mean 100 g and standard deviation 2 g.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the probability that a full box of cola cans weighs between 8.51 kg and 8.52 kg. [6]
\item State an assumption you made in your calculation in part (c). [1]
\end{enumerate}
(Total 19 marks)
\hfill \mbox{\textit{Edexcel S3 2005 Q7 [19]}}