| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Linear transformation to achieve target parameters |
| Difficulty | Standard +0.8 This is a Further Maths statistics question requiring students to work backwards from a target distribution to find unknown coefficients in a linear combination of normal variables. It involves setting up and solving simultaneous equations using properties of mean and variance for independent normals, which goes beyond routine application and requires careful algebraic manipulation of variance terms. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(5a + 8b = 169\) | B1 | Correct equation for the means |
| \(0.16a + 0.01b^2 = 4\) | B1 | Correct equation for the variances (allow \(0.4^2 a + 0.1^2 b^2 = 4\)) |
| \(a = 33.8 - 1.6b \rightarrow 0.16(33.8 - 1.6b) + 0.01b^2 = 4\) | M1 | Attempt to solve two simultaneous equations in \(a\) and \(b\) by eliminating one variable |
| \(0.01b^2 - 0.256b + 1.408 = 0 \rightarrow b = \frac{0.256 \pm \sqrt{0.256^2 - 4(0.01)(1.408)}}{2(0.01)}\) | M1 | Attempt to solve quadratic (must be seen if answers are incorrect) |
| \(b = 8,\ a = 21\) (reject \(b = 17.6,\ a = 5.64\) since \(a \in \mathbb{Z}^+\)) | A1A1 | A1: \(b=8\) or \(a=21\) or both sets without rejecting; A1: choosing correct pair \(b=8,\ a=21\) only |
# Question 3:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $5a + 8b = 169$ | B1 | Correct equation for the means |
| $0.16a + 0.01b^2 = 4$ | B1 | Correct equation for the variances (allow $0.4^2 a + 0.1^2 b^2 = 4$) |
| $a = 33.8 - 1.6b \rightarrow 0.16(33.8 - 1.6b) + 0.01b^2 = 4$ | M1 | Attempt to solve two simultaneous equations in $a$ and $b$ by eliminating one variable |
| $0.01b^2 - 0.256b + 1.408 = 0 \rightarrow b = \frac{0.256 \pm \sqrt{0.256^2 - 4(0.01)(1.408)}}{2(0.01)}$ | M1 | Attempt to solve quadratic (must be seen if answers are incorrect) |
| $b = 8,\ a = 21$ (reject $b = 17.6,\ a = 5.64$ since $a \in \mathbb{Z}^+$) | A1A1 | A1: $b=8$ or $a=21$ or both sets without rejecting; A1: choosing correct pair $b=8,\ a=21$ only |
---\begin{enumerate}
\item The random variable $X \sim \mathrm {~N} \left( 5,0.4 ^ { 2 } \right)$ and the random variable $Y \sim \mathrm {~N} \left( 8,0.1 ^ { 2 } \right)$\\
$X$ and $Y$ are independent random variables.\\
A random sample of $a$ independent observations is taken from the distribution of $X$ and one observation is taken from the distribution of $Y$
\end{enumerate}
The random variable $W = X _ { 1 } + X _ { 2 } + X _ { 3 } + \ldots + X _ { a } + b Y$ and has the distribution $\mathrm { N } \left( 169,2 ^ { 2 } \right)$\\
Find the value of $a$ and the value of $b$
\hfill \mbox{\textit{Edexcel FS2 2022 Q3 [6]}}