| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Functions of random variables |
| Difficulty | Standard +0.8 This Further Statistics question requires systematic algebraic manipulation across multiple constraints (expectation, probability condition, variance) to find unknown parameters, then apply transformations. While methodical rather than requiring deep insight, it involves more sophisticated probability theory than standard A-level and requires careful tracking of multiple equations—typical of AS Further Maths material but not exceptionally challenging within that context. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(x\) | - 1 | 0 | 1 | 2 | 3 |
| \(P ( X = x )\) | \(c\) | \(a\) | \(a\) | \(b\) | \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-4 = 2 - 5\text{E}(X)\) | M1 | For using given information to find expression for \(\text{E}(X)\), i.e. use of \(\text{E}(Y) = 2 - 5\text{E}(X)\) |
| \(\text{E}(X) = 1.2\) | ||
| \(-1\times c + 0\times a + 1\times a + 2\times b + 3\times c = 1.2\) | M1 | For use of \(\sum x\text{P}(X=x) = 1.2\) |
| \(a + 2b + 2c = 1.2\) \(\boxed{1}\) | ||
| \(\text{P}(Y \geqslant -3) = 0.45\) gives \(\text{P}(2-5X \geqslant -3) = 0.45\), i.e. \(\text{P}(X \leqslant 1) = 0.45\) | M1 | For use of \(\text{P}(Y \geqslant -3) = 0.45\) to set up argument by forming equation in \(a\) and \(c\) |
| \(2a + c = 0.45\) \(\boxed{2}\) | ||
| \(2a + b + 2c = 1\) \(\boxed{3}\) | M1 | For use of \(\sum \text{P}(X=x) = 1\) |
| Solving 3 linear equations (matrix or elimination) | M1 | For solving their 3 linear equations |
| \(a = 0.1\), \(b = 0.3\), \(c = 0.25\) | A1, A1 | A1 for any 2 of \(a\), \(b\) or \(c\) correct; A1 for all 3 correct |
| (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(Y) = 75 - (-4)^2\) or \(59\) | M1 | For use of \(\text{Var}(Y) = \text{E}(Y^2) - [\text{E}(Y)]^2\) (may be implied by correct answer) |
| \([\text{Var}(Y) = 5^2\text{Var}(X)\) implies\(]\) \(\text{Var}(X) = 2.36\) | A1 | For use of \(\text{Var}(aX) = a^2\text{Var}(X)\) to reach 2.36 or exact equivalent |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{P}(Y > X) = \text{P}(2-5X > X) \rightarrow \text{P}(X < \frac{1}{3})\) | M1 | For rearranging to the form \(\text{P}(X < k)\) |
| \(\text{P}(X < \frac{1}{3}) = a + c = 0.35\) | A1ft | \(0.1 + 0.25\); provided their \(a\) and \(c\) and their \(a+c\) are all probabilities |
| (2 marks) |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-4 = 2 - 5\text{E}(X)$ | M1 | For using given information to find expression for $\text{E}(X)$, i.e. use of $\text{E}(Y) = 2 - 5\text{E}(X)$ |
| $\text{E}(X) = 1.2$ | | |
| $-1\times c + 0\times a + 1\times a + 2\times b + 3\times c = 1.2$ | M1 | For use of $\sum x\text{P}(X=x) = 1.2$ |
| $a + 2b + 2c = 1.2$ $\boxed{1}$ | | |
| $\text{P}(Y \geqslant -3) = 0.45$ gives $\text{P}(2-5X \geqslant -3) = 0.45$, i.e. $\text{P}(X \leqslant 1) = 0.45$ | M1 | For use of $\text{P}(Y \geqslant -3) = 0.45$ to set up argument by forming equation in $a$ and $c$ |
| $2a + c = 0.45$ $\boxed{2}$ | | |
| $2a + b + 2c = 1$ $\boxed{3}$ | M1 | For use of $\sum \text{P}(X=x) = 1$ |
| Solving 3 linear equations (matrix or elimination) | M1 | For solving their 3 linear equations |
| $a = 0.1$, $b = 0.3$, $c = 0.25$ | A1, A1 | A1 for any 2 of $a$, $b$ or $c$ correct; A1 for all 3 correct |
| **(7 marks)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(Y) = 75 - (-4)^2$ or $59$ | M1 | For use of $\text{Var}(Y) = \text{E}(Y^2) - [\text{E}(Y)]^2$ (may be implied by correct answer) |
| $[\text{Var}(Y) = 5^2\text{Var}(X)$ implies$]$ $\text{Var}(X) = 2.36$ | A1 | For use of $\text{Var}(aX) = a^2\text{Var}(X)$ to reach 2.36 or exact equivalent |
| **(2 marks)** | | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{P}(Y > X) = \text{P}(2-5X > X) \rightarrow \text{P}(X < \frac{1}{3})$ | M1 | For rearranging to the form $\text{P}(X < k)$ |
| $\text{P}(X < \frac{1}{3}) = a + c = 0.35$ | A1ft | $0.1 + 0.25$; provided their $a$ and $c$ and their $a+c$ are all probabilities |
| **(2 marks)** | | |
---\begin{enumerate}
\item The discrete random variable $X$ has probability distribution given by
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 1 & 0 & 1 & 2 & 3 \\
\hline
$P ( X = x )$ & $c$ & $a$ & $a$ & $b$ & $c$ \\
\hline
\end{tabular}
\end{center}
The random variable $Y = 2 - 5 X$\\
Given that $\mathrm { E } ( \mathrm { Y } ) = - 4$ and $\mathrm { P } ( \mathrm { Y } \geqslant - 3 ) = 0.45$\\
(a) find the probability distribution of X .
Given also that $\mathrm { E } \left( \mathrm { Y } ^ { 2 } \right) = 75$\\
(b) find the exact value of $\operatorname { Var } ( \mathrm { X } )$\\
(c) Find $\mathrm { P } ( \mathrm { Y } > \mathrm { X } )$
\section*{Q uestion 2 continued}
\hfill \mbox{\textit{Edexcel FS1 AS Q2 [11]}}