| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Probability distributions with parameters |
| Difficulty | Standard +0.3 Part (a) requires setting up and solving three simultaneous equations using standard probability axioms (sum=1, E(X)=1.2, Var(X)=0.56), which is routine for Further Maths students. Part (b) is a direct application of variance properties. This is a standard textbook exercise with no novel insight required, making it slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a + b + c = 1\) | B1 (AO 3.1a) | Translate into at least one correct equation; \(a+b+c=1\) may be implied by finding \(a\), \(b\), \(c\) which add to \(1\) |
| \(b + 2c = 1.2\) | B1 (AO 1.1b) | At least two correct equations |
| \(b + 4c - 1.2^2 = 0.56 \Rightarrow b + 4c = 2\) | B1 (AO 1.1b) | Three correct equations |
| Attempts to find a value for one of \(a\), \(b\) or \(c\) | M1 (AO 1.1a) | Attempts to solve simultaneous equations |
| One correct value from \(a\), \(b\), \(c\) | A1 (AO 1.1b) | |
| \(a = 0.2,\ b = 0.4,\ c = 0.4\) | A1 (AO 2.2a) | Deduces all three correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(X - 2Y - 11) = \text{Var}(X) + 2^2\text{Var}(Y)\) | M1 (1.1a) | Use the formula \(\text{Var}(X-2Y-11) = \text{Var}(X) + 2^2\text{Var}(Y)\) to obtain \(0.56 + 2^2 \times 15\). Condone one slip leading to either using \(\text{Var}(X) + 2\text{Var}(Y)\) or \(\text{Var}(X) - 2^2\text{Var}(Y)\) |
| \(= 0.56 + 2^2 \times 15 = 60.56\) | A1 (1.1b) | Correct value of \(\text{Var}(X - 2Y - 11)\) |
## Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a + b + c = 1$ | B1 (AO 3.1a) | Translate into at least one correct equation; $a+b+c=1$ may be implied by finding $a$, $b$, $c$ which add to $1$ |
| $b + 2c = 1.2$ | B1 (AO 1.1b) | At least two correct equations |
| $b + 4c - 1.2^2 = 0.56 \Rightarrow b + 4c = 2$ | B1 (AO 1.1b) | Three correct equations |
| Attempts to find a value for one of $a$, $b$ or $c$ | M1 (AO 1.1a) | Attempts to solve simultaneous equations |
| One correct value from $a$, $b$, $c$ | A1 (AO 1.1b) | |
| $a = 0.2,\ b = 0.4,\ c = 0.4$ | A1 (AO 2.2a) | Deduces all three correct values |
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(X - 2Y - 11) = \text{Var}(X) + 2^2\text{Var}(Y)$ | M1 (1.1a) | Use the formula $\text{Var}(X-2Y-11) = \text{Var}(X) + 2^2\text{Var}(Y)$ to obtain $0.56 + 2^2 \times 15$. Condone one slip leading to either using $\text{Var}(X) + 2\text{Var}(Y)$ or $\text{Var}(X) - 2^2\text{Var}(Y)$ |
| $= 0.56 + 2^2 \times 15 = 60.56$ | A1 (1.1b) | Correct value of $\text{Var}(X - 2Y - 11)$ |
---6 The discrete random variable $X$ has probability distribution function
$$\mathrm { P } ( X = x ) = \begin{cases} a & x = 0 \\ b & x = 1 \\ c & x = 2 \\ 0 & \text { otherwise } \end{cases}$$
where $a , b$ and $c$ are constants.\\
The mean of $X$ is 1.2 and the variance of $X$ is 0.56\\
6
\begin{enumerate}[label=(\alph*)]
\item Deduce the values of $a , b$ and $c$\\
6
\item The continuous random variable $Y$ is independent of $X$ and has variance 15
Find $\operatorname { Var } ( X - 2 Y - 11 )$\\[0pt]
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2022 Q6 [8]}}