| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Sum of independent uniforms |
| Difficulty | Easy -1.2 This question tests basic properties of uniform and discrete distributions with straightforward calculations. Part (a) is trivial probability on U(0,10), parts (b)(i) and (b)(ii) require only direct application of E(X+Y) = E(X) + E(Y) and Var(X+Y) = Var(X) + Var(Y) for independent variables—all standard bookwork with no problem-solving or insight required. |
| Spec | 5.02b Expectation and variance: discrete random variables5.03a Continuous random variables: pdf and cdf5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > 4) = 0.6\) | B1 | Uses rectangular distribution model to correctly find \(P(X>4) = 0.6\); oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = 5\) | B1 | Uses rectangular distribution model to obtain \(E(X) = 5\); condone \(E(X) = \frac{9+1}{2} = 5\) |
| \(E(Y) = 3.5\) | M1 | Translates situation for round 2 into discrete uniform distribution model to obtain \(E(Y) = 3.5\); oe |
| Mean total score \(= 5 + 3.5 = 8.5\) | A1F | Obtains mean total score \(= 8.5\); oe; FT their \(E(X)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(X) = \frac{(10-0)^2}{12} = \frac{100}{12}\) | M1 | Substitutes into correct formula for \(\text{Var}(X)\) or \(\text{Var}(Y)\); PI by correct \(\text{Var}(X)\) or \(\text{Var}(Y)\) |
| \(\text{Var}(Y) = \frac{6^2 - 1}{12} = \frac{35}{12}\) | A1 | Obtains correct \(\text{Var}(X)\) and \(\text{Var}(Y)\) (may be unsimplified) |
| \(\text{Var}(X+Y) = \frac{100}{12} + \frac{35}{12} = 11.25\) | A1F | FT their \(\text{Var}(X)\) and \(\text{Var}(Y)\) with answer correct to at least 2 dp |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 4) = 0.6$ | B1 | Uses rectangular distribution model to correctly find $P(X>4) = 0.6$; oe |
**Subtotal: 1 mark**
---
## Question 6(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = 5$ | B1 | Uses rectangular distribution model to obtain $E(X) = 5$; condone $E(X) = \frac{9+1}{2} = 5$ |
| $E(Y) = 3.5$ | M1 | Translates situation for round 2 into discrete uniform distribution model to obtain $E(Y) = 3.5$; oe |
| Mean total score $= 5 + 3.5 = 8.5$ | A1F | Obtains mean total score $= 8.5$; oe; FT their $E(X)$ |
**Subtotal: 3 marks**
---
## Question 6(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(X) = \frac{(10-0)^2}{12} = \frac{100}{12}$ | M1 | Substitutes into correct formula for $\text{Var}(X)$ or $\text{Var}(Y)$; PI by correct $\text{Var}(X)$ or $\text{Var}(Y)$ |
| $\text{Var}(Y) = \frac{6^2 - 1}{12} = \frac{35}{12}$ | A1 | Obtains correct $\text{Var}(X)$ and $\text{Var}(Y)$ (may be unsimplified) |
| $\text{Var}(X+Y) = \frac{100}{12} + \frac{35}{12} = 11.25$ | A1F | FT their $\text{Var}(X)$ and $\text{Var}(Y)$ with answer correct to at least 2 dp |
**Subtotal: 3 marks**
**Question 6 total: 7 marks**
---6 A game consists of two rounds.
The first round of the game uses a random number generator to output the score $X$, a real number between 0 and 10
6
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 4 )$
6
\item The second round of the game uses an unbiased dice, with faces numbered 1 to 6 , to give the score $Y$
The variables $X$ and $Y$ are independent.\\
6 (b) (i) Find the mean total score of the game.\\
6 (b) (ii) Find the variance of the total score of the game.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2023 Q6 [7]}}