| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Uniform Distribution |
| Type | Derive general variance formula |
| Difficulty | Standard +0.8 This question requires proving the variance formula for a discrete uniform distribution using summation formulas (∑r and ∑r²), which is a standard Further Maths Statistics derivation. While it involves multiple steps and algebraic manipulation, it's a well-known result that students are expected to be able to derive, making it moderately challenging but not exceptional for Further AS level. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution |
| 5 |
| ||
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = \sum_{x=1}^n \frac{x}{n} = \frac{1}{n}\sum_{x=1}^n x\) | M1 | Applies formula for \(E(X)\) |
| \(= \frac{\frac{n}{2}(1+n)}{n}\) | A1 | Applies formula for \(\sum x\) |
| \(= \frac{n+1}{2}\) | R1 | Shows \(E(X)=\frac{n+1}{2}\); fully correct solution, no slips, condone missing \(E(X)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^2) = \sum_{i=1}^n \frac{x^2}{n} = \frac{1}{n}\sum_{i=1}^n x^2\) | M1 | Applies formula for \(E(X^2)\) |
| \(= \frac{\frac{1}{6}n(n+1)(2n+1)}{n} = \frac{(n+1)(2n+1)}{6}\) | A1 | Applies formula for \(\sum x^2\) |
| \(\text{Var}(X) = E(X^2) - (E(X))^2\) using their \(E(X^2)\) and \(E(X)=\frac{n+1}{2}\) | M1 | Applies variance formula |
| \(= \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 = \frac{4n^2+6n+2-3n^2-6n-3}{12} = \frac{n^2-1}{12}\) | R1 | Shows \(\text{Var}(X)=\frac{n^2-1}{12}\); needs intermediate line of working; fully correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Dice is unbiased / fair / each score has equal probability of occurring | E1 | States dice is unbiased or fair (condone "chance") |
| \(n=6\) or dice is numbered 1, 2, 3, 4, 5, 6 | E1 | States \(n=6\) or dice numbered 1 to 6 |
## Question 5(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \sum_{x=1}^n \frac{x}{n} = \frac{1}{n}\sum_{x=1}^n x$ | M1 | Applies formula for $E(X)$ |
| $= \frac{\frac{n}{2}(1+n)}{n}$ | A1 | Applies formula for $\sum x$ |
| $= \frac{n+1}{2}$ | R1 | Shows $E(X)=\frac{n+1}{2}$; fully correct solution, no slips, condone missing $E(X)$ |
## Question 5(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^2) = \sum_{i=1}^n \frac{x^2}{n} = \frac{1}{n}\sum_{i=1}^n x^2$ | M1 | Applies formula for $E(X^2)$ |
| $= \frac{\frac{1}{6}n(n+1)(2n+1)}{n} = \frac{(n+1)(2n+1)}{6}$ | A1 | Applies formula for $\sum x^2$ |
| $\text{Var}(X) = E(X^2) - (E(X))^2$ using their $E(X^2)$ and $E(X)=\frac{n+1}{2}$ | M1 | Applies variance formula |
| $= \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 = \frac{4n^2+6n+2-3n^2-6n-3}{12} = \frac{n^2-1}{12}$ | R1 | Shows $\text{Var}(X)=\frac{n^2-1}{12}$; needs intermediate line of working; fully correct |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Dice is unbiased / fair / each score has equal probability of occurring | E1 | States dice is unbiased or fair (condone "chance") |
| $n=6$ or dice is numbered 1, 2, 3, 4, 5, 6 | E1 | States $n=6$ or dice numbered 1 to 6 |5 The discrete random variable $X$ has the following probability distribution function
$$\mathrm { P } ( X = x ) = \begin{cases} \frac { 1 } { n } & x = 1,2 , \ldots , n \\ 0 & \text { otherwise } \end{cases}$$
5
\begin{enumerate}[label=(\alph*)]
\item (i) Prove that $\mathrm { E } ( X ) = \frac { n + 1 } { 2 }$\\[0pt]
[3 marks]\\
5 (a) (ii) Prove that $\operatorname { Var } ( X ) = \frac { n ^ { 2 } - 1 } { 12 }$\\
\begin{center}
\begin{tabular}{|l|l|}
\hline
5
\item & \begin{tabular}{l}
State two conditions under which a discrete uniform distribution can be used to model the score when a cubic dice is rolled. \\[0pt]
[2 marks] \\
\end{tabular} \\
\hline
& \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2019 Q5 [9]}}