| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from discrete frequency table |
| Difficulty | Easy -1.2 This is a straightforward calculation question requiring standard formulas for mean, variance, and standard deviation from a discrete probability distribution, followed by a simple probability lookup. The steps are routine: calculate E(Y), calculate E(Y²), find variance, take square root, then identify which values satisfy the inequality and sum their probabilities. No problem-solving insight or novel techniques required—pure mechanical application of AS-level statistics formulas. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| \(y\) | 15 | 21 | 36 | 43 |
| \(\mathrm { P } ( Y = y )\) | 0.16 | 0.32 | 0.29 | 0.23 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y) = 15 \times 0.16 + 21 \times 0.32 + 36 \times 0.29 + 43 \times 0.23 = 29.45\) | M1 | Uses correct formula for \(E(Y)\) or \(E(Y^2)\); PI by sight of 29.45 or 978.23 |
| \(E(Y^2) = 15^2 \times 0.16 + 21^2 \times 0.32 + 36^2 \times 0.29 + 43^2 \times 0.23 = 978.23\) | A1 | Obtains correct value of \(E(Y)\) or \(E(Y^2)\); PI by correct variance |
| \(s = \sqrt{978.23 - 29.45^2}\), \(s^2 = 110.9275\) | M1 | Uses correct formula for \(s\) or \(s^2\) with their values |
| \(s = 10.53\) correct to two decimal places | R1 | Must see \(s =\) AWRT 10.532 before final rounded answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m = 36\) | B1 | Obtains \(m = 36\) |
| \(P(Y > 36 - 1.5 \times 10.53) = P(Y > 20.2) = P(Y \geq 21)\) | M1 | Obtains correct value of their \(m - 1.5 \times 10.53\); may use more accurate value of \(s\) |
| \(= 0.32 + 0.29 + 0.23 = 0.84\) | A1 | CSO |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y) = 15 \times 0.16 + 21 \times 0.32 + 36 \times 0.29 + 43 \times 0.23 = 29.45$ | M1 | Uses correct formula for $E(Y)$ or $E(Y^2)$; PI by sight of 29.45 or 978.23 |
| $E(Y^2) = 15^2 \times 0.16 + 21^2 \times 0.32 + 36^2 \times 0.29 + 43^2 \times 0.23 = 978.23$ | A1 | Obtains correct value of $E(Y)$ or $E(Y^2)$; PI by correct variance |
| $s = \sqrt{978.23 - 29.45^2}$, $s^2 = 110.9275$ | M1 | Uses correct formula for $s$ or $s^2$ with their values |
| $s = 10.53$ correct to two decimal places | R1 | Must see $s =$ **AWRT** 10.532 before final rounded answer |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = 36$ | B1 | Obtains $m = 36$ |
| $P(Y > 36 - 1.5 \times 10.53) = P(Y > 20.2) = P(Y \geq 21)$ | M1 | Obtains correct value of their $m - 1.5 \times 10.53$; may use more accurate value of $s$ |
| $= 0.32 + 0.29 + 0.23 = 0.84$ | A1 | **CSO** |4 The discrete random variable $Y$ has probability distribution
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$y$ & 15 & 21 & 36 & 43 \\
\hline
$\mathrm { P } ( Y = y )$ & 0.16 & 0.32 & 0.29 & 0.23 \\
\hline
\end{tabular}
\end{center}
The standard deviation of $Y$ is $s$
4
\begin{enumerate}[label=(\alph*)]
\item Show that $s = 10.53$ correct to two decimal places.\\[0pt]
[4 marks]\\
4
\item The median of $Y$ is $m$
Find $\mathrm { P } ( Y > m - 1.5 s )$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2024 Q4 [7]}}