| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Variance of transformed variable |
| Difficulty | Challenging +1.2 This is a multi-part question on continuous probability distributions requiring standard techniques: integration for probability (part a), solving for quartiles (part b), and applying variance transformation rules (part c). While part (c) involves the non-linear transformation Var(44X^{-3}) = 44²Var(X^{-3}) requiring E(X^{-6}) and E(X^{-3}), these are routine integrations for Further Maths students. The question is more computational than conceptual, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > 2) = \int_2^5 \dfrac{3x}{44} + \dfrac{1}{22} \, dx\) | M1 | Condone missing \(dx\); PI by sight of AWRT 0.852 or 0.148 |
| \(= \dfrac{75}{88}\) | A1 | AWRT 0.852 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_1^q \dfrac{3x}{44} + \dfrac{1}{22} \, dx = \dfrac{3}{4}\) | M1 | Integrates a multiple of \(\dfrac{3x}{44} + \dfrac{1}{22}\) to form of \(ax^2 + bx\) |
| \(\left[\dfrac{3x^2}{88} + \dfrac{x}{22}\right]_1^q = \dfrac{3}{4}\) | A1 | Integrates \(k\!\left(\dfrac{3x}{44}+\dfrac{1}{22}\right)\) to obtain \(k\!\left(\dfrac{3x^2}{88}+\dfrac{x}{22}\right)\) |
| \(\dfrac{3q^2}{88} + \dfrac{q}{22} - \dfrac{3}{88} - \dfrac{1}{22} = \dfrac{3}{4}\), leading to \(\dfrac{3q^2}{88} + \dfrac{q}{22} - \dfrac{73}{88} = 0\), \(q = 4.31\) | M1 | Substitutes limits \(q\) and 1, subtracts and sets equal to \(\dfrac{3}{4}\) to form quadratic in \(q\); oe using limits 5 and \(q\) set equal to \(\dfrac{1}{4}\) |
| AWRT 4.31 (reject \(-5.64\)) | A1 | If \(-5.64\) found it must be rejected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(44X^{-3}) = \int_1^5 3x^{-2} + 2x^{-3}\, dx\) | M1 (1.1a) | Uses correct formula for \(kE(X^{-3})\); condone mislabelling |
| \(E(44X^{-3}) = \frac{84}{25}\) | ||
| \(E(44^2X^{-6}) = \int_1^5 132x^{-5} + 88x^{-6}\, dx\) | M1 (1.1a) | Uses correct formula for \(kE(X^{-6})\); condone mislabelling |
| \(E(44^2X^{-6}) = 50.541568\) | ||
| \(E(X^{-3}) = \frac{21}{275}\) oe or AWRT 0.076, or \(E(44X^{-3}) = \frac{84}{25}\) oe, and \(E(X^{-6})\) AWRT 0.026 or \(E(44^2X^{-6})\) AWRT 50.54 | A1 (1.1b) | PI |
| \(\text{Var}(44X^{-3}) = E(44^2X^{-6}) - (E(44X^{-3}))^2\) | M1 (1.1a) | Uses \(\text{Var}(44X^{-3}) = E(44^2X^{-6}) - (E(44X^{-3}))^2\) oe; condone mislabelling; PI |
| \(\text{Var}(44X^{-3}) = 50.541568 - \left(\frac{84}{25}\right)^2\) | ||
| \(\text{Var}(44X^{-3}) = 39.252\) | A1 (1.1b) | AWRT 39.252 |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 2) = \int_2^5 \dfrac{3x}{44} + \dfrac{1}{22} \, dx$ | M1 | Condone missing $dx$; PI by sight of **AWRT** 0.852 or 0.148 |
| $= \dfrac{75}{88}$ | A1 | **AWRT** 0.852 |
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_1^q \dfrac{3x}{44} + \dfrac{1}{22} \, dx = \dfrac{3}{4}$ | M1 | Integrates a multiple of $\dfrac{3x}{44} + \dfrac{1}{22}$ to form of $ax^2 + bx$ |
| $\left[\dfrac{3x^2}{88} + \dfrac{x}{22}\right]_1^q = \dfrac{3}{4}$ | A1 | Integrates $k\!\left(\dfrac{3x}{44}+\dfrac{1}{22}\right)$ to obtain $k\!\left(\dfrac{3x^2}{88}+\dfrac{x}{22}\right)$ |
| $\dfrac{3q^2}{88} + \dfrac{q}{22} - \dfrac{3}{88} - \dfrac{1}{22} = \dfrac{3}{4}$, leading to $\dfrac{3q^2}{88} + \dfrac{q}{22} - \dfrac{73}{88} = 0$, $q = 4.31$ | M1 | Substitutes limits $q$ and 1, subtracts and sets equal to $\dfrac{3}{4}$ to form quadratic in $q$; oe using limits 5 and $q$ set equal to $\dfrac{1}{4}$ |
| **AWRT** 4.31 (reject $-5.64$) | A1 | If $-5.64$ found it must be rejected |
# Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(44X^{-3}) = \int_1^5 3x^{-2} + 2x^{-3}\, dx$ | M1 (1.1a) | Uses correct formula for $kE(X^{-3})$; condone mislabelling |
| $E(44X^{-3}) = \frac{84}{25}$ | | |
| $E(44^2X^{-6}) = \int_1^5 132x^{-5} + 88x^{-6}\, dx$ | M1 (1.1a) | Uses correct formula for $kE(X^{-6})$; condone mislabelling |
| $E(44^2X^{-6}) = 50.541568$ | | |
| $E(X^{-3}) = \frac{21}{275}$ **oe** or **AWRT** 0.076, or $E(44X^{-3}) = \frac{84}{25}$ **oe**, and $E(X^{-6})$ **AWRT** 0.026 or $E(44^2X^{-6})$ **AWRT** 50.54 | A1 (1.1b) | PI |
| $\text{Var}(44X^{-3}) = E(44^2X^{-6}) - (E(44X^{-3}))^2$ | M1 (1.1a) | Uses $\text{Var}(44X^{-3}) = E(44^2X^{-6}) - (E(44X^{-3}))^2$ **oe**; condone mislabelling; PI |
| $\text{Var}(44X^{-3}) = 50.541568 - \left(\frac{84}{25}\right)^2$ | | |
| $\text{Var}(44X^{-3}) = 39.252$ | A1 (1.1b) | **AWRT** 39.252 |
---6 The continuous random variable $X$ has probability density function
$$f ( x ) = \begin{cases} \frac { 3 x } { 44 } + \frac { 1 } { 22 } & 1 \leq x \leq 5 \\ 0 & \text { otherwise } \end{cases}$$
6
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 2 )$\\[0pt]
[2 marks]\\
6
\item Find the upper quartile of $X$
Give your answer to two decimal places.\\
6
\item Find $\operatorname { Var } \left( 44 X ^ { - 3 } \right)$
Give your answer to three decimal places.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2024 Q6 [11]}}