| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Derive or identify E(aX+b) or Var(aX+b) formulas |
| Difficulty | Standard +0.8 This question requires computing E(1/X) and Var(1/X) using integration with a non-linear transformation (reciprocal), then applying linearity properties. The integration of x^(-1) and x^(-2) weighted by 3x^2 is straightforward but non-routine. Part (b) requires recognizing that (5+2X)/X = 5/X + 2 and applying transformation formulas. This goes beyond standard textbook exercises on E(aX+b) by involving a reciprocal transformation and multi-step reasoning, placing it moderately above average difficulty. |
| Spec | 5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E\!\left(\frac{1}{X}\right) = \int_0^1 \frac{1}{x} 3x^2\,dx = \int_0^1 3x\,dx\) | M1 | |
| \(= \left[\frac{3x^2}{2}\right]_0^1 = 1.5\) | A1, A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E\!\left(\frac{1}{X^2}\right) = \int_0^1 \frac{1}{x^2} 3x^2\,dx = \int_0^1 3\,dx\) | M1 | |
| \(= [3x]_0^1 = 3.0\) | A1 | |
| \(\text{Var}\!\left(\frac{1}{X}\right) = 3.0 - (1.5)^2 = 0.75\) | m1, \(A1\checkmark\) | dependent on previous M1; [on their (i)] and Var \(> 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E\!\left(\frac{5+2X}{X}\right) = E\!\left(\frac{5}{X}+2\right)\) | M1 | |
| \(= 5E\!\left(\frac{1}{X}\right) + 2 = 5 \times 1.5 + 2 = 9.5\) | M1, A1 | CAO |
| \(\text{Var}\!\left(\frac{5+2X}{X}\right) = \text{Var}\!\left(\frac{5}{X}+2\right) = 25 \times \text{Var}\!\left(\frac{1}{X}\right)\) | M1 | |
| \(= 25 \times 0.75 = 18.75\) | A1 | CAO |
# Question 6(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E\!\left(\frac{1}{X}\right) = \int_0^1 \frac{1}{x} 3x^2\,dx = \int_0^1 3x\,dx$ | M1 | |
| $= \left[\frac{3x^2}{2}\right]_0^1 = 1.5$ | A1, A1 | CAO |
# Question 6(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E\!\left(\frac{1}{X^2}\right) = \int_0^1 \frac{1}{x^2} 3x^2\,dx = \int_0^1 3\,dx$ | M1 | |
| $= [3x]_0^1 = 3.0$ | A1 | |
| $\text{Var}\!\left(\frac{1}{X}\right) = 3.0 - (1.5)^2 = 0.75$ | m1, $A1\checkmark$ | dependent on previous M1; [on their (i)] and Var $> 0$ |
# Question 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E\!\left(\frac{5+2X}{X}\right) = E\!\left(\frac{5}{X}+2\right)$ | M1 | |
| $= 5E\!\left(\frac{1}{X}\right) + 2 = 5 \times 1.5 + 2 = 9.5$ | M1, A1 | CAO |
| $\text{Var}\!\left(\frac{5+2X}{X}\right) = \text{Var}\!\left(\frac{5}{X}+2\right) = 25 \times \text{Var}\!\left(\frac{1}{X}\right)$ | M1 | |
| $= 25 \times 0.75 = 18.75$ | A1 | CAO |6 The continuous random variable $X$ has the probability density function given by
$$f ( x ) = \left\{ \begin{array} { c c }
3 x ^ { 2 } & 0 < x \leqslant 1 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Determine:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { E } \left( \frac { 1 } { X } \right)$;\\
(3 marks)
\item $\operatorname { Var } \left( \frac { 1 } { X } \right)$.
\end{enumerate}\item Hence, or otherwise, find the mean and the variance of $\left( \frac { 5 + 2 X } { X } \right)$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2007 Q6 [12]}}