| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Symmetry property of PDF |
| Difficulty | Standard +0.3 This question tests understanding of transformations of random variables and their effects on PDFs, plus some algebraic manipulation. Parts (a) and (b) require applying standard results about linear transformations (E(aX+b) and Var(aX+b)), which is routine A-level material. Part (c) involves setting up an integral using symmetry and simplifying to the required cubic equation—straightforward calculus. Part (d) is simple verification by substitution. The multi-part structure adds length but each component uses standard techniques without requiring novel insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Curve of similar shape, \(x = 0\) to \(x = 4\), with highest point \((2, 0.375)\) | B1 | Not straight lines, not bell shaped. Must be correct at \(x=0\) and \(x=4\), highest point must be at \(x=2\), \(y\) value \(\pm \frac{1}{4}\) square. Must not go below the x-axis. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Curve of similar shape, from \(x = 0\) to \(x = 2\), highest point at \(x = 1\) | B1 | Not straight lines, not bell shaped. Must be correct at \(x=0\) and \(x=2\). Highest point must be at \(x=1\). |
| Highest point \((1, 0.75)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{32}\int_{1+a}^{3}(3+2x-x^2)dx = \frac{1}{4}\) or \(\frac{3}{32}\int_{1-a}^{1+a}(3+2x-x^2)dx = \frac{1}{2}\) | M1 | OE Attempt to integrate \(f(x)\) and correct limits with correct RHS. |
| \(\frac{3}{32}\Big[3x+x^2-\frac{x^3}{3}\Big]_{1+a}^{3} = \frac{1}{4}\) or \(\frac{3}{32}\Big[3x+x^2-\frac{x^3}{3}\Big]_{1-a}^{1+a} = \frac{1}{2}\) | A1 | Correct integration. |
| \(a^3 - 12a + 8 = 0\) | A1 | AG Substitute limits and correctly obtain equation. May see \(3/32(6a+4a-6a/3-2a^3/3) = 0.5\). No errors seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.69^3 - 12\times0.69 + 8 = 0.049\) (2 sf) \(> 0\) | B1 | AG Must state either the correct expression and \(> 0\) and \(< 0\) or both answers to 2 sf. Both answers correct and conclusion. Accept equivalent expressions. OR: \(a = 0.695\) (3 sf) which is between 0.69 & 0.70. |
| \(0.70^3 - 12\times0.70 + 8 = -0.057\) (2 sf) \(< 0\) | ||
| Hence \(0.69 < a < 0.70\) |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Curve of similar shape, $x = 0$ to $x = 4$, with highest point $(2, 0.375)$ | B1 | Not straight lines, not bell shaped. Must be correct at $x=0$ and $x=4$, highest point must be at $x=2$, $y$ value $\pm \frac{1}{4}$ square. Must not go below the x-axis. |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Curve of similar shape, from $x = 0$ to $x = 2$, highest point at $x = 1$ | B1 | Not straight lines, not bell shaped. Must be correct at $x=0$ and $x=2$. Highest point must be at $x=1$. |
| Highest point $(1, 0.75)$ | B1 | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{32}\int_{1+a}^{3}(3+2x-x^2)dx = \frac{1}{4}$ **or** $\frac{3}{32}\int_{1-a}^{1+a}(3+2x-x^2)dx = \frac{1}{2}$ | M1 | OE Attempt to integrate $f(x)$ and correct limits with correct RHS. |
| $\frac{3}{32}\Big[3x+x^2-\frac{x^3}{3}\Big]_{1+a}^{3} = \frac{1}{4}$ or $\frac{3}{32}\Big[3x+x^2-\frac{x^3}{3}\Big]_{1-a}^{1+a} = \frac{1}{2}$ | A1 | Correct integration. |
| $a^3 - 12a + 8 = 0$ | A1 | AG Substitute limits and correctly obtain equation. May see $3/32(6a+4a-6a/3-2a^3/3) = 0.5$. No errors seen. |
---
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.69^3 - 12\times0.69 + 8 = 0.049$ (2 sf) $> 0$ | B1 | AG Must state either the correct expression and $> 0$ and $< 0$ or both answers to 2 sf. Both answers correct and conclusion. Accept equivalent expressions. OR: $a = 0.695$ (3 sf) which is between 0.69 & 0.70. |
| $0.70^3 - 12\times0.70 + 8 = -0.057$ (2 sf) $< 0$ | | |
| Hence $0.69 < a < 0.70$ | | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{8f9e5f25-05c1-4a4e-9094-2a1b42416588-08_490_1195_255_475}
The diagram shows the graph of the probability density function of a random variable $X$ that takes values between - 1 and 3 only. It is given that the graph is symmetrical about the line $x = 1$. Between $x = - 1$ and $x = 3$ the graph is a quadratic curve.
The random variable $S$ is such that $\mathrm { E } ( S ) = 2 \times \mathrm { E } ( X )$ and $\operatorname { Var } ( S ) = \operatorname { Var } ( X )$.
\begin{enumerate}[label=(\alph*)]
\item On the grid below, sketch a quadratic graph for the probability density function of $S$.\\
\includegraphics[max width=\textwidth, alt={}, center]{8f9e5f25-05c1-4a4e-9094-2a1b42416588-08_490_1191_1169_479}
The random variable $T$ is such that $\mathrm { E } ( T ) = \mathrm { E } ( X )$ and $\operatorname { Var } ( T ) = \frac { 1 } { 4 } \operatorname { Var } ( X )$.
\item On the grid below, sketch a quadratic graph for the probability density function of $T$.\\
\includegraphics[max width=\textwidth, alt={}, center]{8f9e5f25-05c1-4a4e-9094-2a1b42416588-08_488_1187_1996_479}
It is now given that
$$f ( x ) = \begin{cases} \frac { 3 } { 32 } \left( 3 + 2 x - x ^ { 2 } \right) & - 1 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
\item Given that $\mathrm { P } ( 1 - a < X < 1 + a ) = 0.5$, show that $a ^ { 3 } - 12 a + 8 = 0$.
\item Hence verify that $0.69 < a < 0.70$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q6 [7]}}