| 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 PDF transformations and symmetry properties. Parts (a) and (b) require applying standard linear transformations to sketch PDFs—routine for S2 students. Part (c) involves using symmetry to set up an integral equation and algebraic manipulation, while (d) is simple verification by substitution. The multi-part structure and mark allocation suggest moderate length, but each component uses standard techniques without requiring novel insight or complex problem-solving. |
| 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 |
| \(\dfrac{3}{32}\displaystyle\int_{1+a}^{3}(3+2x-x^2)\,dx = \dfrac{1}{4}\) or \(\dfrac{3}{32}\displaystyle\int_{1-a}^{1+a}(3+2x-x^2)\,dx = \dfrac{1}{2}\) | M1 | OE. Attempt to integrate \(f(x)\) and correct limits with correct RHS. |
| \(\dfrac{3}{32}\Big[3x+x^2-\dfrac{x^3}{3}\Big]_{1+a}^{3} = \dfrac{1}{4}\) or \(\dfrac{3}{32}\Big[3x+x^2-\dfrac{x^3}{3}\Big]_{1-a}^{1+a} = \dfrac{1}{2}\) | 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) \(> 0\) \(\quad 0.70^3 - 12\times0.70 + 8 = -0.057\) (2 sf) \(< 0\) \(\quad\) Hence \(0.69 < a < 0.70\) | 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. |
## 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 |
|--------|-------|----------|
| $\dfrac{3}{32}\displaystyle\int_{1+a}^{3}(3+2x-x^2)\,dx = \dfrac{1}{4}$ **or** $\dfrac{3}{32}\displaystyle\int_{1-a}^{1+a}(3+2x-x^2)\,dx = \dfrac{1}{2}$ | M1 | OE. Attempt to integrate $f(x)$ and correct limits with correct RHS. |
| $\dfrac{3}{32}\Big[3x+x^2-\dfrac{x^3}{3}\Big]_{1+a}^{3} = \dfrac{1}{4}$ **or** $\dfrac{3}{32}\Big[3x+x^2-\dfrac{x^3}{3}\Big]_{1-a}^{1+a} = \dfrac{1}{2}$ | 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) $> 0$ $\quad 0.70^3 - 12\times0.70 + 8 = -0.057$ (2 sf) $< 0$ $\quad$ Hence $0.69 < a < 0.70$ | 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. |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{f45a7c6f-ebb4-43a2-8751-11aab3561d3c-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]{f45a7c6f-ebb4-43a2-8751-11aab3561d3c-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]{f45a7c6f-ebb4-43a2-8751-11aab3561d3c-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]}}