| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| 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 basic properties of PDFs including recognizing symmetry and using standard expectation formulas. Part (a) requires observing that X is non-negative (so P(X≥0)=1), W is symmetric about 0 (so P(W≥0)=0.5), and that g is essentially f extended symmetrically (so q=p/2). Part (b) requires integrating to find E(X) and solving for a. All steps are straightforward applications of standard S2 concepts with no novel problem-solving required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 1 | B1 | No ambiguity |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{1}{2}\) | B1 | No ambiguity |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([q =] \dfrac{1}{2}p\) | B1 | Accept \(2q = p\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(p\displaystyle\int_0^a (a^2 - x^2)\,dx = 1\) | M1 | Attempt to integrate \(f(x)\) and equated to 1 |
| \(\dfrac{2}{3}a^3 p = 1\) | A1 | OE, simplified |
| \(\text{"}\dfrac{3}{2a^3}\text{"}\displaystyle\int_0^a (a^2x - x^3)\,dx = 3\) or \(\text{"}\dfrac{3}{2a^3}\text{"}\displaystyle\int_0^a (a^2x - x^3)\,dx = 3\) | M1 | Attempt to integrate \(xf(x)\), with multiplier \(p\) or \(\dfrac{3}{2a^3}\) or *their* \(p\), and equate to 3 |
| \(p \times \dfrac{a^4}{4} = 3\) | A1 | May be implied by next line |
| \(\text{"}\dfrac{3}{2a^3}\text{"} \times \dfrac{a^4}{4} = 3\) | M1 | OE. Substitute from one equation into the other; FT *their* equations |
| \(a = 8\) | A1 |
## Question 7(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 1 | B1 | No ambiguity |
---
## Question 7(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{1}{2}$ | B1 | No ambiguity |
---
## Question 7(a)(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[q =] \dfrac{1}{2}p$ | B1 | Accept $2q = p$ |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $p\displaystyle\int_0^a (a^2 - x^2)\,dx = 1$ | M1 | Attempt to integrate $f(x)$ and equated to 1 |
| $\dfrac{2}{3}a^3 p = 1$ | A1 | OE, simplified |
| $\text{"}\dfrac{3}{2a^3}\text{"}\displaystyle\int_0^a (a^2x - x^3)\,dx = 3$ or $\text{"}\dfrac{3}{2a^3}\text{"}\displaystyle\int_0^a (a^2x - x^3)\,dx = 3$ | M1 | Attempt to integrate $xf(x)$, with multiplier $p$ or $\dfrac{3}{2a^3}$ or *their* $p$, and equate to 3 |
| $p \times \dfrac{a^4}{4} = 3$ | A1 | May be implied by next line |
| $\text{"}\dfrac{3}{2a^3}\text{"} \times \dfrac{a^4}{4} = 3$ | M1 | OE. Substitute from one equation into the other; FT *their* equations |
| $a = 8$ | A1 | |7 The random variables $X$ and $W$ have probability density functions f and g defined as follows:
$$\begin{gathered}
\mathrm { f } ( x ) = \begin{cases} p \left( a ^ { 2 } - x ^ { 2 } \right) & 0 \leqslant x \leqslant a \\
0 & \text { otherwise } \end{cases} \\
\mathrm { g } ( w ) = \begin{cases} q \left( a ^ { 2 } - w ^ { 2 } \right) & - a \leqslant w \leqslant a \\
0 & \text { otherwise } \end{cases}
\end{gathered}$$
where $a , p$ and $q$ are constants.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the value of $\mathrm { P } ( X \geqslant 0 )$.
\item Write down the value of $\mathrm { P } ( W \geqslant 0 )$.
\item Write down an expression for $q$ in terms of $p$ only.
\end{enumerate}\item Given that $\mathrm { E } ( X ) = 3$, find the value of $a$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q7 [9]}}