| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | March |
| Marks | 8 |
| 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 standard PDF properties with symmetry being the key insight for part (a), routine integration for finding k in (b)(i), and standard variance calculation in (b)(ii). The symmetry argument requires some thought but is a common S2 technique, while the integrations are straightforward polynomial expansions. Slightly above average due to the multi-step nature and need to recognize symmetry, but all techniques are standard A-level statistics. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - 2(a+b)\) or \(1-2a\) or \(0.5-a-b\) or \(1-(a+b)\) or \(a+a+b\) | M1 | OE. Seen or implied – may be on the diagram (or for correct un-simplified final expression). |
| \(P(0.6 \leq X \leq 1.8) = 1 - 2a - b\) | A1 | Accept \(1-(2a+b)\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k\int_0^3 (9x^2 - 6x^3 + x^4)\,dx = 1\) | M1 | Attempt integrate \(f(x)\) ignore limits and \(= 1\). |
| \(k\left[\frac{9x^3}{3} - \frac{6x^4}{4} + \frac{x^5}{5}\right]_0^3 = 1\) | A1 | Correct integration seen, correct limits. |
| \(k \times \frac{81}{10} = 1,\ k = \frac{10}{81}\) | A1 | AG. Convincingly obtained. No errors seen. (Must see integration). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{10}{81}\int_0^3 (9x^4 - 6x^5 + x^6)\,dx\) | M1 | Attempt integrate \(x^2 f(x)\) between 0 and 3, condone missing \(k\). Must see integration or correct answer of 18/7 seen or implied. |
| \(\left[\frac{10}{81}\left(\frac{9x^5}{5} - x^6 + \frac{x^7}{7}\right)\right]_0^3 = \frac{18}{7}\) or \(2.57\ldots\) | ||
| \(\frac{18}{7} - 1.5^2\) | M1 | Their integral of \(x^2 f(x) - 1.5^2\) (or their mean\(^2\)). |
| \(= \frac{9}{28}\) or \(0.321\) | A1 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - 2(a+b)$ or $1-2a$ or $0.5-a-b$ or $1-(a+b)$ or $a+a+b$ | M1 | OE. Seen or implied – may be on the diagram (or for correct un-simplified final expression). |
| $P(0.6 \leq X \leq 1.8) = 1 - 2a - b$ | A1 | Accept $1-(2a+b)$. |
## Question 3(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k\int_0^3 (9x^2 - 6x^3 + x^4)\,dx = 1$ | M1 | Attempt integrate $f(x)$ ignore limits and $= 1$. |
| $k\left[\frac{9x^3}{3} - \frac{6x^4}{4} + \frac{x^5}{5}\right]_0^3 = 1$ | A1 | Correct integration seen, correct limits. |
| $k \times \frac{81}{10} = 1,\ k = \frac{10}{81}$ | A1 | AG. Convincingly obtained. No errors seen. (Must see integration). |
## Question 3(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{10}{81}\int_0^3 (9x^4 - 6x^5 + x^6)\,dx$ | M1 | Attempt integrate $x^2 f(x)$ between 0 and 3, condone missing $k$. Must see integration or correct answer of 18/7 seen or implied. |
| $\left[\frac{10}{81}\left(\frac{9x^5}{5} - x^6 + \frac{x^7}{7}\right)\right]_0^3 = \frac{18}{7}$ or $2.57\ldots$ | | |
| $\frac{18}{7} - 1.5^2$ | M1 | Their integral of $x^2 f(x) - 1.5^2$ (or their mean$^2$). |
| $= \frac{9}{28}$ or $0.321$ | A1 | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{823cc2e5-e408-4b81-ac4d-7e9f584107cc-06_558_1077_260_523}
The diagram shows the graph of the probability density function, f, of a random variable $X$ that takes values between $x = 0$ and $x = 3$ only. The graph is symmetrical about the line $x = 1.5$.
\begin{enumerate}[label=(\alph*)]
\item It is given that $\mathrm { P } ( X < 0.6 ) = a$ and $\mathrm { P } ( 0.6 < X < 1.2 ) = b$.
Find $\mathrm { P } ( 0.6 < X < 1.8 )$ in terms of $a$ and $b$.
\item It is now given that the equation of the probability density function of $X$ is
$$f ( x ) = \begin{cases} k x ^ { 2 } ( 3 - x ) ^ { 2 } & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Show that $k = \frac { 10 } { 81 }$.
\item Find $\operatorname { Var } ( X )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q3 [8]}}