| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | November |
| 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 is a straightforward S2 probability density function question requiring standard techniques: using symmetry properties, integrating a quadratic pdf to apply the total probability condition, and solving simultaneous equations. All steps are routine applications of A-level methods with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | 1 – p or p – 0.5 | M1 |
| [P(–1 < X < 0) =] 2p – 1 | A1 | Clearly as final answer. |
| Answer | Marks |
|---|---|
| 7(b)(i) | 2 2 |
| Answer | Marks | Guidance |
|---|---|---|
| −3 −3 | M1 | OE Attempt integral, with correct limits and RHS. |
| Answer | Marks | Guidance |
|---|---|---|
| 3 2 −3 2 4 3 −3 | A1 | OE Correct integration. |
| Answer | Marks | Guidance |
|---|---|---|
| 4 – 9b = –0.5 leading to 30a – 55b = 6 AG | A1 | Correctly obtained. No errors seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b)(ii) | a – b(9 – 3) = 0 or a – b(4 + 2) = 0 [hence a – 6b = 0] | *M1 |
| Answer | Marks |
|---|---|
| Attempt to solve 30a – 55b = 6 and their a – 6b = 0 | DM1 |
| Answer | Marks |
|---|---|
| 125 125 | A1 |
Question 7:
--- 7(a) ---
7(a) | 1 – p or p – 0.5 | M1 | SOI, e.g. on diagram.
[P(–1 < X < 0) =] 2p – 1 | A1 | Clearly as final answer.
2
--- 7(b)(i) ---
7(b)(i) | 2 2
(a−b(x2 +x))dx = 1 or (ax−b(x3+x2))dx = –0.5
−3 −3 | M1 | OE Attempt integral, with correct limits and RHS.
x3 x2 2 x2 x4 x3 2
ax−b + (= 1) or a −b + (= –0.5)
3 2 −3 2 4 3 −3 | A1 | OE Correct integration.
2a – 8b/3 – 2b +3a – 9b +9b / 2 = 1 or 2a – 4b – 8b / 3 – 9a / 2 + 81b /
4 – 9b = –0.5 leading to 30a – 55b = 6 AG | A1 | Correctly obtained. No errors seen.
3
Question | Answer | Marks | Guidance
--- 7(b)(ii) ---
7(b)(ii) | a – b(9 – 3) = 0 or a – b(4 + 2) = 0 [hence a – 6b = 0] | *M1 | Use f(–3) = 0 or f(2) = 0.
Further attempts at integration M0.
Attempt to solve 30a – 55b = 6 and their a – 6b = 0 | DM1
36 6
a = or 0.288 b = or 0.048
125 125 | A1
3\includegraphics{figure_7}
The diagram shows the graph of the probability density function, f, of a random variable $X$ which takes values between $-3$ and 2 only.
\begin{enumerate}[label=(\alph*)]
\item Given that the graph is symmetrical about the line $x = -0.5$ and that P($X < 0$) = $p$, find P($-1 < X < 0$) in terms of $p$. [2]
\item It is now given that the probability density function shown in the diagram is given by
$$\text{f}(x) = \begin{cases}
a - b(x^2 + x) & -3 \leq x \leq 2, \\
0 & \text{otherwise,}
\end{cases}$$
where $a$ and $b$ are positive constants.
\begin{enumerate}[label=(\roman*)]
\item Show that $30a - 55b = 6$. [3]
\item By substituting a suitable value of $x$ into f($x$), find another equation relating $a$ and $b$ and hence determine the values of $a$ and $b$. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q7 [8]}}