| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Explain why not valid PDF |
| Difficulty | Moderate -0.8 This is a straightforward S2 question testing basic PDF properties: recognizing that area under PDF must equal 1 (part a), applying this condition to find a constant (part b(i)), and using symmetry for expectation (part b(ii)). The variance calculation requires standard integration but is routine. All parts are textbook exercises requiring recall and direct application of definitions rather than problem-solving or insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k = 1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f_2\): area \(> 1\) (area \(\neq 1\)) | B1 | oe |
| \(f_3\): includes negative values of \(f_3\) | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6\int_{-a}^{a}(a^2 - x^2)\,dx = 1\) | M1 | Integ \(f(x) = 1\), ignore limits |
| \(6\left[a^2x - \frac{x^3}{3}\right]_{-a}^{a} = 1\) | A1 | Correct integral and limits |
| \(6(2a^3 - \frac{2a^3}{3}) = 1\), \(\frac{24a^3}{3} = 1\) or \(8a^3 = 1\), \(a = \frac{1}{2}\) AG | A1 | Correctly obtained. No errors seen. (SR Verification scores M1A1 only max 2/3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6\int_{-0.5}^{0.5}\left(\frac{x^2}{4} - x^4\right)dx\) | M1 | attempt int \(x^2f(x)\) & correct limits |
| \(= 6\left[\frac{x^3}{12} - \frac{x^5}{5}\right]_{-0.5}^{0.5} = 0.05\), \(\text{Var} = 0.05 - 0^2\) | ||
| \(= 0.05\) oe | A1 | cao; allow omission of \(-0^2\) |
## Question 5(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k = 1$ | B1 | |
## Question 5(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f_2$: area $> 1$ (area $\neq 1$) | B1 | oe |
| $f_3$: includes negative values of $f_3$ | B1 | oe |
## Question 5(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6\int_{-a}^{a}(a^2 - x^2)\,dx = 1$ | M1 | Integ $f(x) = 1$, ignore limits |
| $6\left[a^2x - \frac{x^3}{3}\right]_{-a}^{a} = 1$ | A1 | Correct integral and limits |
| $6(2a^3 - \frac{2a^3}{3}) = 1$, $\frac{24a^3}{3} = 1$ or $8a^3 = 1$, $a = \frac{1}{2}$ **AG** | A1 | Correctly obtained. No errors seen. (SR Verification scores M1A1 only max 2/3) |
## Question 5(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0$ | B1 | |
## Question 5(b)(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6\int_{-0.5}^{0.5}\left(\frac{x^2}{4} - x^4\right)dx$ | M1 | attempt int $x^2f(x)$ & correct limits |
| $= 6\left[\frac{x^3}{12} - \frac{x^5}{5}\right]_{-0.5}^{0.5} = 0.05$, $\text{Var} = 0.05 - 0^2$ | | |
| $= 0.05$ oe | A1 | cao; allow omission of $-0^2$ |
---5
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{61ba010c-d6a2-4c19-9998-0ae048244a32-06_292_517_264_338}\\
\includegraphics[max width=\textwidth, alt={}, center]{61ba010c-d6a2-4c19-9998-0ae048244a32-06_289_518_264_858}\\
\includegraphics[max width=\textwidth, alt={}, center]{61ba010c-d6a2-4c19-9998-0ae048244a32-06_273_510_365_1377}
The diagram shows the graphs of three functions, $f _ { 1 } , f _ { 2 }$ and $f _ { 3 }$. The function $f _ { 1 }$ is a probability density function.
\begin{enumerate}[label=(\roman*)]
\item State the value of $k$.
\item For each of the functions $\mathrm { f } _ { 2 }$ and $\mathrm { f } _ { 3 }$, state why it cannot be a probability density function.
\end{enumerate}\item The probability density function g is defined by
$$g ( x ) = \begin{cases} 6 \left( a ^ { 2 } - x ^ { 2 } \right) & - a \leqslant x \leqslant a \\ 0 & \text { otherwise } \end{cases}$$
where $a$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Show that $a = \frac { 1 } { 2 }$.
\item State the value of $\mathrm { E } ( X )$.
\item Find $\operatorname { Var } ( X )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2017 Q5 [9]}}