| 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 | Find median or percentiles |
| Difficulty | Standard +0.3 This is a straightforward continuous probability distribution question requiring standard techniques: integration for expectation (given as 'show that'), variance calculation using E(X²) - [E(X)]², and using the median property that P(X < m) = 0.5. All steps are routine applications of formulas with no conceptual challenges, making it slightly easier than average. |
| 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 |
| \(\frac{3}{16}\int_2^4 (4x^2 - x^3)\,dx\) | M1 | Attempt to integrate \(xf(x)\); ignore limits (must see a power increase for attempted integration) |
| \(= \frac{3}{16}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_2^4\) | M1 | Attempt to integrate \(xf(x)\) with correct limits (must see a power increase for attempted integration) |
| \(= \frac{3}{16}\left(\frac{256}{3} - 64 - \left(\frac{32}{3} - 4\right)\right) = \frac{11}{4}\) (AG) | A1 | Correct substitution of correct limits (at least 2 terms seen) and answer seen; no errors seen i.e. NO recovery of errors and no non-exact decimals (e.g. 21.33) seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{16}\int_2^4 (4x^3 - x^4)\,dx\) | *M1 | Attempt to integrate \(x^2f(x)\) with correct limits (integration must be seen not implied; must see a power increase for attempted integration) |
| \(= \frac{3}{16}\left[x^4 - \frac{x^5}{5}\right]_2^4 \left[= \frac{39}{5}\ \text{or}\ 7.8\right]\); \(\text{Var}(X) = '\frac{39}{5}' - \left(\frac{11}{4}\right)^2\) | DM1 | *their* \(\int x^2f(x)\,dx - \left(\frac{11}{4}\right)^2\), with \(\int x^2f(x)\,dx\) evaluated, not necessarily simplified |
| \(= \frac{19}{80}\) or \(0.2375\) (or \(0.238\) (3 sf)) | A1 | SC if M0 then score B1 for \(\frac{39}{5}\) and B1 for \(\frac{19}{80}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{16}\int_2^3 (4x - x^2)\,dx\) | M1 | Attempt to integrate correct integral and limits; must see a power increase for attempted integration; OE (integrate 3 to 4) OR alternative method: integrate from \(m\) to 4 or 2 to \(m\) and equate to \(0.5\) to obtain cubic (\(m^3 - 6m^2 + 24 = 0\), oe); (NB integrating from \(m\) to 3 and equating to 0.5M0) |
| \(= \frac{3}{16}\left[2x^2 - \frac{x^3}{3}\right]_2^3 \left[= \frac{3}{16}\left(18 - 9 - \left(8 - \frac{8}{3}\right)\right)\right]\left[= \frac{11}{16}\right]\); \('\frac{11}{16}' - \frac{1}{2}\) | M1 | *Their* \(\int f(x)\,dx - \frac{1}{2}\), oe \((1/2 - 5/16)\); OR alternative method: \(m\) obtained from cubic (\(m = 2.69459\)) and attempt to integrate \(f(x)\) from '*their* \(m\)' (\(2 < m < 4\)) to 3; must see a power increase for attempted integration and limits substituted |
| \(\frac{3}{16}\) or \(0.1875\) | A1 | Condone \(0.187\) or \(0.188\) |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{16}\int_2^4 (4x^2 - x^3)\,dx$ | M1 | Attempt to integrate $xf(x)$; ignore limits (must see a power increase for attempted integration) |
| $= \frac{3}{16}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_2^4$ | M1 | Attempt to integrate $xf(x)$ with correct limits (must see a power increase for attempted integration) |
| $= \frac{3}{16}\left(\frac{256}{3} - 64 - \left(\frac{32}{3} - 4\right)\right) = \frac{11}{4}$ (AG) | A1 | Correct substitution of correct limits (at least 2 terms seen) and answer seen; no errors seen i.e. NO recovery of errors and no non-exact decimals (e.g. 21.33) seen |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{16}\int_2^4 (4x^3 - x^4)\,dx$ | *M1 | Attempt to integrate $x^2f(x)$ with correct limits (integration must be seen not implied; must see a power increase for attempted integration) |
| $= \frac{3}{16}\left[x^4 - \frac{x^5}{5}\right]_2^4 \left[= \frac{39}{5}\ \text{or}\ 7.8\right]$; $\text{Var}(X) = '\frac{39}{5}' - \left(\frac{11}{4}\right)^2$ | DM1 | *their* $\int x^2f(x)\,dx - \left(\frac{11}{4}\right)^2$, with $\int x^2f(x)\,dx$ evaluated, not necessarily simplified |
| $= \frac{19}{80}$ or $0.2375$ (or $0.238$ (3 sf)) | A1 | SC if M0 then score B1 for $\frac{39}{5}$ and B1 for $\frac{19}{80}$ |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{16}\int_2^3 (4x - x^2)\,dx$ | M1 | Attempt to integrate correct integral and limits; must see a power increase for attempted integration; OE (integrate 3 to 4) OR alternative method: integrate from $m$ to 4 or 2 to $m$ and equate to $0.5$ to obtain cubic ($m^3 - 6m^2 + 24 = 0$, oe); (NB integrating from $m$ to 3 and equating to 0.5M0) |
| $= \frac{3}{16}\left[2x^2 - \frac{x^3}{3}\right]_2^3 \left[= \frac{3}{16}\left(18 - 9 - \left(8 - \frac{8}{3}\right)\right)\right]\left[= \frac{11}{16}\right]$; $'\frac{11}{16}' - \frac{1}{2}$ | M1 | *Their* $\int f(x)\,dx - \frac{1}{2}$, oe $(1/2 - 5/16)$; OR alternative method: $m$ obtained from cubic ($m = 2.69459$) and attempt to integrate $f(x)$ from '*their* $m$' ($2 < m < 4$) to 3; must see a power increase for attempted integration and limits substituted |
| $\frac{3}{16}$ or $0.1875$ | A1 | Condone $0.187$ or $0.188$ |5 A random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 3 } { 16 } \left( 4 x - x ^ { 2 } \right) & 2 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { E } ( X ) = \frac { 11 } { 4 }$.
\item Find $\operatorname { Var } ( X )$.
\item Given that the median of $X$ is $m$, find $\mathrm { P } ( m < X < 3 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q5 [9]}}