| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find parameter from expectation |
| Difficulty | Standard +0.3 This is a straightforward application of standard probability density function properties. Part (a) requires computing E(X) by integration and equating to ln 2, part (b) uses the normalization condition ∫f(x)dx=1, and part (c) finds the median by solving ∫f(x)dx=0.5. All steps are routine calculus with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{a}^{b} \frac{x}{x^2}\, dx\) | M1 | Attempt to integrate \(x f(x)\) from \(a\) to \(b\). |
| \(= [\ln x]_a^b\) or \(\ln b - \ln a\) | A1 | |
| \(\ln\frac{b}{a} = \ln 2\) or \(\ln 2a = \ln b\) oe; Hence \(b = 2a\) (AG) | A1 | Must see both statements. No errors seen. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{a}^{b} \frac{1}{x^2}\, dx = 1\) | M1 | Attempt to integrate \(f(x)\) and equate to 1. Ignore limits. |
| \(\left[-\frac{1}{x}\right]_a^b = 1\) or \(\frac{1}{a} - \frac{1}{b} = 1\); \(\frac{1}{a} - \frac{1}{2a} = 1\) | A1 | Integrate with correct limits and substitute \(b = 2a\). |
| \(\frac{1}{2a} = 1\) or \(2 + (-1) = 2a\); Hence \(a = \frac{1}{2}\) (AG) | A1 | Obtain convincingly (at least one step from previous answer), no errors seen (ignore \(a = 0\)). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{0.5}^{m} \frac{1}{x^2}\, dx = \frac{1}{2}\) or \(\int_{m}^{1} \frac{1}{x^2}\, dx = \frac{1}{2}\) | M1 | Attempt integrate \(f(x)\), with correct limits, and equate to \(\frac{1}{2}\). |
| \(\left[-\frac{1}{x}\right]_{0.5}^{m} = \frac{1}{2}\) or \(\left[-\frac{1}{x}\right]_{m}^{1} = \frac{1}{2}\) | ||
| \(2 - \frac{1}{m} = \frac{1}{2}\) or \(\frac{1}{m} - 1 = \frac{1}{2}\) | A1 | oe. Correct equation after substituting limits. |
| \(m = \frac{2}{3}\) or \(0.667\) (3 sf) | A1 |
## Question 5(a):
$\int_{a}^{b} \frac{x}{x^2}\, dx$ | M1 | Attempt to integrate $x f(x)$ from $a$ to $b$.
$= [\ln x]_a^b$ or $\ln b - \ln a$ | A1 |
$\ln\frac{b}{a} = \ln 2$ or $\ln 2a = \ln b$ oe; Hence $b = 2a$ (AG) | A1 | Must see both statements. No errors seen.
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## Question 5(b):
$\int_{a}^{b} \frac{1}{x^2}\, dx = 1$ | M1 | Attempt to integrate $f(x)$ and equate to 1. Ignore limits.
$\left[-\frac{1}{x}\right]_a^b = 1$ or $\frac{1}{a} - \frac{1}{b} = 1$; $\frac{1}{a} - \frac{1}{2a} = 1$ | A1 | Integrate with correct limits and substitute $b = 2a$.
$\frac{1}{2a} = 1$ or $2 + (-1) = 2a$; Hence $a = \frac{1}{2}$ (AG) | A1 | Obtain convincingly (at least one step from previous answer), no errors seen (ignore $a = 0$).
---
## Question 5(c):
$\int_{0.5}^{m} \frac{1}{x^2}\, dx = \frac{1}{2}$ or $\int_{m}^{1} \frac{1}{x^2}\, dx = \frac{1}{2}$ | M1 | Attempt integrate $f(x)$, with correct limits, and equate to $\frac{1}{2}$.
$\left[-\frac{1}{x}\right]_{0.5}^{m} = \frac{1}{2}$ or $\left[-\frac{1}{x}\right]_{m}^{1} = \frac{1}{2}$ |
$2 - \frac{1}{m} = \frac{1}{2}$ or $\frac{1}{m} - 1 = \frac{1}{2}$ | A1 | oe. Correct equation after substituting limits.
$m = \frac{2}{3}$ or $0.667$ (3 sf) | A1 |
---5 The random variable $X$ has probability density function, f, given by
$$f ( x ) = \begin{cases} \frac { 1 } { x ^ { 2 } } & a < x < b \\ 0 & \text { otherwise } \end{cases}$$
where $a$ and $b$ are positive constants.
\begin{enumerate}[label=(\alph*)]
\item It is given that $\mathrm { E } ( X ) = \ln 2$.
Show that $b = 2 a$.
\item Show that $a = \frac { 1 } { 2 }$.
\item Find the median of $X$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q5 [9]}}