| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2002 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Compare mean and median using probability |
| Difficulty | Moderate -0.3 This is a straightforward continuous probability distribution question requiring standard techniques: integrating the pdf to find k, calculating E(X) using the definition, finding a probability, and interpreting the relationship between mean and median. All steps are routine applications of formulas with no conceptual challenges, making it slightly easier than average for A-level. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(k\int_{20}^{28} \frac{1}{x^2} dx = 1\) | M1 | For equating to 1 and attempt to integrate. |
| \(k\left[-\frac{1}{x}\right]_{20}^{28} = 1\) | A1 | For correct integration. |
| \(k\left[\frac{1}{20} - \frac{1}{28}\right] = 1 \Rightarrow k = 70\) | A1 (3 marks) | For given answer correctly obtained (no decimals seen). |
| (ii) \(E(X) = k\int_{20}^{28} \frac{1}{x} dx = k[\ln x] = 23.6, 23.5, 70\ln 1.4, 70\ln(7/5)\) | M1, A1, A1 (3 marks) | For attempt to evaluate \(\int_{20}^{28} \frac{70}{x} dx\). For correct integration. For correct answer. |
| (iii) \(P(X < E(X)) = \int_{20}^{23.55} \frac{70}{x^2} dx = 0.528\) (accept \(0.534\) from 23.6; \(0.521, 23.5\)) | M1, A1 (2 marks) | For attempt to evaluate \(\int \frac{70}{x^2}\) between their limits \((<28)\). For correct answer. |
| (iv) Greater. Prob in (iii) is \(> 0.5\) | B1R, B1R (2 marks) | For correct statement. For correct reason. Follow through from (iii) or calculating med. = 23.3. |
(i) $k\int_{20}^{28} \frac{1}{x^2} dx = 1$ | M1 | For equating to 1 and attempt to integrate.
$k\left[-\frac{1}{x}\right]_{20}^{28} = 1$ | A1 | For correct integration.
$k\left[\frac{1}{20} - \frac{1}{28}\right] = 1 \Rightarrow k = 70$ | A1 (3 marks) | For given answer correctly obtained (no decimals seen).
(ii) $E(X) = k\int_{20}^{28} \frac{1}{x} dx = k[\ln x] = 23.6, 23.5, 70\ln 1.4, 70\ln(7/5)$ | M1, A1, A1 (3 marks) | For attempt to evaluate $\int_{20}^{28} \frac{70}{x} dx$. For correct integration. For correct answer.
(iii) $P(X < E(X)) = \int_{20}^{23.55} \frac{70}{x^2} dx = 0.528$ (accept $0.534$ from 23.6; $0.521, 23.5$) | M1, A1 (2 marks) | For attempt to evaluate $\int \frac{70}{x^2}$ between their limits $(<28)$. For correct answer.
(iv) Greater. Prob in (iii) is $> 0.5$ | B1R, B1R (2 marks) | For correct statement. For correct reason. Follow through from (iii) or calculating med. = 23.3.
---The average speed of a bus, $x$ km h$^{-1}$, on a certain journey is a continuous random variable $X$ with probability density function given by
$$\text{f}(x) = \begin{cases}
\frac{k}{x^2} & 20 \leq x \leq 28, \\
0 & \text{otherwise}.
\end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Show that $k = 70$. [3]
\item Find E$(X)$. [3]
\item Find P$(X < \text{E}(X))$. [2]
\item Hence determine whether the mean is greater or less than the median. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2002 Q6 [10]}}