| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Moderate -0.3 This is a straightforward continuous probability distribution question requiring standard techniques: finding k by integration (∫f(t)dt=1), calculating P(T>3) by integration, and finding the median by solving ∫f(t)dt=0.5. The integration of k/(t+1) is routine (natural log), and all parts follow directly from the definition of a pdf with no problem-solving insight required. Slightly easier than average due to the simple function form and standard bookwork nature. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(\int_0^4 \frac{k}{t+1} dt = 1\) | M1 |
| \([k \ln(t+1)]_0^4 = 1\) | A1 | \(\ln(t+1)\) seen (or \(\ln(x+1)\)). Correct use of limits 0 and 4 |
| \(k = 1/\ln 5\) | M1 A1 [4] | Correct given answer legit obtained |
| AG | ||
| (ii) | \(P(T > 3) = \int_3^4 \frac{k}{t+1} dt\) | M1 |
| \(= [k \ln(t+1)]_3^4\) | A1 | Correct integration with correct limits seen (o.e.) |
| \(= 1 - \ln 4/\ln 5 = 0.139\) | A1 [3] | Correct answer |
| (iii) | \(\int_0^m \frac{k}{t+1} dt = 0.5\) | M1 |
| \([k \ln(t+1)]_0^m = 0.5\) | M1 | Attempt to solve equation with at least k, ln and \(m\) on LHS and 0.5 on RHS |
| \(k \ln(m+1) = 0.5\) | ||
| \(m = 1.24\) min | A1 [3] | Correct answer |
(i) | $\int_0^4 \frac{k}{t+1} dt = 1$ | M1 | Equating to 1 and attempt to integrate
$[k \ln(t+1)]_0^4 = 1$ | A1 | $\ln(t+1)$ seen (or $\ln(x+1)$). Correct use of limits 0 and 4
$k = 1/\ln 5$ | M1 A1 [4] | Correct given answer legit obtained
| | | AG
(ii) | $P(T > 3) = \int_3^4 \frac{k}{t+1} dt$ | M1 | Attempt to integrate with one limit 3
$= [k \ln(t+1)]_3^4$ | A1 | Correct integration with correct limits seen (o.e.)
$= 1 - \ln 4/\ln 5 = 0.139$ | A1 [3] | Correct answer
(iii) | $\int_0^m \frac{k}{t+1} dt = 0.5$ | M1 | Equating to 0.5 and attempt to integrate
$[k \ln(t+1)]_0^m = 0.5$ | M1 | Attempt to solve equation with at least k, ln and $m$ on LHS and 0.5 on RHS
$k \ln(m+1) = 0.5$ | |
$m = 1.24$ min | A1 [3] | Correct answer7 If Usha is stung by a bee she always develops an allergic reaction. The time taken in minutes for Usha to develop the reaction can be modelled using the probability density function given by
$$\mathrm { f } ( t ) = \begin{cases} \frac { k } { t + 1 } & 0 \leqslant t \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 1 } { \ln 5 }$.\\
(ii) Find the probability that it takes more than 3 minutes for Usha to develop a reaction.\\
(iii) Find the median time for Usha to develop a reaction.
\hfill \mbox{\textit{CAIE S2 2008 Q7 [10]}}