| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| 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 | Standard +0.3 This is a standard S2 probability density function question requiring routine integration to find k, mean, and probability, plus a conceptual understanding of quartiles. All techniques are straightforward applications of formulas with no novel problem-solving required, 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 integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int_3^6 k(6t - t^2)dt = 1\) | M1 | For equating to 1 and a sensible attempt to integrate |
| Answer | Marks | Guidance |
|---|---|---|
| \(k([108 - 216/3] - [27 - 9]) = 1\) | A1 | Correct integration and correct limits |
| \(k = 1/18\) AG | A1 | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) mean \(= \int_3^6 t \cdot k(6t^2 - t^3)dt\) | M1 | Attempt to evaluate the integral of \(t(t)\) (t or x) |
| \(= k\left(2t^3 - \frac{t^4}{4}\right)_3^6 = k(432 - 324) - k(54 - 81/4)\) | A1 | Correct integral and correct limits (condone loss of k) |
| \(= \frac{33}{8}\) (4.13) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) \(\int_5^6 k(6t - t^2)dt\) | M1 | Attempt to evaluate the integral between 5 and 6 or |
| \(= k\left[3t^2 - \frac{t^3}{3}\right]_5^6 = k\left(36 - \frac{100}{3}\right)\) | A1 | Correct answer |
| \(= \frac{4}{27}\) (0.148) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| (iv) the area on the left is > 0.75 or (iii) is < 0.25, UQ is less than 5 | M1 | sensible reason |
| A1ft | fit their (iii); SR B1ft correct but 0.25/0.75 implied |
(i) $\int_3^6 k(6t - t^2)dt = 1$ | M1 | For equating to 1 and a sensible attempt to integrate
$k\left[3t^2 - t^3/3\right]_3^6 = 1$
$k([108 - 216/3] - [27 - 9]) = 1$ | A1 | Correct integration and correct limits
$k = 1/18$ AG | A1 | Given answer correctly obtained
**[3]**
(ii) mean $= \int_3^6 t \cdot k(6t^2 - t^3)dt$ | M1 | Attempt to evaluate the integral of $t(t)$ (t or x)
$= k\left(2t^3 - \frac{t^4}{4}\right)_3^6 = k(432 - 324) - k(54 - 81/4)$ | A1 | Correct integral and correct limits (condone loss of k)
$= \frac{33}{8}$ (4.13) | A1 | Correct answer
**[3]**
(iii) $\int_5^6 k(6t - t^2)dt$ | M1 | Attempt to evaluate the integral between 5 and 6 or
$= k\left[3t^2 - \frac{t^3}{3}\right]_5^6 = k\left(36 - \frac{100}{3}\right)$ | A1 | Correct answer
$= \frac{4}{27}$ (0.148) | A1 | Correct answer
**[2]**
(iv) the area on the left is > 0.75 or (iii) is < 0.25, UQ is less than 5 | M1 | sensible reason
| A1ft | fit their (iii); SR B1ft correct but 0.25/0.75 implied
**[2]**5 The time in minutes taken by candidates to answer a question in an examination has probability density function given by
$$\mathrm { f } ( t ) = \begin{cases} k \left( 6 t - t ^ { 2 } \right) & 3 \leqslant t \leqslant 6 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 1 } { 18 }$.\\
(ii) Find the mean time.\\
(iii) Find the probability that a candidate, chosen at random, takes longer than 5 minutes to answer the question.\\
(iv) Is the upper quartile of the times greater than 5 minutes, equal to 5 minutes or less than 5 minutes? Give a reason for your answer.
\hfill \mbox{\textit{CAIE S2 2009 Q5 [10]}}