| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Parameter interpretation in context |
| Difficulty | Standard +0.3 This is a straightforward S2 probability density function question requiring standard techniques: interpreting context (trivial), finding k by integration (routine), calculating E(X) with a simple power integration, solving an exponential equation, and finding a probability by integration. All steps are textbook exercises 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 integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) All cars stayed between 1 and 9 hours | B1 | Or equivalent |
| (ii) \(\int_1^9 kx^{3/2} dx = 1\) | M1 | Equating to 1 and attempting to integrate |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = 3/4\) | A1 | Correct answer, legitimately obtained |
| (iii) \(\int_1 0.75 x^{-1/2} dx = [-1.5x^{1/2}]_1 = 4.5 - 1.5 = 3\) hours | M1 | Attempt to integrate \(xf(x)\) |
| A1 | Correct integral with correct limits | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.25 > e^x\) | M1\(^*\) | Equality or inequality involving \(e^x\) and \(0.75\) |
| M1dep\(^*\) | Solving attempt by logs or trial and error | |
| \(x > 1.39\) hours (oe) | A1 | For \(1.39\) no errors seen |
| (v) \(P(X > 1.386) = \int_{1.386}^9 0.75 x^{-3/2} dx\) | M1 | Attempting to integrate from their (iv) to 9, or from 1 to their (iv) |
| \(= [-1.5x^{-1/2}]_{1.386}^9 = -0.5 - 1.274 = 0.774\) | A1 | Correct answer (Accept 0.772) |
**(i)** All cars stayed between 1 and 9 hours | B1 | Or equivalent | 1 mark
**(ii)** $\int_1^9 kx^{3/2} dx = 1$ | M1 | Equating to 1 and attempting to integrate
$[-2kx^{-1/2}]_1^9 = 1$
$-2k/3 - (-2k) = 1$
$k = 3/4$ | A1 | Correct answer, legitimately obtained | 2 marks total
**(iii)** $\int_1 0.75 x^{-1/2} dx = [-1.5x^{1/2}]_1 = 4.5 - 1.5 = 3$ hours | M1 | Attempt to integrate $xf(x)$
| A1 | Correct integral with correct limits | 3 marks total
**(iv)** $1 - e^x > 0.75$
$0.25 > e^x$ | M1$^*$ | Equality or inequality involving $e^x$ and $0.75$
| M1dep$^*$ | Solving attempt by logs or trial and error
$x > 1.39$ hours (oe) | A1 | For $1.39$ no errors seen | 3 marks total
**(v)** $P(X > 1.386) = \int_{1.386}^9 0.75 x^{-3/2} dx$ | M1 | Attempting to integrate from their (iv) to 9, or from 1 to their (iv)
$= [-1.5x^{-1/2}]_{1.386}^9 = -0.5 - 1.274 = 0.774$ | A1 | Correct answer (Accept 0.772) | 2 marks total7 At a town centre car park the length of stay in hours is denoted by the random variable $X$, which has probability density function given by
$$f ( x ) = \begin{cases} k x ^ { - \frac { 3 } { 2 } } & 1 \leqslant x \leqslant 9 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Interpret the inequalities $1 \leqslant x \leqslant 9$ in the definition of $\mathrm { f } ( x )$ in the context of the question.\\
(ii) Show that $k = \frac { 3 } { 4 }$.\\
(iii) Calculate the mean length of stay.
The charge for a length of stay of $x$ hours is $\left( 1 - \mathrm { e } ^ { - x } \right)$ dollars.\\
(iv) Find the length of stay for the charge to be at least 0.75 dollars\\
(v) Find the probability of the charge being at least 0.75 dollars.
\hfill \mbox{\textit{CAIE S2 2006 Q7 [11]}}