| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Symmetry property of PDF |
| Difficulty | Standard +0.3 This is a straightforward continuous probability distribution question requiring standard techniques: finding k by integration (though the factorized form makes this easy), recognizing symmetry to state E(X) without calculation, computing Var(X) using standard formulas, and applying binomial probability. The symmetry observation is a nice touch but the overall problem is slightly easier than average A-level statistics work. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-k\int_2^3(x^2 - 5x + 6)\,dx = 1\) | M1 | Integ \(= 1\); ignore limits; also \(-6\int_2^3(x^2-5x+6)\,dx\) ignore limits |
| \(\left(-k\left(\frac{3^3}{3} - 5\times\frac{3^2}{2} + 6\times3 - \left[\frac{2^3}{3} - 5\times\frac{2^2}{2} + 6\times2\right]\right) = 1\right)\) | ||
| \(-k \times \left(-\frac{1}{6}\right) = 1\) or \(k \times \frac{1}{6} = 1\); \((k=6\) AG) | A1[2] | Correctly obtain \(-\frac{1}{6}\) or \(\frac{1}{6}\); CWO no rounded decimals; Correctly obtain 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = 2.5\) | B1 | Condone 25000 |
| \(-6\int_2^3(x^4 - 5x^3 + 6x^2)\,dx \quad (= -6\times(-1.05))\) | M1* | Integ \(x^2 f(x)\); ignore limits |
| \(- \text{``}2.5\text{''}^2\) | Dep M1* | Subtr \(\mu^2\) |
| \(= 0.05\) | A1[4] | ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-6\int_2^{2.2}(x^2 - 5x + 6)\,dx \quad (= 0.104)\) | M1 | Integ with limits 2, 2.2 or 2.2, 3 |
| \(1 - (1 - \text{``}0.104\text{''})^4 = 0.355/0.356\) | M1, A1[3] | Or equivalent |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-k\int_2^3(x^2 - 5x + 6)\,dx = 1$ | M1 | Integ $= 1$; ignore limits; also $-6\int_2^3(x^2-5x+6)\,dx$ ignore limits |
| $\left(-k\left(\frac{3^3}{3} - 5\times\frac{3^2}{2} + 6\times3 - \left[\frac{2^3}{3} - 5\times\frac{2^2}{2} + 6\times2\right]\right) = 1\right)$ | | |
| $-k \times \left(-\frac{1}{6}\right) = 1$ or $k \times \frac{1}{6} = 1$; $(k=6$ **AG**)| A1[2] | Correctly obtain $-\frac{1}{6}$ or $\frac{1}{6}$; CWO no rounded decimals; Correctly obtain 1 |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 2.5$ | B1 | Condone 25000 |
| $-6\int_2^3(x^4 - 5x^3 + 6x^2)\,dx \quad (= -6\times(-1.05))$ | M1* | Integ $x^2 f(x)$; ignore limits |
| $- \text{``}2.5\text{''}^2$ | Dep M1* | Subtr $\mu^2$ |
| $= 0.05$ | A1[4] | ISW |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-6\int_2^{2.2}(x^2 - 5x + 6)\,dx \quad (= 0.104)$ | M1 | Integ with limits 2, 2.2 or 2.2, 3 |
| $1 - (1 - \text{``}0.104\text{''})^4 = 0.355/0.356$ | M1, A1[3] | Or equivalent |
---6 The distance travelled, in kilometres, by a Grippo brake pad before it needs to be replaced is modelled by $10000 X$, where $X$ is a random variable having the probability density function
$$f ( x ) = \begin{cases} - k \left( x ^ { 2 } - 5 x + 6 \right) & 2 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
The graph of $y = \mathrm { f } ( x )$ is shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{c1dcf0f5-e971-4afd-81ca-4d860732825c-3_439_1100_580_520}\\
(i) Show that $k = 6$.\\
(ii) State the value of $\mathrm { E } ( X )$ and find $\operatorname { Var } ( X )$.\\
(iii) Sami fits four new Grippo brake pads on his car. Find the probability that at least one of these brake pads will need to be replaced after travelling less than 22000 km .
\hfill \mbox{\textit{CAIE S2 2011 Q6 [9]}}