| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find parameter from probability condition |
| Difficulty | Standard +0.3 This is a standard S2 probability density function question requiring routine integration techniques: finding k using the total probability condition, solving P(X≥d)=0.2 by integration and equation solving, and computing E(X). All three parts follow textbook procedures with straightforward polynomial integration, 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 |
|---|---|---|
| \(\int k(x-2)^2 dx = 1\) | M1 | Attempt to integrate f(x) with correct limits and = 1 |
| \(\left(\frac{k(x-2)^3}{3}\right)_0^2 = 1\) | ||
| \(k[0 - (-\frac{8}{3})] = 1\) | A1 | Must see this line or better, e.g. \(k \times \frac{8}{3} = 1\) |
| \(k = \frac{3}{8}\) AG | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3}{8}\int_d^2 (x-2)^2 dx = 0.2\) | M1 | f(x)dx with limits d and 2 or 0 and d, and = 0.2 or 0.8; Condone missing 'k' |
| \(\left(\frac{3}{8} \frac{(x-2)^3}{3}\right)_d^2 = 0.2\) oe | ||
| \(\frac{3}{8}\left[0 - \frac{(d-2)^3}{3}\right] = 0.2\) oe | M1 | Reasonable attempt to integrate from a correct expression, with limits substituted to give expression in d'. Condone missing 'k' |
| \(((d-2)^3 = -1.6)\) | ||
| \(d = 0.83(0)\) (3 s.f.) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3}{8}\int_0^2 x(x-2)^2 dx\) | M1 | Attempt integ xf(x); ignore limits, condone missing k |
| \(\left(= \frac{3}{8}\int_0 (x^3 - 4x^2 + 4x)dx\right)\) | \(\left(\left\{\frac{3}{8}\left[x x \frac{(x-2)^3}{3} - \int \frac{(x-2)^3}{3} dx\right]_0^2\right)\right)\) | |
| \(= \frac{3}{8}\left[\frac{x^4}{4} - \frac{4x^3}{3} + 2x^2\right]_0^2\) | \(= \frac{3}{8}\left[x x \frac{(x-2)^3}{3} - \frac{(x-2)^3}{12}\right]_0^2\) | |
| \(= \frac{1}{2}\) | A1 | Correct integration & limits, condone missing k |
**(i)**
$\int k(x-2)^2 dx = 1$ | M1 | Attempt to integrate f(x) with correct limits and = 1
$\left(\frac{k(x-2)^3}{3}\right)_0^2 = 1$ | |
$k[0 - (-\frac{8}{3})] = 1$ | A1 | Must see this line or better, e.g. $k \times \frac{8}{3} = 1$
$k = \frac{3}{8}$ AG | [2]
**(ii)**
$\frac{3}{8}\int_d^2 (x-2)^2 dx = 0.2$ | M1 | f(x)dx with limits d and 2 or 0 and d, and = 0.2 or 0.8; Condone missing 'k'
$\left(\frac{3}{8} \frac{(x-2)^3}{3}\right)_d^2 = 0.2$ oe | |
$\frac{3}{8}\left[0 - \frac{(d-2)^3}{3}\right] = 0.2$ oe | M1 | Reasonable attempt to integrate from a correct expression, with limits substituted to give expression in d'. Condone missing 'k'
$((d-2)^3 = -1.6)$ | |
$d = 0.83(0)$ (3 s.f.) | A1 | [3]
**(iii)**
$\frac{3}{8}\int_0^2 x(x-2)^2 dx$ | M1 | Attempt integ xf(x); ignore limits, condone missing k
$\left(= \frac{3}{8}\int_0 (x^3 - 4x^2 + 4x)dx\right)$ | | $\left(\left\{\frac{3}{8}\left[x x \frac{(x-2)^3}{3} - \int \frac{(x-2)^3}{3} dx\right]_0^2\right)\right)$
$= \frac{3}{8}\left[\frac{x^4}{4} - \frac{4x^3}{3} + 2x^2\right]_0^2$ | | $= \frac{3}{8}\left[x x \frac{(x-2)^3}{3} - \frac{(x-2)^3}{12}\right]_0^2$
$= \frac{1}{2}$ | A1 | Correct integration & limits, condone missing k
**Total: [3]**5 The volume, in $\mathrm { cm } ^ { 3 }$, of liquid left in a glass by people when they have finished drinking all they want is modelled by the random variable $X$ with probability density function given by
$$f ( x ) = \begin{cases} k ( x - 2 ) ^ { 2 } & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 3 } { 8 }$.\\
(ii) $20 \%$ of people leave at least $d \mathrm {~cm} ^ { 3 }$ of liquid in a glass. Find $d$.\\
(iii) Find $\mathrm { E } ( X )$.
\hfill \mbox{\textit{CAIE S2 2013 Q5 [8]}}