| 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 a probability equation for a parameter, and calculating an expected value. 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 |
|---|---|---|
| \(\int_0^2 k(x-2)^2 dx = 1\) | 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 | \(\int f(x)dx\) with limits \(d\) and 2 or 0 and \(d\), and \(= 0.2\) or \(0.8\) Condone missing '\(k\)' |
| \(\left(\frac{3}{8}\frac{(x-2)^3}{3}\right)_d^2 = 0.2)\) | ||
| \(\frac{3}{8}\left[0 - \frac{(d-2)^3}{3}\right] = 0.2\) o.e | 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 |
| \((= \frac{3}{8}\int_0 (x^3 - 4x^2 + 4x)dx)\) | ||
| \(= \frac{3}{8}\left[\frac{x^4}{4} - \frac{4x^3}{3} + 2x^2\right]_0^2\) | A1 | \(\left(\frac{3}{8}\left[x \times \frac{(x-2)^3}{3} - \int\frac{(x-2)^3}{3}dx\right]_0^2\right)\) |
| \(= \frac{3}{8}\left[x \times \frac{(x-2)^3}{3} - \frac{(x-2)^3}{12}\right]_0^2\) | ||
| \(= \frac{1}{2}\) | A1 [3] | Correct integration & limits, condone missing k |
**(i)**
$\int_0^2 k(x-2)^2 dx = 1$ | 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 | $\int f(x)dx$ with limits $d$ and 2 or 0 and $d$, and $= 0.2$ or $0.8$ Condone missing '$k$'
$\left(\frac{3}{8}\frac{(x-2)^3}{3}\right)_d^2 = 0.2)$ | |
$\frac{3}{8}\left[0 - \frac{(d-2)^3}{3}\right] = 0.2$ o.e | 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
$(= \frac{3}{8}\int_0 (x^3 - 4x^2 + 4x)dx)$ | |
$= \frac{3}{8}\left[\frac{x^4}{4} - \frac{4x^3}{3} + 2x^2\right]_0^2$ | A1 | $\left(\frac{3}{8}\left[x \times \frac{(x-2)^3}{3} - \int\frac{(x-2)^3}{3}dx\right]_0^2\right)$
| | | $= \frac{3}{8}\left[x \times \frac{(x-2)^3}{3} - \frac{(x-2)^3}{12}\right]_0^2$
$= \frac{1}{2}$ | A1 [3] | Correct integration & limits, condone missing k5 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]}}