| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Moderate -0.3 This is a straightforward continuous probability distribution question requiring standard techniques: finding k by integration (given answer), calculating E(X) with a simple power function, and finding P(X > 20). The integration is routine (x^{-1/2} integrates easily), and part (iv) requires basic contextual reasoning. Slightly easier than average due to the guided structure and simple algebraic form. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_4^{25} kx^{-\frac{1}{2}}\,dx = 1\) | M1 | Attempt integrate & \(= 1\). Ignore limits |
| \(\left[\frac{kx^{\frac{1}{2}}}{\frac{1}{2}}\right]_4^{25} = 1\) | ||
| \(2k(5-2) = 1\) | A1 [2] | or equiv correct subst of correct limits |
| \((k = \frac{1}{6}\) AG) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{6}\int_4^{25} x^{\frac{1}{2}}\,dx\) | M1 | Attempt integ \(xf(x)\). Ignore limits |
| \(= \frac{1}{6}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_4^{25}\ \left(= \frac{1}{9}(125-8)\right)\) | A1 | Correct integrand and limits |
| \(= 13\) | A1 [3] | Or \(117/9\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{6}\int_{20}^{25} x^{-\frac{3}{2}}\,dx\) | M1 | Attempt integ \(f(x)\) from 20 to 25; or \(1 - \int_4^{20}\) |
| \(\left(= \frac{1}{6}\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_{20}^{25} = \frac{1}{3}(5 - \sqrt{20})\right)\) | ||
| \(= 0.176\) (3 sfs) | A1 [2] | Accept surd form |
| Answer | Marks | Guidance |
|---|---|---|
| Wkly demand may be \(> 25\) (or \(< 4\)) | B1 [1] | or other sensible |
## Question 6:
**(i)**
$\int_4^{25} kx^{-\frac{1}{2}}\,dx = 1$ | M1 | Attempt integrate & $= 1$. Ignore limits
$\left[\frac{kx^{\frac{1}{2}}}{\frac{1}{2}}\right]_4^{25} = 1$ | |
$2k(5-2) = 1$ | A1 [2] | or equiv correct subst of correct limits
$(k = \frac{1}{6}$ **AG**)| |
**(ii)**
$\frac{1}{6}\int_4^{25} x^{\frac{1}{2}}\,dx$ | M1 | Attempt integ $xf(x)$. Ignore limits
$= \frac{1}{6}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_4^{25}\ \left(= \frac{1}{9}(125-8)\right)$ | A1 | Correct integrand and limits
$= 13$ | A1 [3] | Or $117/9$
**(iii)**
$\frac{1}{6}\int_{20}^{25} x^{-\frac{3}{2}}\,dx$ | M1 | Attempt integ $f(x)$ from 20 to 25; or $1 - \int_4^{20}$
$\left(= \frac{1}{6}\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_{20}^{25} = \frac{1}{3}(5 - \sqrt{20})\right)$ | |
$= 0.176$ (3 sfs) | A1 [2] | Accept surd form
**(iv)**
Wkly demand may be $> 25$ (or $< 4$) | B1 [1] | or other sensible
---6 At a certain shop the weekly demand, in kilograms, for flour is modelled by the random variable $X$ with probability density function given by
$$f ( x ) = \begin{cases} k x ^ { - \frac { 1 } { 2 } } & 4 \leqslant x \leqslant 25 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 1 } { 6 }$.\\
(ii) Calculate the mean weekly demand for flour at the shop.\\
(iii) At the beginning of one week, the shop has 20 kg of flour in stock. Find the probability that this will not be enough to meet the demand for that week.\\
(iv) Give a reason why the model may not be realistic.
\hfill \mbox{\textit{CAIE S2 2012 Q6 [8]}}