| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.3 This is a standard S2 probability density function question requiring routine integration techniques: finding k by integrating to 1, computing E(X), deriving F(x) by integration, and locating the median by solving F(x)=0.5. All steps 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 integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_2^3 kx(x-2)\,dx = 1\) | M1 | \(\int f(x) = 1\) |
| \(\left[\frac{1}{3}kx^3 - kx^2\right]_2^3 = 1\) | M1 | attempt \(\int\); need either \(x^3\) or \(x^2\) |
| correct integral | A1 | |
| \((9k - 9k) - \left(\frac{8k}{3} - 4k\right) = 1\), \(\quad k = \frac{3}{4} = 0.75\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \int_2^3 \frac{3}{4}x^2(x-2)\,dx\) | M1 | attempt \(\int xf(x)\) |
| \(= \left[\frac{3}{16}x^4 - \frac{1}{2}x^3\right]_2^3\) | A1 | correct integral |
| \(= 2.6875 = 2\frac{11}{16}\) | A1 | awrt 2.69 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(x) = \int_2^x \frac{3}{4}(t^2 - 2t)\,dt\) | M1 | \(\int f(x)\) with variable limit or \(+C\) |
| \(= \left[\frac{3}{4}\left(\frac{1}{3}t^3 - t^2\right)\right]_2^x\) | A1 | correct integral |
| lower limit of 2 or \(F(2)=0\) or \(F(3)=1\) | A1 | |
| \(= \frac{1}{4}(x^3 - 3x^2 + 4)\) | A1 | |
| \(F(x) = \begin{cases} 0 & x \leq 2 \\ \frac{1}{4}(x^3 - 3x^2 + 4) & 2 < x < 3 \\ 1 & x \geq 3 \end{cases}\) | B1\(\checkmark\), B1 | middle, ends |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(x) = \frac{1}{2}\), \(\quad \frac{1}{4}(x^3 - 3x^2 + 4) = \frac{1}{2}\) | M1 | their \(F(x) = \frac{1}{2}\) |
| \(x^3 - 3x^2 + 2 = 0\); check \(x=2.75\): positive; \(x=2.70\): negative \(\Rightarrow\) root between 2.70 and 2.75 | M1 |
# Question 5:
## Part (a)
| $\int_2^3 kx(x-2)\,dx = 1$ | M1 | $\int f(x) = 1$ |
| $\left[\frac{1}{3}kx^3 - kx^2\right]_2^3 = 1$ | M1 | attempt $\int$; need either $x^3$ or $x^2$ |
| correct integral | A1 | |
| $(9k - 9k) - \left(\frac{8k}{3} - 4k\right) = 1$, $\quad k = \frac{3}{4} = 0.75$ | A1 | cso |
## Part (b)
| $E(X) = \int_2^3 \frac{3}{4}x^2(x-2)\,dx$ | M1 | attempt $\int xf(x)$ |
| $= \left[\frac{3}{16}x^4 - \frac{1}{2}x^3\right]_2^3$ | A1 | correct integral |
| $= 2.6875 = 2\frac{11}{16}$ | A1 | awrt 2.69 |
## Part (c)
| $F(x) = \int_2^x \frac{3}{4}(t^2 - 2t)\,dt$ | M1 | $\int f(x)$ with variable limit or $+C$ |
| $= \left[\frac{3}{4}\left(\frac{1}{3}t^3 - t^2\right)\right]_2^x$ | A1 | correct integral |
| lower limit of 2 or $F(2)=0$ or $F(3)=1$ | A1 | |
| $= \frac{1}{4}(x^3 - 3x^2 + 4)$ | A1 | |
| $F(x) = \begin{cases} 0 & x \leq 2 \\ \frac{1}{4}(x^3 - 3x^2 + 4) & 2 < x < 3 \\ 1 & x \geq 3 \end{cases}$ | B1$\checkmark$, B1 | middle, ends |
## Part (d)
| $F(x) = \frac{1}{2}$, $\quad \frac{1}{4}(x^3 - 3x^2 + 4) = \frac{1}{2}$ | M1 | their $F(x) = \frac{1}{2}$ |
| $x^3 - 3x^2 + 2 = 0$; check $x=2.75$: positive; $x=2.70$: negative $\Rightarrow$ root between 2.70 and 2.75 | M1 | |
---5. A continuous random variable $X$ has probability density function $\mathrm { f } ( x )$ where
$$f ( x ) = \begin{cases} k x ( x - 2 ) , & 2 \leq x \leq 3 \\ 0 , & \text { otherwise } \end{cases}$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 3 } { 4 }$.
Find
\item $\mathrm { E } ( X )$,
\item the cumulative distribution function $\mathrm { F } ( x )$.
\item Show that the median value of $X$ lies between 2.70 and 2.75.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2006 Q5 [15]}}