| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with multiple regions |
| Difficulty | Moderate -0.3 This is a standard S2 probability density function question covering routine techniques: integration to find k, deriving CDF, calculating expectation, finding mode via differentiation, and locating median. All parts follow textbook procedures with no novel problem-solving required, though the multi-part nature and cumulative 18 marks make it slightly more substantial than a minimal exercise, placing it just below average difficulty. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^4 k(5-x)dx = 1\) | Limits required M1 | |
| \(k\left[\frac{5x^2}{2} - \frac{x^3}{3}\right]_0^4 = 1\) | A1 | |
| Sub in limits and solve to give \(k = \frac{3}{56}\) | Correct solution A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(x_0) = \int_0^{x_0} f(x)dx = \int_0^{x_0} \frac{3}{56}(5-x)dx = \frac{3}{56}\left[\frac{5x^2}{2} - \frac{x^3}{3}\right]_0^{x_0}\) | Variable upper limit required M1 | |
| \(= \frac{x_0^2}{112}(15 - 2x_0)\) | A1 | |
| \[F(x) = \begin{cases} 0 & x < 0 \\ \frac{x^2}{112}(15-2x) & 0 \leq x \leq 4 \\ 1 & x > 4 \end{cases}\] | Ends, middle. | B1, B1J |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(x) = \int_0^4 \frac{3}{56}x^2(5-x)dx = \frac{3}{56}\left[\frac{5x^3}{3} - \frac{x^4}{4}\right]_0^4 = 2.29\) | M1A1A1 | |
| \(\int xf(x)dx\), \(\left[\frac{5x^3}{3} - \frac{x^4}{4}\right]\), 3sf \((2\frac{2}{7})\) | M1A1A1 |
| Answer | Marks |
|---|---|
| \(f'(x) = \frac{3}{56}(5-2x) = 0 \Rightarrow \text{Mode} = 2.5\) | Attempt \(f'(x)\), \((5-2x) = 0\), 2.5 M1A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(2.3) = 0.491\), \(F(2.5) = 0.558\) | Their F, awrt 0.491 & 0.558 or 0.984 & -6.5 cso M1, A1 | |
| \(F(m) = 0.5 \Rightarrow m\) lies between 2.3 and 2.5 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Mean (2.29) < Median (2.3-2.5) < Mode (2.5) | B1 | |
| Negative skew | B1 dep |
## Part (a)
$\int_0^4 k(5-x)dx = 1$ | Limits required M1 |
$k\left[\frac{5x^2}{2} - \frac{x^3}{3}\right]_0^4 = 1$ | | A1 |
Sub in limits and solve to give $k = \frac{3}{56}$ | Correct solution A1 |
(3 marks)
## Part (b)
$F(x_0) = \int_0^{x_0} f(x)dx = \int_0^{x_0} \frac{3}{56}(5-x)dx = \frac{3}{56}\left[\frac{5x^2}{2} - \frac{x^3}{3}\right]_0^{x_0}$ | Variable upper limit required M1 |
$= \frac{x_0^2}{112}(15 - 2x_0)$ | | A1 |
$$F(x) = \begin{cases} 0 & x < 0 \\ \frac{x^2}{112}(15-2x) & 0 \leq x \leq 4 \\ 1 & x > 4 \end{cases}$$ | Ends, middle. | B1, B1J |
(4 marks)
## Part (c)
$E(x) = \int_0^4 \frac{3}{56}x^2(5-x)dx = \frac{3}{56}\left[\frac{5x^3}{3} - \frac{x^4}{4}\right]_0^4 = 2.29$ | | M1A1A1 |
$\int xf(x)dx$, $\left[\frac{5x^3}{3} - \frac{x^4}{4}\right]$, 3sf $(2\frac{2}{7})$ | | M1A1A1 |
(3 marks)
## Part (d)
$f'(x) = \frac{3}{56}(5-2x) = 0 \Rightarrow \text{Mode} = 2.5$ | Attempt $f'(x)$, $(5-2x) = 0$, 2.5 M1A1A1 |
(Or Sketch M1, $x = 0$ & 5 A1, Mode=2.5 A1)
(3 marks)
## Part (e)
$F(2.3) = 0.491$, $F(2.5) = 0.558$ | Their F, awrt 0.491 & 0.558 or 0.984 & -6.5 cso M1, A1 |
$F(m) = 0.5 \Rightarrow m$ lies between 2.3 and 2.5 | | A1 |
(3 marks)
## Part (f)
Mean (2.29) < Median (2.3-2.5) < Mode (2.5) | | B1 |
Negative skew | | B1 dep |
(2 marks)
**Total 18 Marks**The continuous random variable $X$ has probability density function
$$\text{f}(x) = \begin{cases}
kx(5 - x), & 0 \leq x \leq 4, \\
0, & \text{otherwise,}
\end{cases}$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac{3}{56}$. [3]
\item Find the cumulative distribution function F($x$) for all values of $x$. [4]
\item Evaluate E($X$). [3]
\item Find the modal value of $X$. [3]
\item Verify that the median value of $X$ lies between 2.3 and 2.5. [3]
\item Comment on the skewness of $X$. Justify your answer. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2004 Q7 [18]}}