| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2001 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF to PDF derivation |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard CDF-to-PDF differentiation, basic calculus (finding mode via f'(x)=0), sketching, and computing mean via integration. All parts follow routine procedures with no novel problem-solving required. The verification steps in (e) and (f) involve simple substitution. Slightly above average difficulty only due to the multi-part nature and integration required for the mean. |
| 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 |
|---|---|---|
| (a) \(f(x) = \frac{d}{dx}F(x) = \frac{1}{27}(-3x^2 + 12x)\) | M1, A1 | Attempt \(\frac{d}{dx}\); \(\frac{1}{27}(-3x^2+12x)\) (3) |
| (b) \(\frac{d}{dx}[f(x)] = 0 \Rightarrow -6x + 12 = 0 \Rightarrow x = 2\) is mode | M1, A1 | (2) |
| (c) Sketch of \(f(x)\): \(x\), \(f(x)\) axes marked, at least points 1 and 4 shown, correct shape with maximum | B1, B1, B1 | (3) |
| (d) \(\mu = \int_1^4 \left(\frac{4x^2}{9} - \frac{x^3}{9}\right)dx\) | M1 (attempt \(\int x f(x)\,dx\)) | |
| \(= \frac{1}{9}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_1^4 = \left(\frac{256}{27} - \frac{256}{36}\right) - \left(\frac{4}{27} - \frac{1}{36}\right)\) | M1 (correct limits) | |
| \(= \underline{2.25}\) or \(\frac{9}{4}\) | A1 | (3) |
| (e) \(F(2.25) = \frac{1}{27}(-2.25^3 + 6 \times 2.25^2 - 5) = 0.517\) | B1 | (AWRT \(0.52\)) (1) |
| (f) \(F(\mu) > 0.5 \Rightarrow \mu >\) median | B1✓ (from (e)) | |
| \(F(2) = \frac{1}{27}(-8 + 24 - 5) = \frac{11}{27} = 0.407 \Rightarrow\) mode \(<\) median | B1 | (2) — (14) |
## Question 6:
**(a)** $f(x) = \frac{d}{dx}F(x) = \frac{1}{27}(-3x^2 + 12x)$ | M1, A1 | Attempt $\frac{d}{dx}$; $\frac{1}{27}(-3x^2+12x)$ **(3)**
**(b)** $\frac{d}{dx}[f(x)] = 0 \Rightarrow -6x + 12 = 0 \Rightarrow x = 2$ is mode | M1, A1 | **(2)**
**(c)** Sketch of $f(x)$: $x$, $f(x)$ axes marked, at least points 1 and 4 shown, correct shape with maximum | B1, B1, B1 | **(3)**
**(d)** $\mu = \int_1^4 \left(\frac{4x^2}{9} - \frac{x^3}{9}\right)dx$ | M1 (attempt $\int x f(x)\,dx$) |
$= \frac{1}{9}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_1^4 = \left(\frac{256}{27} - \frac{256}{36}\right) - \left(\frac{4}{27} - \frac{1}{36}\right)$ | M1 (correct limits) |
$= \underline{2.25}$ or $\frac{9}{4}$ | A1 | **(3)**
**(e)** $F(2.25) = \frac{1}{27}(-2.25^3 + 6 \times 2.25^2 - 5) = 0.517$ | B1 | (AWRT $0.52$) **(1)**
**(f)** $F(\mu) > 0.5 \Rightarrow \mu >$ median | B1✓ (from (e)) |
$F(2) = \frac{1}{27}(-8 + 24 - 5) = \frac{11}{27} = 0.407 \Rightarrow$ mode $<$ median | B1 | **(2)** — **(14)**
---6. The continuous random variable X has cumulative distribution function $\mathrm { F } ( x )$ given by
$$\mathrm { F } ( x ) = \left\{ \begin{array} { l r }
0 , & x < 1 \\
\frac { 1 } { 27 } \left( - x ^ { 3 } + 6 x ^ { 2 } - 5 \right) , & 1 \leq x \leq 4 \\
1 , & x > 4
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Find the probability density function $\mathrm { f } ( x )$.
\item Find the mode of $X$.
\item Sketch $\mathrm { f } ( x )$ for all values of $x$.
\item Find the mean $\mu$ of X .
\item Show that $\mathrm { F } ( \mu ) > 0.5$.
\item Show that the median of $X$ lies between the mode and the mean.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2001 Q6 [14]}}