| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Mode from PDF |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard CDF/PDF concepts. Part (a) is direct substitution, (b) requires differentiation of a polynomial (chain rule practice), (c) involves finding stationary points by differentiation and testing, and (d) is a simple interpretation. All techniques are routine for S2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(P(X < 2) = F(2) = \frac{1}{432} \times 4 \times (4 - 32 + 72) = \frac{11}{27}\) | M1 A1 |
| (b) | \(F(x) = \frac{1}{432}(x^4 - 16x^3 + 72x^2)\) | M1 |
| \(f(x) = F'(x) = \frac{1}{432}(4x^3 - 48x^2 + 144x)\) | M1 A1 | |
| \(\therefore f(x) = \begin{cases} \frac{1}{108}(x^3 - 12x^2 + 36x), & 0 \leq x \leq 6, \\ 0, & \text{otherwise.} \end{cases}\) | A1 | \(\left[\text{or } \frac{1}{108}x(x-6)^2\right]\) |
| (c) | \(f'(x) = \frac{1}{108}(3x^2 - 24x + 36)\) | M1 |
| for S.P. \(= 0\) giving \(x^2 - 8x + 12 = 0\) | M1 A1 | |
| \(\therefore (x - 6)(x - 2) = 0\) so \(x = 2\) or \(6\) | M1 | |
| some justification, e.g. +ve cubic / \(f(x) = 0\) at 0 and 6 \(\therefore\) mode \(= 2\) | M1 A1 | |
| (d) | median higher as \(P(X < 2)\) is less than \(\frac{1}{2}\) | B1 |
(a) | $P(X < 2) = F(2) = \frac{1}{432} \times 4 \times (4 - 32 + 72) = \frac{11}{27}$ | M1 A1 |
(b) | $F(x) = \frac{1}{432}(x^4 - 16x^3 + 72x^2)$ | M1 |
| $f(x) = F'(x) = \frac{1}{432}(4x^3 - 48x^2 + 144x)$ | M1 A1 |
| $\therefore f(x) = \begin{cases} \frac{1}{108}(x^3 - 12x^2 + 36x), & 0 \leq x \leq 6, \\ 0, & \text{otherwise.} \end{cases}$ | A1 | $\left[\text{or } \frac{1}{108}x(x-6)^2\right]$ |
(c) | $f'(x) = \frac{1}{108}(3x^2 - 24x + 36)$ | M1 |
| for S.P. $= 0$ giving $x^2 - 8x + 12 = 0$ | M1 A1 |
| $\therefore (x - 6)(x - 2) = 0$ so $x = 2$ or $6$ | M1 |
| some justification, e.g. +ve cubic / $f(x) = 0$ at 0 and 6 $\therefore$ mode $= 2$ | M1 A1 |
(d) | median higher as $P(X < 2)$ is less than $\frac{1}{2}$ | B1 | (13)The continuous random variable $X$ has the following cumulative distribution function:
$$\text{F}(x) = \begin{cases}
0, & x < 0, \\
\frac{1}{432} x^2(x^2 - 16x + 72), & 0 \leq x \leq 6, \\
1, & x > 6.
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find P($X < 2$). [2 marks]
\item Find and specify fully the probability density function f($x$) of $X$. [4 marks]
\item Show that the mode of $X$ is 2. [6 marks]
\item State, with a reason, whether the median of $X$ is higher or lower than the mode of $X$. [1 mark]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q5 [13]}}