| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF with additional constraints |
| Difficulty | Standard +0.8 This S2 question requires understanding that the mode occurs where the PDF is maximized, necessitating differentiation of the CDF to find f(x), then setting f'(x)=0. Students must also apply continuity conditions F(2)=0 and F(3)=1 to find k. The multi-step algebraic manipulation and the conceptual leap connecting mode to PDF maximum makes this harder than typical S2 CDF questions, though still within standard A-level scope. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = ak + 2bkx - 3kx^2\) | M1 | Attempting to differentiate \(F(x)\), at least one \(x^n \to x^{n-1}\) |
| \(\frac{df(x)}{dx} = 2kb - 6kx\) | M1dA1 | Attempting to differentiate \(f(x)\), at least one \(x^n \to x^{n-1}\); dependent on previous M; A1 condone missing \(\frac{df(x)}{dx}\) |
| \(2kb - 6kx = 0 \Rightarrow k(2b - 6x) = 0 \Rightarrow 2b - 6x = 0\) | M1d | Putting 2nd differential \(= 0\); dependent on previous method mark |
| \(2b - 6 \times \frac{8}{3} = 0\) | M1d | Substituting \(x = \frac{8}{3}\); allow with \(k\) in; dependent on previous method mark |
| \(b = 8\)* | A1cso | Answer given so must have been awarded all previous marks with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-3k\!\left(x^2 - \frac{2bx}{3} - \frac{a}{3}\right)\) | M1 | Factorising by taking \(-3k\) out |
| \(-3k\!\left(\!\left(x - \frac{b}{3}\right)^2 - \left(\frac{b}{3}\right)^2 - \frac{a}{3}\right)\) or quoting \(\frac{-b}{2a}\) | M1d | Attempting to complete the square; dependent on previous M; \(\left(x - \frac{b}{3}\right)^2 \pm c\) |
| \(-3k\!\left(x - \frac{b}{3}\right)^2 + \frac{b^2k}{3} + ak\) | A1 | Correct completed square form |
| Max at \(x = \frac{b}{3}\) | M1d | Selecting their \(b/3\); dependent on previous method mark |
| \(\frac{b}{3} = \frac{8}{3}\) | M1d | Putting their \(\frac{b}{3} = \frac{8}{3}\); dependent on previous method mark |
| \(b = 8\)* | A1cso | Answer given; all steps must have been shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F(2) = 0\): \(k(2a + 32 - 8) = 0\) or \(k(2a + 4b - 8) = 0\) oe | M1 | Attempting to form an equation using \(F(2) = 0\), or \(F(3) = 1\) or \(F(3) - F(2) = 1\); need to substitute \(x\) value and equate |
| \(a = -12\) | A1 | May be implied by \(k = 1/9\); do not award if M1 not given |
| \(F(3) = 1\): \(k(-36 + 72 - 27) = 1\) or \(k(-36 + 9b - 27) = 1\) oe | M1 | Forming an equation using two of \(F(2) = 0\), \(F(3) = 1\), or \(F(3) - F(2) = 1\) |
| \(k = \frac{1}{9}\) | A1 | Allow equivalent fractions or awrt 0.111 |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = ak + 2bkx - 3kx^2$ | M1 | Attempting to differentiate $F(x)$, at least one $x^n \to x^{n-1}$ |
| $\frac{df(x)}{dx} = 2kb - 6kx$ | M1dA1 | Attempting to differentiate $f(x)$, at least one $x^n \to x^{n-1}$; dependent on previous M; A1 condone missing $\frac{df(x)}{dx}$ |
| $2kb - 6kx = 0 \Rightarrow k(2b - 6x) = 0 \Rightarrow 2b - 6x = 0$ | M1d | Putting 2nd differential $= 0$; dependent on previous method mark |
| $2b - 6 \times \frac{8}{3} = 0$ | M1d | Substituting $x = \frac{8}{3}$; allow with $k$ in; dependent on previous method mark |
| $b = 8$* | A1cso | Answer given so must have been awarded all previous marks with no errors |
**Alternative method – completing the square:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-3k\!\left(x^2 - \frac{2bx}{3} - \frac{a}{3}\right)$ | M1 | Factorising by taking $-3k$ out |
| $-3k\!\left(\!\left(x - \frac{b}{3}\right)^2 - \left(\frac{b}{3}\right)^2 - \frac{a}{3}\right)$ or quoting $\frac{-b}{2a}$ | M1d | Attempting to complete the square; dependent on previous M; $\left(x - \frac{b}{3}\right)^2 \pm c$ |
| $-3k\!\left(x - \frac{b}{3}\right)^2 + \frac{b^2k}{3} + ak$ | A1 | Correct completed square form |
| Max at $x = \frac{b}{3}$ | M1d | Selecting their $b/3$; dependent on previous method mark |
| $\frac{b}{3} = \frac{8}{3}$ | M1d | Putting their $\frac{b}{3} = \frac{8}{3}$; dependent on previous method mark |
| $b = 8$* | A1cso | Answer given; all steps must have been shown |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(2) = 0$: $k(2a + 32 - 8) = 0$ or $k(2a + 4b - 8) = 0$ oe | M1 | Attempting to form an equation using $F(2) = 0$, or $F(3) = 1$ or $F(3) - F(2) = 1$; need to substitute $x$ value and equate |
| $a = -12$ | A1 | May be implied by $k = 1/9$; do not award if M1 not given |
| $F(3) = 1$: $k(-36 + 72 - 27) = 1$ or $k(-36 + 9b - 27) = 1$ oe | M1 | Forming an equation using two of $F(2) = 0$, $F(3) = 1$, or $F(3) - F(2) = 1$ |
| $k = \frac{1}{9}$ | A1 | Allow equivalent fractions or awrt 0.111 |
---4. A continuous random variable $X$ has cumulative distribution function $\mathrm { F } ( x )$ given by
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c c }
0 & x < 2 \\
k \left( a x + b x ^ { 2 } - x ^ { 3 } \right) & 2 \leqslant x \leqslant 3 \\
1 & x > 3
\end{array} \right.$$
Given that the mode of $X$ is $\frac { 8 } { 3 }$
\begin{enumerate}[label=(\alph*)]
\item show that $b = 8$
\item find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2016 Q4 [10]}}