| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF with additional constraints |
| Difficulty | Challenging +1.2 This is a multi-part CDF question requiring differentiation to find the pdf, using the mode condition to establish relationships between constants, and applying continuity conditions. While it involves several steps and algebraic manipulation, the techniques are standard S2 material (differentiating to get pdf, setting f'(x)=0 for mode, using boundary conditions). The 'show that' format provides targets to work towards, reducing problem-solving demand compared to open-ended questions. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = [k](a + 3bx^2 - 4x^3)\) | M1 | For attempting to differentiate \(x^n \to x^{n-1}\). Condone missing \(k\) (May be implied by 2nd M1) |
| \([k](6bx - 12x^2) = 0\) | M1 | For correctly differentiating twice and equating to zero. Condone missing \(k\) |
| \(9b - 27 = 0 \Rightarrow b = 3\) or \(6\times3\times1.5 - 12\times1.5^2 = 0 \Rightarrow \therefore b = 3\) | A1* | Substituting \(x=1.5\) leading to a correct linear equation in \(b\) leading to \(b=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a + 3 - 1 - 4 = 0 \Rightarrow [a = 2]\) | B1* | For correctly using \(F(1)=0\) to form an equation in \(a\) (May be seen in part (a)) and substitution of \(b=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k\left(2\times2 + 3\times2^3 - 2^4 - 4\right) = 1 \Rightarrow k = \frac{1}{8}\) | M1 | For using \(F(2)=1\) to form a correct equation in terms of \(k\) only. May be seen in any part of the question |
| \(F(1.4) = 0.3988\ldots\), \(F(1.5) = 0.5078\ldots\) | M1A1 | For a calculation of \(F(1.4)\) or \(F(1.5)\) correct to 2 sf (If \(F(x)=0\) used then allow 1 sf or better). Allow \(F(1.4) =\) awrt \(3.190k\) or \(F(1.5) =\) awrt \(4.063k\) |
| \(0.399 < 0.5 < 0.508\), therefore the median lies between 1.4 and 1.5 | A1 | For a calculation of \(F(1.4)\) and \(F(1.5)\) correct to 2 sf. Dependent on previous A1. For a correct comparison and conclusion. Allow comparisons in words e.g. For \(F(X)=0\) a comment about a change in sign implies a comparison with 0 |
| Alternative: \(x_1 = 2.91\ldots\), \(x_2 = 1.49\ldots\), \(x_3 = -0.70\ldots\) So \(x = 1.49\ldots\) as \(1 \leqslant x \leqslant 2\) | M1 A1 | For solving the given equation. May be implied by \(2.91\ldots\) or \(1.49\ldots\) or \(-0.70\ldots\) For \(x=1.49\ldots\) identified as being in the range specified by the CDF. May be implied by rejecting the other solutions |
| \(1.4 < 1.49\ldots < 1.5\) [therefore, the median lies between 1.4 and 1.5] | dA1 | Dependent on previous A1. For a correct comparison and conclusion |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = [k](a + 3bx^2 - 4x^3)$ | M1 | For attempting to differentiate $x^n \to x^{n-1}$. Condone missing $k$ (May be implied by 2nd M1) |
| $[k](6bx - 12x^2) = 0$ | M1 | For correctly differentiating twice and equating to zero. Condone missing $k$ |
| $9b - 27 = 0 \Rightarrow b = 3$ or $6\times3\times1.5 - 12\times1.5^2 = 0 \Rightarrow \therefore b = 3$ | A1* | Substituting $x=1.5$ leading to a correct linear equation in $b$ leading to $b=3$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a + 3 - 1 - 4 = 0 \Rightarrow [a = 2]$ | B1* | For correctly using $F(1)=0$ to form an equation in $a$ (May be seen in part (a)) and substitution of $b=3$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k\left(2\times2 + 3\times2^3 - 2^4 - 4\right) = 1 \Rightarrow k = \frac{1}{8}$ | M1 | For using $F(2)=1$ to form a correct equation in terms of $k$ only. May be seen in any part of the question |
| $F(1.4) = 0.3988\ldots$, $F(1.5) = 0.5078\ldots$ | M1A1 | For a calculation of $F(1.4)$ **or** $F(1.5)$ correct to 2 sf (If $F(x)=0$ used then allow 1 sf or better). Allow $F(1.4) =$ awrt $3.190k$ or $F(1.5) =$ awrt $4.063k$ |
| $0.399 < 0.5 < 0.508$, therefore the median lies between 1.4 and 1.5 | A1 | For a calculation of $F(1.4)$ **and** $F(1.5)$ correct to 2 sf. Dependent on previous A1. For a correct comparison and conclusion. Allow comparisons in words e.g. For $F(X)=0$ a comment about a change in sign implies a comparison with 0 |
| **Alternative:** $x_1 = 2.91\ldots$, $x_2 = 1.49\ldots$, $x_3 = -0.70\ldots$ So $x = 1.49\ldots$ as $1 \leqslant x \leqslant 2$ | M1 A1 | For solving the given equation. May be implied by $2.91\ldots$ or $1.49\ldots$ or $-0.70\ldots$ For $x=1.49\ldots$ identified as being in the range specified by the CDF. May be implied by rejecting the other solutions |
| $1.4 < 1.49\ldots < 1.5$ [therefore, the median lies between 1.4 and 1.5] | dA1 | Dependent on previous A1. For a correct comparison and conclusion |\begin{enumerate}
\item A continuous random variable $X$ has cumulative distribution function $\mathrm { F } ( x )$ given by
\end{enumerate}
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c c }
0 & x < 1 \\
k \left( a x + b x ^ { 3 } - x ^ { 4 } - 4 \right) & 1 \leqslant x \leqslant 2 \\
1 & x > 2
\end{array} \right.$$
where $a$, $b$ and $k$ are non-zero constants.\\
Given that the mode of $X$ is 1.5\\
(a) show that $b = 3$\\
(b) Hence show that $a = 2$\\
(c) Show that the median of $X$ lies between 1.4 and 1.5
\hfill \mbox{\textit{Edexcel S2 2024 Q7 [8]}}