| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF with additional constraints |
| Difficulty | Standard +0.8 This is a multi-part S2 question requiring systematic application of CDF properties (continuity, F(k)=1), derivation of pdf by differentiation, calculation of E(X) and Var(X) using integration, and solving a system of three equations in three unknowns. While each individual step is standard, the extended algebraic manipulation across five connected parts and the need to carefully track relationships between a, b, and k elevates this above routine exercises. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([F(k)=1 \Rightarrow] ak + bk^2 = 1 \Rightarrow ak = 1 - bk^2\) | B1* | Answer is given so no incorrect working can be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(f(x) = a + 2bx\) | B1 | For a correct expression for \(f(x)\) (may be implied by correct \(E(X)\)) |
| \(E(X) = \int_0^k (ax + 2bx^2)\, dx \Rightarrow \left[\frac{ax^2}{2} + \frac{2bx^3}{3}\right]_0^k = \frac{6}{5}\) | M1 | Attempt to integrate \(x\cdot f(x)\); at least one \(x^n \to x^{n+1}\); f.t. their \(f(x)\). Integrating \(x\cdot F(x)\) is M0 |
| \(\frac{ak^2}{2} + \frac{2bk^3}{3} = \frac{6}{5}\) | dM1, A1 | Dependent on previous M1; equating to \(\frac{6}{5}\) and substituting \(k\) (no need to see substitution of lower limit 0) |
| \(15ak^2 + 20bk^3 = 36\) | ||
| \(15k(1-bk^2) + 20bk^3 = 36\) | M1 | For substitution of \(ak = 1 - bk^2\) into their equation |
| \(5bk^3 = 36 - 15k\) | A1* | Answer is given so no incorrect working can be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(E(X^2) = \int_0^k (ax^2 + 2bx^3)\, dx \Rightarrow \left[\frac{ax^3}{3} + \frac{bx^4}{2}\right]_0^k\) | M1 | Attempt to integrate \(x^2 f(x)\); at least one \(x^n \to x^{n+1}\); f.t. their \(f(x)\). \(x^2 F(x)\) is M0 |
| \(\text{Var}(X) = \frac{ak^3}{3} + \frac{bk^4}{2} - \frac{36}{25} = \frac{22}{75}\) | dM1, A1 | Dependent on previous M1; substitution of correct limits and subtraction of \(\frac{36}{25} = \frac{22}{75}\) |
| \(10ak^3 + 15bk^4 = 52\) | ||
| \(10k^2(1-bk^2) + 15bk^4 = 52\) | M1 | For substitution of \(ak = 1 - bk^2\) into their equation |
| \(5bk^4 = 52 - 10k^2\) | A1* | Answer is given so no incorrect working can be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{k} = \frac{36-15k}{52-10k^2}\) | M1 | For solving simultaneously to set up an equation in \(k\) only |
| \(5k^2 - 36k + 52 = 0\) | A1 | For a correct 3-term quadratic |
| \((k-2)(5k-26) = 0\) | M1 | For solving their 3-term quadratic by factorising, completing the square or using formula. \(k=5.2\) implies M1A1M1 |
| \(k = 2\) | A1 | 2 only cao. Correct answer on its own scores 4 out of 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \('40'b = 36 - '30' \Rightarrow b = \frac{3}{20}\) or \('80'b = 52 - '40' \Rightarrow b = \frac{3}{20}\) | B1ft | For \(b = \frac{3}{20}\) f.t. their \(k\); \(b = \frac{36-15k}{5k^3}\). Common f.t. answer \(b = \frac{-525}{8788} \approx -0.0597\) from \(k=5.2\) |
| \(2a + \frac{3}{5} = 1 \Rightarrow a = \frac{1}{5}\) | B1ft | For \(a = \frac{1}{5}\) f.t. their \(k\) and their \(b\); \(a = \frac{1-bk^2}{k}\). Common f.t. answer \(a = \frac{85}{169} \approx 0.503\) from \(k=5.2\) |
# Question 6:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[F(k)=1 \Rightarrow] ak + bk^2 = 1 \Rightarrow ak = 1 - bk^2$ | B1* | Answer is given so no incorrect working can be seen |
**Total: (1)**
---
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $f(x) = a + 2bx$ | B1 | For a correct expression for $f(x)$ (may be implied by correct $E(X)$) |
| $E(X) = \int_0^k (ax + 2bx^2)\, dx \Rightarrow \left[\frac{ax^2}{2} + \frac{2bx^3}{3}\right]_0^k = \frac{6}{5}$ | M1 | Attempt to integrate $x\cdot f(x)$; at least one $x^n \to x^{n+1}$; f.t. their $f(x)$. Integrating $x\cdot F(x)$ is M0 |
| $\frac{ak^2}{2} + \frac{2bk^3}{3} = \frac{6}{5}$ | dM1, A1 | Dependent on previous M1; equating to $\frac{6}{5}$ and substituting $k$ (no need to see substitution of lower limit 0) |
| $15ak^2 + 20bk^3 = 36$ | | |
| $15k(1-bk^2) + 20bk^3 = 36$ | M1 | For substitution of $ak = 1 - bk^2$ into their equation |
| $5bk^3 = 36 - 15k$ | A1* | Answer is given so no incorrect working can be seen |
**Total: (6)**
---
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $E(X^2) = \int_0^k (ax^2 + 2bx^3)\, dx \Rightarrow \left[\frac{ax^3}{3} + \frac{bx^4}{2}\right]_0^k$ | M1 | Attempt to integrate $x^2 f(x)$; at least one $x^n \to x^{n+1}$; f.t. their $f(x)$. $x^2 F(x)$ is M0 |
| $\text{Var}(X) = \frac{ak^3}{3} + \frac{bk^4}{2} - \frac{36}{25} = \frac{22}{75}$ | dM1, A1 | Dependent on previous M1; substitution of correct limits and subtraction of $\frac{36}{25} = \frac{22}{75}$ |
| $10ak^3 + 15bk^4 = 52$ | | |
| $10k^2(1-bk^2) + 15bk^4 = 52$ | M1 | For substitution of $ak = 1 - bk^2$ into their equation |
| $5bk^4 = 52 - 10k^2$ | A1* | Answer is given so no incorrect working can be seen |
**Total: (5)**
---
## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{k} = \frac{36-15k}{52-10k^2}$ | M1 | For solving simultaneously to set up an equation in $k$ only |
| $5k^2 - 36k + 52 = 0$ | A1 | For a correct 3-term quadratic |
| $(k-2)(5k-26) = 0$ | M1 | For solving their 3-term quadratic by factorising, completing the square or using formula. $k=5.2$ implies M1A1M1 |
| $k = 2$ | A1 | 2 only cao. Correct answer on its own scores 4 out of 4 |
**Total: (4)**
---
## Part (e):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $'40'b = 36 - '30' \Rightarrow b = \frac{3}{20}$ or $'80'b = 52 - '40' \Rightarrow b = \frac{3}{20}$ | B1ft | For $b = \frac{3}{20}$ f.t. their $k$; $b = \frac{36-15k}{5k^3}$. Common f.t. answer $b = \frac{-525}{8788} \approx -0.0597$ from $k=5.2$ |
| $2a + \frac{3}{5} = 1 \Rightarrow a = \frac{1}{5}$ | B1ft | For $a = \frac{1}{5}$ f.t. their $k$ and their $b$; $a = \frac{1-bk^2}{k}$. Common f.t. answer $a = \frac{85}{169} \approx 0.503$ from $k=5.2$ |
**Total: (2)**
**Question Total: 18 marks**\begin{enumerate}
\item The continuous random variable $X$ has cumulative distribution function
\end{enumerate}
$$\mathrm { F } ( x ) = \left\{ \begin{array} { l r }
0 & x < 0 \\
a x + b x ^ { 2 } & 0 \leqslant x \leqslant k \\
1 & x > k
\end{array} \right.$$
where $a , b$ and $k$ are positive constants.\\
(a) Show that $a k = 1 - b k ^ { 2 }$
Using part (a) and given that $\mathrm { E } ( X ) = \frac { 6 } { 5 }$\\
(b) show that $5 b k ^ { 3 } = 36 - 15 k$
Using part (a) and given that $\mathrm { E } ( X ) = \frac { 6 } { 5 }$ and $\operatorname { Var } ( X ) = \frac { 22 } { 75 }$\\
(c) show that $5 b k ^ { 4 } = 52 - 10 k ^ { 2 }$
Given that $k < 3$\\
(d) find the value of $k$\\
(e) Hence find the value of $a$ and the value of $b$
\hfill \mbox{\textit{Edexcel S2 2023 Q6 [18]}}