| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Compare mean and median using probability |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics question requiring standard techniques: finding the mode by differentiating the pdf, comparing median to a value using cumulative probability (no calculation needed, just reasoning), and computing variance using the standard formula Var(aY) = a²Var(Y). Part (a) involves routine calculus, part (b) is conceptual reasoning about the median, and part (c) applies the variance formula with given E(Y²). All techniques are standard for FS2, with no novel problem-solving required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sketch or differentiation to find mode of \(Y\); \(f'(y) = \frac{1}{24}(2-2y) = 0\) | M1 | Attempt to find mode e.g. differentiation, complete the square or sketch |
| Mode occurs at \(Y = 1\) | A1* | Fully correct justification, must show or explain why a max e.g. reference to negative quadratic (and 1 is in \([0,3]\)). Allow "decreasing function" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| By symmetry \(P(Y < 2) = 2 \times \frac{13}{36} = \frac{13}{18}\) (or 0.72 or better) | M1 | Use of symmetry or other method e.g. calculator to determine \(P(Y<2)\) |
| Median is less than 2 since \(\frac{13}{18} > \frac{1}{2}\) | A1 | For median is less than 2 with correct reasoning. ALT: \(F(y) = \frac{1}{24}\left(-\frac{y^3}{3} + y^2 + 8y\right)\) and \(F(y) = \frac{1}{2}\); awrt 1.37 with evidence that \(F(y)=\frac{1}{2}\) attempted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y) = \int_0^3 \frac{1}{24} y(y+2)(4-y)\,dy\) or \(\int_0^3 \frac{1}{24}(8y + 2y^2 - y^3)\,dy\) | M1 | Attempt to integrate \(yf(y)\) – ignore limits. Must see algebraic integration, need at least one \(y^n \to y^{n+1}\) |
| \(E(Y) = \frac{1}{24}\left[4y^2 + \frac{2}{3}y^3 - \frac{y^4}{4}\right]_0^3 \to \frac{1}{24}\left[(36+18-\frac{81}{4})-(0)\right] = \frac{45}{32}\) | M1 | Substitution of correct limits into integral of \(yf(y)\). Don't need to see limits if intention is clear and answer correct |
| \(\text{Var}(Y) = E(Y^2) - [E(Y)]^2 = \frac{213}{80} - \left(\frac{45}{32}\right)^2 = \frac{3507}{5120}\) | M1 | Attempt to find variance using \(E(Y^2) - [E(Y)]^2\) |
| \(\text{Var}(2Y) = 4\text{Var}(Y)\) | M1 | Use of \(\text{Var}(2Y) = 2^2\text{Var}(Y)\). Independent of 1st–3rd M1 |
| \(= \frac{3507}{1280} = 2.73984\ldots\) awrt 2.74 | A1 | Dep on all Ms |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketch or differentiation to find mode of $Y$; $f'(y) = \frac{1}{24}(2-2y) = 0$ | M1 | Attempt to find mode e.g. differentiation, complete the square or sketch |
| Mode occurs at $Y = 1$ | A1* | Fully correct justification, must show or explain why a max e.g. reference to negative quadratic (and 1 is in $[0,3]$). Allow "decreasing function" |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| By symmetry $P(Y < 2) = 2 \times \frac{13}{36} = \frac{13}{18}$ (or 0.72 or better) | M1 | Use of symmetry or other method e.g. calculator to determine $P(Y<2)$ |
| Median is less than 2 since $\frac{13}{18} > \frac{1}{2}$ | A1 | For median is less than 2 with correct reasoning. ALT: $F(y) = \frac{1}{24}\left(-\frac{y^3}{3} + y^2 + 8y\right)$ and $F(y) = \frac{1}{2}$; awrt 1.37 with evidence that $F(y)=\frac{1}{2}$ attempted |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y) = \int_0^3 \frac{1}{24} y(y+2)(4-y)\,dy$ or $\int_0^3 \frac{1}{24}(8y + 2y^2 - y^3)\,dy$ | M1 | Attempt to integrate $yf(y)$ – ignore limits. Must see algebraic integration, need at least one $y^n \to y^{n+1}$ |
| $E(Y) = \frac{1}{24}\left[4y^2 + \frac{2}{3}y^3 - \frac{y^4}{4}\right]_0^3 \to \frac{1}{24}\left[(36+18-\frac{81}{4})-(0)\right] = \frac{45}{32}$ | M1 | Substitution of correct limits into integral of $yf(y)$. Don't need to see limits if intention is clear and answer correct |
| $\text{Var}(Y) = E(Y^2) - [E(Y)]^2 = \frac{213}{80} - \left(\frac{45}{32}\right)^2 = \frac{3507}{5120}$ | M1 | Attempt to find variance using $E(Y^2) - [E(Y)]^2$ |
| $\text{Var}(2Y) = 4\text{Var}(Y)$ | M1 | Use of $\text{Var}(2Y) = 2^2\text{Var}(Y)$. Independent of 1st–3rd M1 |
| $= \frac{3507}{1280} = 2.73984\ldots$ awrt **2.74** | A1 | Dep on all Ms |
---\begin{enumerate}
\item The continuous random variable $Y$ has probability density function
\end{enumerate}
$$f ( y ) = \left\{ \begin{array} { c c }
\frac { 1 } { 24 } ( y + 2 ) ( 4 - y ) & 0 \leqslant y \leqslant 3 \\
0 & \text { otherwise }
\end{array} \right.$$
(a) Show that the mode of $Y$ is 1 , justifying your reasoning.
Given that $\mathrm { P } ( Y < 1 ) = \frac { 13 } { 36 }$\\
(b) determine whether the median of $Y$ is less than, equal to, or greater than 2 Give a reason for your answer.
Given that $\mathrm { E } \left( Y ^ { 2 } \right) = \frac { 213 } { 80 }$\\
(c) find, using algebraic integration, $\operatorname { Var } ( 2 Y )$
\hfill \mbox{\textit{Edexcel FS2 AS 2024 Q3 [9]}}