| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Maximum or minimum of uniforms |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics 2 question testing basic uniform distribution properties. Part (a) is direct probability calculation, part (b) requires finding w from E(H)=10 then multiplying independent probabilities (routine), and part (c) needs finding area as a function of k then taking expectation. All steps are standard applications of formulas with no novel insight required, though the geometric interpretation in (c) adds minor complexity. |
| Spec | 5.02a Discrete probability distributions: general5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(G>12) = \frac{3}{18} \left[= \frac{1}{6}\right]\) | B1 | Allow 0.167 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w = 18\) | B1 | 18 |
| Probability both greater than \(12 = \frac{1}{6} \times \frac{18-12}{18-2}\) | M1 | Correct calculation shown using their value for \(w\); may be implied by e.g. \(\frac{1}{6} \times \frac{3}{8}\) |
| \(= \frac{1}{16}\) | A1* | Allow 0.0625 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct triangle identified (isosceles, symmetrical about \(y\)-axis, highest point on positive \(y\)-axis, base vertices below \(x\)-axis) | M1 | Implied by all three coordinates or calculation to find area / awrt 28.1 |
| \(x\) coordinate of base of triangle \(= \pm\left(\frac{k+3}{4}\right)\) | M1 | For finding either \(x\) coordinate; do not accept numerical value for \(k\) but condone \(E(K)\) |
| Area \(= \frac{1}{2} \times 2 \times \frac{k+3}{4} \times (k+3)\) | M1 | Finding area of triangle in terms of \(k\); condone \(E(K)\) |
| \(E(A) = \int_5^{10} \frac{1}{5} \times \frac{1}{4}(k+3)^2\, dk\) | M1 | Using model correctly: \(\int_5^{10} f(k) \times \text{their area}\, dk\); or using \(E(K)=7.5\) and \(E(K^2)=\text{Var}(K)+(E(K))^2\) with area \(= \frac{1}{4}(k^2+6k+9)\) |
| \(= \frac{337}{12}\) | A1cso | awrt 28.1 |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(G>12) = \frac{3}{18} \left[= \frac{1}{6}\right]$ | B1 | Allow 0.167 or better |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = 18$ | B1 | 18 |
| Probability both greater than $12 = \frac{1}{6} \times \frac{18-12}{18-2}$ | M1 | Correct calculation shown using their value for $w$; may be implied by e.g. $\frac{1}{6} \times \frac{3}{8}$ |
| $= \frac{1}{16}$ | A1* | Allow 0.0625 oe |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct triangle identified (isosceles, symmetrical about $y$-axis, highest point on positive $y$-axis, base vertices below $x$-axis) | M1 | Implied by all three coordinates or calculation to find area / awrt 28.1 |
| $x$ coordinate of base of triangle $= \pm\left(\frac{k+3}{4}\right)$ | M1 | For finding either $x$ coordinate; do not accept numerical value for $k$ but condone $E(K)$ |
| Area $= \frac{1}{2} \times 2 \times \frac{k+3}{4} \times (k+3)$ | M1 | Finding area of triangle in terms of $k$; condone $E(K)$ |
| $E(A) = \int_5^{10} \frac{1}{5} \times \frac{1}{4}(k+3)^2\, dk$ | M1 | Using model correctly: $\int_5^{10} f(k) \times \text{their area}\, dk$; or using $E(K)=7.5$ and $E(K^2)=\text{Var}(K)+(E(K))^2$ with area $= \frac{1}{4}(k^2+6k+9)$ |
| $= \frac{337}{12}$ | A1cso | awrt 28.1 |
---\begin{enumerate}
\item The random variable $G$ has a continuous uniform distribution over the interval $[ - 3,15 ]$\\
(a) Calculate $\mathrm { P } ( G > 12 )$
\end{enumerate}
The random variable $H$ has a continuous uniform distribution over the interval [2, w] The random variables $G$ and $H$ are independent and $\mathrm { E } ( H ) = 10$\\
(b) Show that the probability that $G$ and $H$ are both greater than 12 is $\frac { 1 } { 16 }$
The random variable $A$ is the area on a coordinate grid bounded by
$$\begin{aligned}
& y = - 3 \\
& y = - 4 | x | + k
\end{aligned}$$
where $k$ is a value from the continuous uniform distribution over the interval [5,10]\\
(c) Calculate the expected value of $A$
\hfill \mbox{\textit{Edexcel FS2 2024 Q4 [9]}}