| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | E(g(X)) and Var(g(X)) by integration |
| Difficulty | Challenging +1.2 This question requires setting up E(g(X)) where g(L) = 2L + 80/L (perimeter formula using the area constraint), then integrating over [4,10]. While it involves a non-linear transformation and algebraic integration of a rational function, the setup is straightforward and the integration is routine (polynomial + logarithm). It's moderately harder than average due to the multi-step reasoning and algebraic manipulation required, but remains a standard Further Stats application. |
| Spec | 5.02a Discrete probability distributions: general5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P = 2\left(L + \frac{40}{L}\right)\) | M1 | 3.1a — Finding expression for perimeter in terms of \(L\) |
| \(f(l) = \frac{1}{6}\) | B1 | 1.1b — Correct distribution for \(L\) (may be implied by \(E(L)=7\)) |
| \(E(P) = E\!\left(2\!\left(L+\frac{40}{L}\right)\right) = \int_4^{10} \frac{1}{6}\!\left(2l + \frac{80}{l}\right)\mathrm{d}l\) | M1 | 2.1 — Setting up integral for expectation of perimeter |
| \(= \left[\frac{1}{6}l^2 + \frac{40}{3}\ln\lvert l\rvert\right]_4^{10}\) | M1A1 | 1.1b — Correct integration |
| \(= \left(\frac{1}{6}(100) + \frac{40}{3}\ln 10\right) - \left(\frac{1}{6}(16) + \frac{40}{3}\ln 4\right)\) | M1 | 1.1b — depM1: use of correct limits 10 and 4 |
| \(= 26.217\ldots\) awrt \(26.2\) | A1 | 1.1b — Exact value is \(14 + \frac{40\ln(2.5)}{3}\) |
# Question 7:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P = 2\left(L + \frac{40}{L}\right)$ | M1 | 3.1a — Finding expression for perimeter in terms of $L$ |
| $f(l) = \frac{1}{6}$ | B1 | 1.1b — Correct distribution for $L$ (may be implied by $E(L)=7$) |
| $E(P) = E\!\left(2\!\left(L+\frac{40}{L}\right)\right) = \int_4^{10} \frac{1}{6}\!\left(2l + \frac{80}{l}\right)\mathrm{d}l$ | M1 | 2.1 — Setting up integral for expectation of perimeter |
| $= \left[\frac{1}{6}l^2 + \frac{40}{3}\ln\lvert l\rvert\right]_4^{10}$ | M1A1 | 1.1b — Correct integration |
| $= \left(\frac{1}{6}(100) + \frac{40}{3}\ln 10\right) - \left(\frac{1}{6}(16) + \frac{40}{3}\ln 4\right)$ | M1 | 1.1b — depM1: use of correct limits 10 and 4 |
| $= 26.217\ldots$ awrt $26.2$ | A1 | 1.1b — Exact value is $14 + \frac{40\ln(2.5)}{3}$ |
---\begin{enumerate}
\item A rectangle is to have an area of $40 \mathrm {~cm} ^ { 2 }$
\end{enumerate}
The length of the rectangle, $L \mathrm {~cm}$, follows a continuous uniform distribution over the interval [4, 10]
Find the expected value of the perimeter of the rectangle.\\
Use algebraic integration, rather than your calculator, to evaluate any definite integrals.
\hfill \mbox{\textit{Edexcel FS2 2022 Q7 [7]}}