| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Probability |
| Type | Random cut distribution identification |
| Difficulty | Standard +0.3 This is a straightforward S2 geometric probability question requiring identification of a uniform distribution and basic probability calculations. Part (a) is simple recognition that L ~ U(20,40). Parts (b)-(d) involve routine probability calculations with linear transformations, requiring only standard A-level techniques without novel insight or complex multi-step reasoning. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(L\sim U[20,40]\) or \(f(l)=\begin{cases}\frac{1}{20} & 20\leq l\leq 40\\ 0 & \text{otherwise}\end{cases}\) | B1 B1 | 1st B1: Uniform/rectangular or \(f(l)=\frac{1}{20}\); 2nd B1: for \([20,40]\) or correct pdf fully specified |
| Answer | Marks | Guidance |
|---|---|---|
\(P(27.5| M1 A1 |
M1 for finding \(P(27.5 | |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{L}{4}\right)^2<64<\left(\frac{L}{4}\right)<8\); \(P(L<32)=\frac{32-20}{40-20}=0.6\) | M1, M1 A1oe | 1st M1: correct expression for area and comparison with 64; 2nd M1: using \(P(L<\text{'32'})\) with uniform distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{L}{4}\right)^2-\left(\frac{40-L}{4}\right)^2>81\); \(\frac{-1600+80L}{16}>81\); \(L>36.2\); \(P(L>36.2)=\frac{40-36.2}{40-20}=0.19\) | M1, A1oe, M1 A1oe | 1st M1: correct difference of two areas compared with 81; A1: \(L>36.2\); 2nd M1: using \(P(L>\text{'36.2'})\) with uniform distribution |
# Question 5:
## Part (a)
| $L\sim U[20,40]$ or $f(l)=\begin{cases}\frac{1}{20} & 20\leq l\leq 40\\ 0 & \text{otherwise}\end{cases}$ | B1 B1 | 1st B1: Uniform/rectangular or $f(l)=\frac{1}{20}$; 2nd B1: for $[20,40]$ or correct pdf fully specified |
## Part (b)
| $P(27.5<L<28.5)=\frac{28.5-27.5}{40-20}=\frac{1}{20}$ | M1 A1 | M1 for finding $P(27.5<L<28.5)$ from uniform distribution |
## Part (c)
| $\left(\frac{L}{4}\right)^2<64<\left(\frac{L}{4}\right)<8$; $P(L<32)=\frac{32-20}{40-20}=0.6$ | M1, M1 A1oe | 1st M1: correct expression for area and comparison with 64; 2nd M1: using $P(L<\text{'32'})$ with uniform distribution |
## Part (d)
| $\left(\frac{L}{4}\right)^2-\left(\frac{40-L}{4}\right)^2>81$; $\frac{-1600+80L}{16}>81$; $L>36.2$; $P(L>36.2)=\frac{40-36.2}{40-20}=0.19$ | M1, A1oe, M1 A1oe | 1st M1: correct difference of two areas compared with 81; A1: $L>36.2$; 2nd M1: using $P(L>\text{'36.2'})$ with uniform distribution |
---\begin{enumerate}
\item A string of length 40 cm is cut into 2 pieces at a random point. The continuous random variable $L$ represents the length of the longer piece of string.\\
(a) Write down the distribution of $L$\\
(b) Find the probability that the length of the longer piece of string is 28 cm to the nearest cm
\end{enumerate}
Each piece of string is used to form the perimeter of a square.\\
(c) Calculate the probability that the area of the larger square is less than $64 \mathrm {~cm} ^ { 2 }$\\
(d) Calculate the probability that the difference in area between the two squares is greater than $81 \mathrm {~cm} ^ { 2 }$\\
\hfill \mbox{\textit{Edexcel S2 2016 Q5 [11]}}