| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | State or write down basic properties |
| Difficulty | Easy -1.8 This is a trivial one-step calculation using E(X) = np for a binomial distribution. It requires only basic algebraic manipulation (5 = 0.04n, so n = 125) with no conceptual difficulty or problem-solving required. |
| Spec | 5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(P(N \geq 10) = 1 - P(N \leq 9)\) | M1 | Using or writing \(1 - P(N \leq 9)\) or \(1 - P(N < 10)\) |
| \(= 0.4126\) | A1 | awrt \(0.413\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(Y\) represents number of owls per \(200 \text{ km}^2 \Rightarrow Y \sim Po(1.8)\) | B1 | Using or writing \(Po(1.8)\) |
| \(P(Y=2) = \dfrac{e^{-1.8}1.8^2}{2!}\) | M1 A1 | M1: for a single term of the form \(\dfrac{e^{-\lambda}\lambda^2}{2!}\) with any value for \(\lambda\), or \(P(X \leq 2) - P(X \leq 1)\) |
| \(= 0.2678\) | A1: awrt \(0.268\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| Normal approximation | M1 | Using or writing normal approximation with mean \(= 450\) |
| \(\mu = 50 \times 9 = 450 \quad \sigma^2 = 450\) | M1 | Using or writing the mean \(=\) variance. Does not need to be \(450\). May be seen in the standardisation calculation. |
| \(P\!\left(X \geq 470\right) \approx 1 - P\!\left(Z < \dfrac{469.5 - 450}{\sqrt{450}}\right)\) | M1 | \(\pm\left(\dfrac{(470 \text{ or } 469.5 \text{ or } 470.5) - \textit{their mean}}{\textit{their sd}}\right)\); may be implied by a correct answer or \(z =\) awrt \(0.92\) |
| dM1 A1 | dep on previous M mark. Using continuity correction \(470 \pm 0.5\). May be implied by correct answer or \(z =\) awrt \(0.92\). A1: correct standardisation \(\dfrac{469.5 - 450}{\sqrt{450}}\) or awrt \(0.92\) or correct answer | |
| \(= 0.1788\) | A1 | awrt \(0.179\) (6) |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $P(N \geq 10) = 1 - P(N \leq 9)$ | M1 | Using or writing $1 - P(N \leq 9)$ or $1 - P(N < 10)$ |
| $= 0.4126$ | A1 | awrt $0.413$ |
---
### Part (b)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $Y$ represents number of owls per $200 \text{ km}^2 \Rightarrow Y \sim Po(1.8)$ | B1 | Using or writing $Po(1.8)$ |
| $P(Y=2) = \dfrac{e^{-1.8}1.8^2}{2!}$ | M1 A1 | M1: for a single term of the form $\dfrac{e^{-\lambda}\lambda^2}{2!}$ with any value for $\lambda$, or $P(X \leq 2) - P(X \leq 1)$ |
| $= 0.2678$ | | A1: awrt $0.268$ |
---
### Part (c)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| Normal approximation | M1 | Using or writing normal approximation with mean $= 450$ |
| $\mu = 50 \times 9 = 450 \quad \sigma^2 = 450$ | M1 | Using or writing the mean $=$ variance. Does not need to be $450$. May be seen in the standardisation calculation. |
| $P\!\left(X \geq 470\right) \approx 1 - P\!\left(Z < \dfrac{469.5 - 450}{\sqrt{450}}\right)$ | M1 | $\pm\left(\dfrac{(470 \text{ or } 469.5 \text{ or } 470.5) - \textit{their mean}}{\textit{their sd}}\right)$; may be implied by a correct answer or $z =$ awrt $0.92$ |
| | dM1 A1 | dep on previous M mark. Using continuity correction $470 \pm 0.5$. May be implied by correct answer or $z =$ awrt $0.92$. A1: correct standardisation $\dfrac{469.5 - 450}{\sqrt{450}}$ or awrt $0.92$ or correct answer |
| $= 0.1788$ | A1 | awrt $0.179$ **(6)** |\begin{enumerate}
\item It is estimated that $4 \%$ of people have green eyes. In a random sample of size $n$, the expected number of people with green eyes is 5 .\\
(a) Calculate the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q1}}