| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard techniques: (a) uses cumulative tables/calculator, (b) is direct probability calculation, (c) requires simple rate adjustment (9/6=1.5 per 10 min), and (d) applies normal approximation for large λ. All parts are routine S2 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim Po(9)\), \(P(X \leqslant 10) = 0.7060\) | ||
| \(v = 11\) | B1 | Not \(v > 11\) or any inequality |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(4 \leqslant X \leqslant 11) = P(X \leqslant 11) - P(X \leqslant 3)\) | M1 | For writing or using \(P(X \leqslant 11) - P(X \leqslant 3)\); may be implied by correct answer |
| \(0.8030 - 0.0212 = 0.7818\) | A1 | awrt \(\mathbf{0.782}\) |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Po(1.5)\) | B1 | May be implied by \(P(X=0)=0.2231\) or \(P(X=1)=0.3346...\) or \(P(X \leqslant 1)=0.5578\) |
| \(P(Y > 1) = 1 - P(Y \leqslant 1)\) or \(1 - 0.5578\) | M1 | For writing \(1 - P(Y \leqslant 1)\) or \(1 - 0.5578\); condone use of \(X\) or any other letter |
| \(= 0.4422\) | A1 | awrt \(\mathbf{0.442}\); correct answer only scores 3/3 |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(W\) = number of visits to school website in 8 hours | ||
| \(W \sim Po(72)\) approximately \(N(72, 72)\) | M1 A1 | 1st M1: using normal approximation with \(\mu = 72\); 1st A1: for \(\mu = 72\) and \(\sigma^2 = 72\) or \(\sigma = \sqrt{72}\) |
| \(P(W > 80) = P\!\left(Z > \dfrac{80.5 - 72}{\sqrt{72}}\right)\) | M1 M1 | 2nd M1: using \(80.5\) or \(79.5\); 3rd M1: standardising using \(79.5\), \(80.5\) or \(80\) with their mean and standard deviation |
| \(= P(Z > 1.00...)\) | ||
| \(= 0.1587\) (or \(0.15824\) from calculator) | A1 | awrt \(\mathbf{0.159/0.158}\) |
| (5) | ||
| [11] |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim Po(9)$, $P(X \leqslant 10) = 0.7060$ | | |
| $v = 11$ | **B1** | Not $v > 11$ or any inequality |
| | **(1)** | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(4 \leqslant X \leqslant 11) = P(X \leqslant 11) - P(X \leqslant 3)$ | **M1** | For writing or using $P(X \leqslant 11) - P(X \leqslant 3)$; may be implied by correct answer |
| $0.8030 - 0.0212 = 0.7818$ | **A1** | awrt $\mathbf{0.782}$ |
| | **(2)** | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Po(1.5)$ | **B1** | May be implied by $P(X=0)=0.2231$ or $P(X=1)=0.3346...$ or $P(X \leqslant 1)=0.5578$ |
| $P(Y > 1) = 1 - P(Y \leqslant 1)$ or $1 - 0.5578$ | **M1** | For writing $1 - P(Y \leqslant 1)$ or $1 - 0.5578$; condone use of $X$ or any other letter |
| $= 0.4422$ | **A1** | awrt $\mathbf{0.442}$; correct answer only scores 3/3 |
| | **(3)** | |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $W$ = number of visits to school website in 8 hours | | |
| $W \sim Po(72)$ approximately $N(72, 72)$ | **M1 A1** | 1st M1: using normal approximation with $\mu = 72$; 1st A1: for $\mu = 72$ and $\sigma^2 = 72$ or $\sigma = \sqrt{72}$ |
| $P(W > 80) = P\!\left(Z > \dfrac{80.5 - 72}{\sqrt{72}}\right)$ | **M1 M1** | 2nd M1: using $80.5$ or $79.5$; 3rd M1: standardising using $79.5$, $80.5$ or $80$ with their mean and standard deviation |
| $= P(Z > 1.00...)$ | | |
| $= 0.1587$ (or $0.15824$ from calculator) | **A1** | awrt $\mathbf{0.159/0.158}$ |
| | **(5)** | |
| | **[11]** | |\begin{enumerate}
\item During a typical day, a school website receives visits randomly at a rate of 9 per hour.
\end{enumerate}
The probability that the school website receives fewer than $v$ visits in a randomly selected one hour period is less than 0.75\\
(a) Find the largest possible value of $v$\\
(b) Find the probability that in a randomly selected one hour period, the school website receives at least 4 but at most 11 visits.\\
(c) Find the probability that in a randomly selected 10 minute period, the school website receives more than 1 visit.\\
(d) Using a suitable approximation, find the probability that in a randomly selected 8 hour period the school website receives more than 80 visits.\\
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\hfill \mbox{\textit{Edexcel S2 2016 Q1 [11]}}