| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - large lambda direct |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard techniques: direct probability calculation, rate adjustment for different time periods, and normal approximation for large λ. All methods are routine S2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02i 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 |
|---|---|
| (a) Let \(X =\) no. of failures per hour \(\therefore X \sim \text{Po}(3)\); \(P(X = 0) = 0.0498\) | A1 |
| (b) Let \(Y =\) no. of failures per half-hour \(\therefore Y \sim \text{Po}(1.5)\); \(P(Y > 4) = 1 - P(Y \leq 4) = 1 - 0.9814 = 0.0186\) | M1 M1 A1 |
| (c) Let \(F =\) no. of failures per 24 hrs \(\therefore F \sim \text{Po}(72)\); \(N\) approx. \(\sim N(72, 72)\); \(P(F < 60) = P(G < 59.5) = P(Z < \frac{59.5 - 72}{\sqrt{72}}) = P(Z < -1.47) = 1 - 0.9292 = 0.0708\) | M1 M1 M1 A1 A1 |
| (d) \(P(F = 72) = P(71.5 < G < 72.5) = P(Z < \frac{72.5 - 72}{\sqrt{72}}) - P(Z < \frac{71.5 - 72}{\sqrt{72}}) = P(Z < 0.06) - P(Z < -0.06) = 0.5239 - 0.4761 = 0.0478\) | M1 M1 A1 A1 |
(a) Let $X =$ no. of failures per hour $\therefore X \sim \text{Po}(3)$; $P(X = 0) = 0.0498$ | A1 |
(b) Let $Y =$ no. of failures per half-hour $\therefore Y \sim \text{Po}(1.5)$; $P(Y > 4) = 1 - P(Y \leq 4) = 1 - 0.9814 = 0.0186$ | M1 M1 A1 |
(c) Let $F =$ no. of failures per 24 hrs $\therefore F \sim \text{Po}(72)$; $N$ approx. $\sim N(72, 72)$; $P(F < 60) = P(G < 59.5) = P(Z < \frac{59.5 - 72}{\sqrt{72}}) = P(Z < -1.47) = 1 - 0.9292 = 0.0708$ | M1 M1 M1 A1 A1 |
(d) $P(F = 72) = P(71.5 < G < 72.5) = P(Z < \frac{72.5 - 72}{\sqrt{72}}) - P(Z < \frac{71.5 - 72}{\sqrt{72}}) = P(Z < 0.06) - P(Z < -0.06) = 0.5239 - 0.4761 = 0.0478$ | M1 M1 A1 A1 |
---It is believed that the number of sets of traffic lights that fail per hour in a particular large city follows a Poisson distribution with a mean of 3.
Find the probability that
\begin{enumerate}[label=(\alph*)]
\item there will be no failures in a one-hour period, [1 mark]
\item there will be more than 4 failures in a 30-minute period. [3 marks]
\end{enumerate}
Using a suitable approximation, find the probability that in a 24-hour period there will be
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item less than 60 failures, [5 marks]
\item exactly 72 failures. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q4 [13]}}