| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Standard +0.8 This multi-part Poisson question requires understanding of rate scaling across different time periods, combining Poisson with binomial probability (part c), and applying normal approximation to Poisson (part d). While individual parts use standard techniques, the variety of approaches and the conceptual leap in part (c) to recognize the binomial structure elevate this above routine S2 questions. |
| 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 probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| \(X\) = no. of customers arriving in 10 minute period; \(X \sim Po(3)\) | ||
| \(P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.6472 = 0.3528\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(Y\) = no. of customers in 30 minute period; \(Y \sim Po(9)\) | B1 | |
| \(P(Y \leq 7) = 0.3239\) | M1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p\) = probability of no customers in 5 minute period \(= e^{-1.5}\) | B1 | |
| \(C\) = number of 5 minute periods with no customers; \(C \sim B(6, p)\) | M1 | |
| \(P(C \leq 1) = (1-p)^6 + 6(1-p)^5 p = 0.59866\ldots\) | M1, M1 A1, A1 | accept awrt 0.599 (6) |
| Answer | Marks | Guidance |
|---|---|---|
| \(W\) = no. of customers on Wednesday morning; \(3\frac{1}{2}\) hours \(= 210\) minutes \(\therefore W \sim Po(63)\) | B1 | '63' |
| Normal approximation: \(W \approx \sim N(63, (\sqrt{63})^2)\) | M1 A1 | |
| \(P(W > 49) \approx P(W \geq 49.5)\) | M1 | \(\pm\frac{1}{2}\) |
| \(= P\left(Z \geq \frac{49.5 - 63}{\sqrt{63}}\right)\) | M1 | standardising |
| \(= P(Z \geq -1.7008)\) | A1 | |
| \(= 0.9554\) (tables) | A1 | accept awrt 0.955 or 0.956 (7) |
## Question 6:
### Part (a):
| $X$ = no. of customers arriving in 10 minute period; $X \sim Po(3)$ | | |
|---|---|---|
| $P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.6472 = 0.3528$ | M1 A1 | (2) |
### Part (b):
| $Y$ = no. of customers in 30 minute period; $Y \sim Po(9)$ | B1 | |
|---|---|---|
| $P(Y \leq 7) = 0.3239$ | M1 A1 | (3) |
### Part (c):
| $p$ = probability of no customers in 5 minute period $= e^{-1.5}$ | B1 | |
|---|---|---|
| $C$ = number of 5 minute periods with no customers; $C \sim B(6, p)$ | M1 | |
| $P(C \leq 1) = (1-p)^6 + 6(1-p)^5 p = 0.59866\ldots$ | M1, M1 A1, A1 | accept awrt 0.599 (6) |
### Part (d):
| $W$ = no. of customers on Wednesday morning; $3\frac{1}{2}$ hours $= 210$ minutes $\therefore W \sim Po(63)$ | B1 | '63' |
|---|---|---|
| Normal approximation: $W \approx \sim N(63, (\sqrt{63})^2)$ | M1 A1 | |
| $P(W > 49) \approx P(W \geq 49.5)$ | M1 | $\pm\frac{1}{2}$ |
| $= P\left(Z \geq \frac{49.5 - 63}{\sqrt{63}}\right)$ | M1 | standardising |
| $= P(Z \geq -1.7008)$ | A1 | |
| $= 0.9554$ (tables) | A1 | accept awrt 0.955 or 0.956 (7) |6. On a typical weekday morning customers arrive at a village post office independently and at a rate of 3 per 10 minute period.
Find the probability that
\begin{enumerate}[label=(\alph*)]
\item at least 4 customers arrive in the next 10 minutes,
\item no more than 7 customers arrive between 11.00 a.m. and 11.30 a.m.
The period from 11.00 a.m. to 11.30 a.m. next Tuesday morning will be divided into 6 periods of 5 minutes each.
\item Find the probability that no customers arrive in at most one of these periods.
The post office is open for $3 \frac { 1 } { 2 }$ hours on Wednesday mornings.
\item Using a suitable approximation, estimate the probability that more than 49 customers arrive at the post office next Wednesday morning.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [18]}}