| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Consecutive non-overlapping periods |
| Difficulty | Challenging +1.2 Part (a) is routine Poisson probability calculation. Part (b) requires recognizing a binomial distribution of Poisson events, which is standard S2 material. Part (c) is the challenging element—it requires conditional probability insight with Poisson processes, recognizing that given a total of 4 muffins, their distribution across time intervals follows a multinomial pattern. This conceptual leap beyond standard textbook exercises elevates it above average difficulty, though the actual calculation once set up is manageable. |
| Spec | 5.02i 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 modelling |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(R \sim \text{Po}(8)\) | B1 | B1 writing or using Po(8) (may be implied by one correct probability from 0.5925, 0.0424 0.4530 or 0.0996) |
| \(P(4 \leqslant R \leqslant 8) = P(R \leqslant 8) - P(R \leqslant 3) = 0.5925 - 0.0424\) | M1 | M1 writing or using \(P(R \leqslant 8) - P(R \leqslant 3)\) |
| \(= 0.5501 = \text{awrt } 0.550\) | A1 | A1 awrt 0.550 (calc: 0.55016...) correct answer scores 3 out of 3 |
| (3) | ||
| (b) \(H \sim \text{Po}(4)\) | B1 | 1st B1 writing or using Po(4) |
| \(P(H \leqslant 2) = 0.2381\) | B1 | 2nd B1 awrt 0.238 |
| \(Y \sim B(5, "0.2381")\) | M1 | 1st M1 choosing binomial distribution with \(n = 5\) and their \(p\) |
| \(P(Y = 2) = {}^5C_2(''0.2381'')^2(1 - ''0.2381'')^3\) | M1 | 2nd M1 \({}^5C_2 p^2(1 - p)^3\) with \(0 < p < 1\) |
| \(= 0.25073... = \text{awrt } 0.251\) | A1 | A1 awrt 0.251 |
| (5) | ||
| (c) \(W =\) number sold in first fifteen minutes. \(X =\) number sold in last forty five minutes. | M1 M1 | 1st M1 attempt at either correct product \(P(W = 4)P(X = 0)\) or \(P(W = 3)P(X = 1)\) from \(W \sim \text{Po}(2)\) and \(X \sim \text{Po}(6)\). implied by awrt 0.0902 ×awrt 0.0025 or awrt 0.180×awrt 0.0149 or awrt 0.0029 |
| \(F =\) number of muffins sold in first 15 minutes. \(F \sim B(4, 0.25)\). \(P(F > 2) = P(F = 3) + P(F = 4)\) | ||
| \(P(W > X | R = 4) = \frac{P(W = 4)P(X = 0) + P(W = 3)P(X = 1)}{P(R = 4)}\) | |
| \(= \frac{\frac{e^{-2}2^4}{4!} + \frac{0!}{0!} + \frac{e^{-2}2^3}{3!} + \frac{e^{-6}6^0}{0!}}{e^{-8}8^4/4!}\) | M1 | 2nd M1 conditional probability with \(P(R = 4)\) from R ~ Po(8) on denominator. implied by awrt 0.0573 seen in the denominator of a probability expression |
| \(= \frac{13}{256}\) (awrt 0.0508 or awrt 0.0509) | A1 | 3rd M1 complete expression for the required probability. implied (awrt 0.0902×awrt 0.0025+awrt 0.180×awrt 0.0149)/awrt 0.0573 for 3rd M1. A1 allow awrt 0.0508 or awrt 0.0509 from use of tables |
| (4) | ||
| [Total 12] | ||
| ALT | 1st M1 identifying \(B(4, 0.25)\); 2nd M1 \(P(F = 3) + P(F = 4)\) from \(B(4, 0.25)\); 3rd M1 \(4p^3q + p^4\) from \(B(4, 0.25)\) |
**(a)** $R \sim \text{Po}(8)$ | B1 | B1 writing or using Po(8) (may be implied by one correct probability from 0.5925, 0.0424 0.4530 or 0.0996)
$P(4 \leqslant R \leqslant 8) = P(R \leqslant 8) - P(R \leqslant 3) = 0.5925 - 0.0424$ | M1 | M1 writing or using $P(R \leqslant 8) - P(R \leqslant 3)$
$= 0.5501 = \text{awrt } 0.550$ | A1 | A1 awrt 0.550 (calc: 0.55016...) correct answer scores 3 out of 3
| | (3)
**(b)** $H \sim \text{Po}(4)$ | B1 | 1st B1 writing or using Po(4)
$P(H \leqslant 2) = 0.2381$ | B1 | 2nd B1 awrt 0.238
$Y \sim B(5, "0.2381")$ | M1 | 1st M1 choosing binomial distribution with $n = 5$ and their $p$
$P(Y = 2) = {}^5C_2(''0.2381'')^2(1 - ''0.2381'')^3$ | M1 | 2nd M1 ${}^5C_2 p^2(1 - p)^3$ with $0 < p < 1$
$= 0.25073... = \text{awrt } 0.251$ | A1 | A1 awrt 0.251
| | (5)
**(c)** $W =$ number sold in first fifteen minutes. $X =$ number sold in last forty five minutes. | M1 M1 | 1st M1 attempt at either correct product $P(W = 4)P(X = 0)$ or $P(W = 3)P(X = 1)$ from $W \sim \text{Po}(2)$ and $X \sim \text{Po}(6)$. implied by awrt 0.0902 ×awrt 0.0025 or awrt 0.180×awrt 0.0149 or awrt 0.0029
| $F =$ number of muffins sold in first 15 minutes. $F \sim B(4, 0.25)$. $P(F > 2) = P(F = 3) + P(F = 4)$
$P(W > X | R = 4) = \frac{P(W = 4)P(X = 0) + P(W = 3)P(X = 1)}{P(R = 4)}$
$= \frac{\frac{e^{-2}2^4}{4!} + \frac{0!}{0!} + \frac{e^{-2}2^3}{3!} + \frac{e^{-6}6^0}{0!}}{e^{-8}8^4/4!}$ | M1 | 2nd M1 conditional probability with $P(R = 4)$ from R ~ Po(8) on denominator. implied by awrt 0.0573 seen in the denominator of a probability expression
$= \frac{13}{256}$ (awrt 0.0508 or awrt 0.0509) | A1 | 3rd M1 complete expression for the required probability. implied (awrt 0.0902×awrt 0.0025+awrt 0.180×awrt 0.0149)/awrt 0.0573 for 3rd M1. A1 allow awrt 0.0508 or awrt 0.0509 from use of tables
| | (4)
| | [Total 12]
**ALT** |1st M1 identifying $B(4, 0.25)$; 2nd M1 $P(F = 3) + P(F = 4)$ from $B(4, 0.25)$; 3rd M1 $4p^3q + p^4$ from $B(4, 0.25)$\begin{enumerate}
\item A bakery sells muffins individually at an average rate of 8 muffins per hour.\\
(a) Find the probability that, in a randomly selected one-hour period, the bakery sells at least 4 but not more than 8 muffins.
\end{enumerate}
A sample of 5 non-overlapping half-hour periods is selected at random.\\
(b) Find the probability that the bakery sells fewer than 3 muffins in exactly 2 of these periods.
Given that 4 muffins were sold in a one-hour period,\\
(c) find the probability that more muffins were sold in the first 15 minutes than in the last 45 minutes.
\hfill \mbox{\textit{Edexcel S2 2023 Q7 [12]}}