| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Frequency distribution and Poisson fit |
| Difficulty | Standard +0.3 This is a standard S2 Poisson distribution question covering routine calculations: mean/variance from frequency data, justifying Poisson fit (mean ≈ variance), basic probability calculations, normal approximation to Poisson, and binomial probability. All techniques are textbook exercises with no novel problem-solving required. The multi-part structure and normal approximation requirement add some length but not conceptual difficulty. |
| Spec | 2.04b Binomial distribution: as model B(n,p)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 probabilities |
| Number of complaints per day | 0 | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) |
| Frequency | 12 | 28 | 37 | 38 | 29 | 17 | 19 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \([\text{Mean} =]\ 2.95\) | B1 | cao, allow exact equivalents |
| \([\text{Variance} =]\ \frac{2091}{180} - ("2.95")^2\) | M1 | ft their mean. Using \(\frac{\sum fx^2}{180} - (\text{their mean})^2\) or \(\frac{180}{179}\left(\frac{\sum fx^2}{180} - (\text{their mean})^2\right)\) |
| \(= 2.914\ldots \quad (s^2 = 2.930\ldots)\) awrt 2.91 (2.93) | A1cso | awrt 2.91 (2.93). Allow with a square root — may be implied by awrt 1.71 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| The mean is close to the variance | B1 | cao — Allow equivalent wording. Allow mean = variance. If no values/non compatible values calculated, then B0. Condone the use of 'closed' for 'close' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(W \sim \text{Po}(3)\) | — | — |
| \([P(W \geq 3) =]\ 1 - P(W \leq 2) = 0.5768\) awrt 0.577 | M1 A1 | M1 for \(1 - P(W, 2)\) or \(1 - 0.4232\); A1 awrt 0.577 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \([P(4 < W < 8) =]\ P(W \leq 7) - P(W \leq 4)\) or \(P(W=5)+P(W=6)+P(W=7)\) | M1 | for \(P(W \leq 7) - P(W \leq 4)\) or \(P(W=5)+P(W=6)+P(W=7)\); or \(0.9881 - 0.8153\) or \(0.1008 + 0.0504 + 0.0216\) |
| \(= 0.1728\ldots\) awrt 0.173 | A1 | awrt 0.173 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(X \sim N(21, 21)\) | B1 | for writing or using \(N(21,21)\). May be seen in a standardisation expression |
| \([P(X < 19) =]\ P\!\left(Z \leq \frac{18.5-21}{\sqrt{21}}\right) [= -0.5455\ldots]\) | M1 | for standardisation \((\pm)\) using their mean and sd. Allow 17.5, 18, 18.5, 19, 19.5, 22.5, 23, 23.5, 24, 24.5 |
| \([P(X > 23) =]\ P\!\left(Z \geq \frac{23.5-21}{\sqrt{21}}\right) [= 0.5455\ldots]\) | M1 | for using \(19 \pm 0.5\) or \(23 \pm 0.5\) |
| \(= 0.2912\) (calc \(0.29268\ldots\))* | A1 | for a fully correct standardisation expression. Implied by awrt \(\pm 0.546\) |
| — | A1*cso | awrt 0.291 or 0.293 from correct working seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(Y \sim B(13,\ "0.29")\) | M1 | for writing or using \(B(13, 0.29)\) ft their 0.29 (Must be 2 sf or better) or for \((p)^5(1-p)^8\) ft their 0.29 (Must be 2 sf or better). Condone \(B(0.29, 13)\) |
| \([P(Y=5) =]\ {}^{13}C_5\,("0.29")^5\,(1-"0.29")^8 = 0.170465\ldots\) awrt 0.17 | M1 A1 | M1 for \({}^{13}C_5\,(p)^5\,(1-p)^8\) oe with \(0 < p < 1\). Allow 1287 for \({}^{13}C_5\); A1 awrt 0.17 (0.17168 from using 0.2912) |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $[\text{Mean} =]\ 2.95$ | B1 | cao, allow exact equivalents |
| $[\text{Variance} =]\ \frac{2091}{180} - ("2.95")^2$ | M1 | ft their mean. Using $\frac{\sum fx^2}{180} - (\text{their mean})^2$ or $\frac{180}{179}\left(\frac{\sum fx^2}{180} - (\text{their mean})^2\right)$ |
| $= 2.914\ldots \quad (s^2 = 2.930\ldots)$ awrt 2.91 (2.93) | A1cso | awrt 2.91 (2.93). Allow with a square root — may be implied by awrt 1.71 |
## Part (b)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| The mean is close to the variance | B1 | cao — Allow equivalent wording. Allow mean = variance. If no values/non compatible values calculated, then B0. Condone the use of 'closed' for 'close' |
## Part (c)(i)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $W \sim \text{Po}(3)$ | — | — |
| $[P(W \geq 3) =]\ 1 - P(W \leq 2) = 0.5768$ awrt 0.577 | M1 A1 | M1 for $1 - P(W, 2)$ or $1 - 0.4232$; A1 awrt 0.577 |
## Part (c)(ii)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $[P(4 < W < 8) =]\ P(W \leq 7) - P(W \leq 4)$ or $P(W=5)+P(W=6)+P(W=7)$ | M1 | for $P(W \leq 7) - P(W \leq 4)$ **or** $P(W=5)+P(W=6)+P(W=7)$; or $0.9881 - 0.8153$ **or** $0.1008 + 0.0504 + 0.0216$ |
| $= 0.1728\ldots$ awrt 0.173 | A1 | awrt 0.173 |
## Part (d)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $X \sim N(21, 21)$ | B1 | for writing or using $N(21,21)$. May be seen in a standardisation expression |
| $[P(X < 19) =]\ P\!\left(Z \leq \frac{18.5-21}{\sqrt{21}}\right) [= -0.5455\ldots]$ | M1 | for standardisation $(\pm)$ using their mean and sd. Allow 17.5, 18, 18.5, 19, 19.5, 22.5, 23, 23.5, 24, 24.5 |
| $[P(X > 23) =]\ P\!\left(Z \geq \frac{23.5-21}{\sqrt{21}}\right) [= 0.5455\ldots]$ | M1 | for using $19 \pm 0.5$ or $23 \pm 0.5$ |
| $= 0.2912$ (calc $0.29268\ldots$)* | A1 | for a fully correct standardisation expression. Implied by awrt $\pm 0.546$ |
| — | A1*cso | awrt 0.291 or 0.293 from correct working seen |
## Part (e)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $Y \sim B(13,\ "0.29")$ | M1 | for writing or using $B(13, 0.29)$ ft their 0.29 (Must be 2 sf or better) or for $(p)^5(1-p)^8$ ft their 0.29 (Must be 2 sf or better). Condone $B(0.29, 13)$ |
| $[P(Y=5) =]\ {}^{13}C_5\,("0.29")^5\,(1-"0.29")^8 = 0.170465\ldots$ awrt 0.17 | M1 A1 | M1 for ${}^{13}C_5\,(p)^5\,(1-p)^8$ oe with $0 < p < 1$. Allow 1287 for ${}^{13}C_5$; A1 awrt 0.17 (0.17168 from using 0.2912) |\begin{enumerate}
\item The manager of a supermarket is investigating the number of complaints per day received from customers.
\end{enumerate}
A random sample of 180 days is taken and the results are shown in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
Number of complaints per day & 0 & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
Frequency & 12 & 28 & 37 & 38 & 29 & 17 & 19 & 0 \\
\hline
\end{tabular}
\end{center}
(a) Calculate the mean and the variance of these data.\\
(b) Explain why the results in part (a) suggest that a Poisson distribution may be a suitable model for the number of complaints per day.
The manager uses a Poisson distribution with mean 3 to model the number of complaints per day.\\
(c) For a randomly selected day find, using the manager's model, the probability that there are\\
(i) at least 3 complaints,\\
(ii) more than 4 complaints but less than 8 complaints.
A week consists of 7 consecutive days.\\
(d) Using the manager's model and a suitable approximation, show that the probability that there are less than 19 complaints in a randomly selected week is 0.29 to 2 decimal places.\\
Show your working clearly.\\
(Solutions relying on calculator technology are not acceptable.)
A period of 13 weeks is selected at random.\\
(e) Find the probability that in this period there are exactly 5 weeks that have less than 19 complaints.\\
Show your working clearly.
\hfill \mbox{\textit{Edexcel S2 2024 Q1 [16]}}