| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Frequency distribution and Poisson fit |
| Difficulty | Standard +0.3 This is a straightforward S2 Poisson distribution question requiring standard calculations: mean/variance from frequency table, recognizing Poisson conditions (mean ≈ variance), basic Poisson probability calculations, and a simple binomial application. All techniques are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Number of cherries | 0 | 1 | 2 | 3 | 4 | 5 | \(\geqslant 6\) |
| Frequency | 24 | 37 | 21 | 12 | 4 | 2 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Mean \(= 1.41\) | B1 | cao, allow 141/100 |
| Variance \(= \frac{343}{100} - 1.41^2\) | M1 | Using \(\frac{\sum fx^2}{100} - (\text{mean})^2\) or \(\frac{100}{99}\left(\frac{\sum fx^2}{100} - (\text{mean})^2\right)\) oe; allow square root for M mark; if no working shown for \(\sum fx^2\) must see 343, 3.43 or correct answer |
| \(= 1.4419\) \((s^2 = 1.456)\) | A1 | awrt 1.44 or 1.46 for \(s^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| The mean is close to the variance | B1 | cao - allow alternative wording; allow "mean equals variance" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(X \sim \text{Po}(1.5)\), \(P(X=2) = \frac{e^{-1.5}1.5^2}{2!}\) | M1 | Writing or using \(\frac{e^{-\lambda}\lambda^2}{2!}\) or \(P(X \leq 2) - P(X \leq 1)\) |
| \(= 0.2510\) | A1 | awrt 0.251 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(P(X \geq 1) = 1 - P(X=0) = 1 - e^{-1.5}\) or \(1 - 0.2231\) | M1 | Writing or using \(1 - P(X=0)\) oe |
| \(= 0.77686\ldots\) | A1 | awrt 0.777 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(Y \sim \text{Po}(7.5)\) | B1 | Writing Po(7.5) |
| \(P(Y \geq 11) = 1 - P(Y \leq 10)\) | M1 | Writing \(P(Y \geq 11)\) or \(1 - P(Y \leq 10)\) oe |
| \(= 1 - 0.8622\) | ||
| \(= 0.1378\) * | A1cso | Seeing \(1 - 0.8622\) leading to 0.1378 cso (both B1 and M1 awarded) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(A \sim B(12, 0.1378)\) | M1 | Using \((p)^n(1-p)^{12-n}\) where \(p = 0.1378\) or 0.138; condone missing \(nCr\) |
| \(P(A=3) = \binom{12}{3}(0.1378)^3(0.8622)^9\) | M1 | \(\binom{12}{3}(p)^3(1-p)^9\), with \(0 < p < 1\); allow 220 or 12C3 instead of \(\binom{12}{3}\) |
| \(= 0.1516\) | A1 | awrt 0.152 |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Mean $= 1.41$ | B1 | cao, allow 141/100 |
| Variance $= \frac{343}{100} - 1.41^2$ | M1 | Using $\frac{\sum fx^2}{100} - (\text{mean})^2$ or $\frac{100}{99}\left(\frac{\sum fx^2}{100} - (\text{mean})^2\right)$ oe; allow square root for M mark; if no working shown for $\sum fx^2$ must see 343, 3.43 or correct answer |
| $= 1.4419$ $(s^2 = 1.456)$ | A1 | awrt 1.44 or 1.46 for $s^2$ |
### Part (b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| The mean is close to the variance | B1 | cao - allow alternative wording; allow "mean equals variance" |
### Part (c)(i):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $X \sim \text{Po}(1.5)$, $P(X=2) = \frac{e^{-1.5}1.5^2}{2!}$ | M1 | Writing or using $\frac{e^{-\lambda}\lambda^2}{2!}$ or $P(X \leq 2) - P(X \leq 1)$ |
| $= 0.2510$ | A1 | awrt 0.251 |
### Part (c)(ii):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $P(X \geq 1) = 1 - P(X=0) = 1 - e^{-1.5}$ or $1 - 0.2231$ | M1 | Writing or using $1 - P(X=0)$ oe |
| $= 0.77686\ldots$ | A1 | awrt 0.777 |
### Part (d):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $Y \sim \text{Po}(7.5)$ | B1 | Writing Po(7.5) |
| $P(Y \geq 11) = 1 - P(Y \leq 10)$ | M1 | Writing $P(Y \geq 11)$ or $1 - P(Y \leq 10)$ oe |
| $= 1 - 0.8622$ | | |
| $= 0.1378$ * | A1cso | Seeing $1 - 0.8622$ leading to 0.1378 cso (both B1 and M1 awarded) |
### Part (e):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $A \sim B(12, 0.1378)$ | M1 | Using $(p)^n(1-p)^{12-n}$ where $p = 0.1378$ or 0.138; condone missing $nCr$ |
| $P(A=3) = \binom{12}{3}(0.1378)^3(0.8622)^9$ | M1 | $\binom{12}{3}(p)^3(1-p)^9$, with $0 < p < 1$; allow 220 or 12C3 instead of $\binom{12}{3}$ |
| $= 0.1516$ | A1 | awrt 0.152 |
**Total: 14 marks**\begin{enumerate}
\item A student is investigating the numbers of cherries in a Rays fruit cake. A random sample of Rays fruit cakes is taken and the results are shown in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
Number of cherries & 0 & 1 & 2 & 3 & 4 & 5 & $\geqslant 6$ \\
\hline
Frequency & 24 & 37 & 21 & 12 & 4 & 2 & 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 cherries in a Rays fruit cake.
The number of cherries in a Rays fruit cake follows a Poisson distribution with mean 1.5 A Rays fruit cake is to be selected at random.
Find the probability that it contains\\
(c) (i) exactly 2 cherries,\\
(ii) at least 1 cherry.
Rays fruit cakes are sold in packets of 5\\
(d) Show that the probability that there are more than 10 cherries, in total, in a randomly selected packet of Rays fruit cakes, is 0.1378 correct to 4 decimal places.
Twelve packets of Rays fruit cakes are selected at random.\\
(e) Find the probability that exactly 3 packets contain more than 10 cherries.
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\hfill \mbox{\textit{Edexcel S2 2016 Q1 [14]}}