AQA S2 2007 January — Question 2 13 marks

Exam BoardAQA
ModuleS2 (Statistics 2)
Year2007
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSum of Poisson processes
TypeMulti-period repeated application
DifficultyModerate -0.3 This is a straightforward S2 question testing standard Poisson distribution techniques: direct probability calculations, sum of independent Poisson variables (which students learn forms another Poisson), and binomial probability with the result from part (a). The mean/variance comparison in part (c) is textbook material. All parts are routine applications with no novel problem-solving required, making it slightly easier than average.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson

2 The number of computers, \(A\), bought during one day from the Amplebuy computer store can be modelled by a Poisson distribution with a mean of 3.5. The number of computers, \(B\), bought during one day from the Bestbuy computer store can be modelled by a Poisson distribution with a mean of 5.0 .
    1. Calculate \(\mathrm { P } ( A = 4 )\).
    2. Determine \(\mathrm { P } ( B \leqslant 6 )\).
    3. Find the probability that a total of fewer than 10 computers is bought from these two stores on one particular day.
  1. Calculate the probability that a total of fewer than 10 computers is bought from these two stores on at least 4 out of 5 consecutive days.
  2. The numbers of computers bought from the Choicebuy computer store over a 10-day period are recorded as $$\begin{array} { l l l l l l l l l l } 8 & 12 & 6 & 6 & 9 & 15 & 10 & 8 & 6 & 12 \end{array}$$
    1. Calculate the mean and variance of these data.
    2. State, giving a reason based on your results in part (c)(i), whether or not a Poisson distribution provides a suitable model for these data.

Question 2:
Part (a)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(A=4) = \dfrac{e^{-3.5} \times (3.5)^4}{4!} = 0.189\)M1A1 Total: 2
Part (a)(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(B \leq 6) = 0.762\)B1 Total: 1
Part (a)(iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(T = A + B \sim \text{Po}(8.5)\)
\(P(T \text{ fewer than } 10) = P(T < 10)\)M1 Use of Po\((8.5)\)
\(= P(T \leq 9)\)M1 \(T \leq 9\) attempted
\(= 0.653\)A1 CAO, Total: 3
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(X \sim B(5, 0.653)\)B1 \(X \sim B(5, \text{their } p)\)
\(P(X \geq 4) = \dbinom{5}{4}(0.653)^4(0.347) + (0.653)^5\)M1
\(= 0.31547 + 0.11873 = 0.434\)A1\(\checkmark\) On their \(p\) from (a)(iii), Total: 3
Part (c)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\bar{x} = 9.2\)B1
\(s^2 = 9.29\)B1 \(\sigma^2 = 8.36\), Total: 2
Part (c)(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Mean and variance have similar values which suggests that Poisson distribution may be appropriateB1\(\checkmark\) B1\(\checkmark\) Total: 2
## Question 2:

### Part (a)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A=4) = \dfrac{e^{-3.5} \times (3.5)^4}{4!} = 0.189$ | M1A1 | **Total: 2** |

### Part (a)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B \leq 6) = 0.762$ | B1 | **Total: 1** |

### Part (a)(iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = A + B \sim \text{Po}(8.5)$ | | |
| $P(T \text{ fewer than } 10) = P(T < 10)$ | M1 | Use of Po$(8.5)$ |
| $= P(T \leq 9)$ | M1 | $T \leq 9$ attempted |
| $= 0.653$ | A1 | CAO, **Total: 3** |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim B(5, 0.653)$ | B1 | $X \sim B(5, \text{their } p)$ |
| $P(X \geq 4) = \dbinom{5}{4}(0.653)^4(0.347) + (0.653)^5$ | M1 | |
| $= 0.31547 + 0.11873 = 0.434$ | A1$\checkmark$ | On their $p$ from (a)(iii), **Total: 3** |

### Part (c)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = 9.2$ | B1 | |
| $s^2 = 9.29$ | B1 | $\sigma^2 = 8.36$, **Total: 2** |

### Part (c)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean and variance have similar values which suggests that Poisson distribution may be appropriate | B1$\checkmark$ B1$\checkmark$ | **Total: 2** |

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2 The number of computers, $A$, bought during one day from the Amplebuy computer store can be modelled by a Poisson distribution with a mean of 3.5.

The number of computers, $B$, bought during one day from the Bestbuy computer store can be modelled by a Poisson distribution with a mean of 5.0 .
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Calculate $\mathrm { P } ( A = 4 )$.
\item Determine $\mathrm { P } ( B \leqslant 6 )$.
\item Find the probability that a total of fewer than 10 computers is bought from these two stores on one particular day.
\end{enumerate}\item Calculate the probability that a total of fewer than 10 computers is bought from these two stores on at least 4 out of 5 consecutive days.
\item The numbers of computers bought from the Choicebuy computer store over a 10-day period are recorded as

$$\begin{array} { l l l l l l l l l l } 
8 & 12 & 6 & 6 & 9 & 15 & 10 & 8 & 6 & 12
\end{array}$$
\begin{enumerate}[label=(\roman*)]
\item Calculate the mean and variance of these data.
\item State, giving a reason based on your results in part (c)(i), whether or not a Poisson distribution provides a suitable model for these data.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA S2 2007 Q2 [13]}}