CAIE S2 2012 June — Question 4 8 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2012
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSum of Poisson processes
TypeBasic sum of two Poissons
DifficultyStandard +0.3 This is a straightforward application of Poisson distribution with scaled parameters. Part (i) requires recognizing that independent Poisson variables sum to another Poisson (λ = 0.32×8 + 0.45×12 = 8.0) and computing P(X ≥ 3). Part (ii) is a direct calculation with λ = 0.32×155 = 49.6. Both parts are standard textbook exercises requiring only recall of properties and calculator work, making this 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

4 Bacteria of a certain type are randomly distributed in the water in two ponds, \(A\) and \(B\). The average numbers of bacteria per \(\mathrm { cm } ^ { 3 }\) in \(A\) and \(B\) are 0.32 and 0.45 respectively.
  1. Samples of \(8 \mathrm {~cm} ^ { 3 }\) of water from \(A\) and \(12 \mathrm {~cm} ^ { 3 }\) of water from \(B\) are taken at random. Find the probability that the total number of bacteria in these samples is at least 3 .
  2. Find the probability that in a random sample of \(155 \mathrm {~cm} ^ { 3 }\) of water from \(A\), the number of bacteria is less than 35 .

Question 4:
(i)
AnswerMarks Guidance
\(\lambda = 8 \times 0.32 + 12 \times 0.45\ (= 7.96)\)M1
\(1 - e^{-7.96}(1 + 7.96 + \frac{7.96^2}{2})\)M1 \(1 - P(X \leq 2)\), any \(\lambda\) allow one end error
\(= 0.986\) (3 sfs)A1 [3]
(ii)
AnswerMarks Guidance
\(\lambda = 155 \times 0.32 = 49.6\)B1
\(N(49.6,\ 49.6)\)M1 \(N(\lambda,\ \lambda)\) any \(\lambda\). May be implied
\(\frac{34.5 - 49.6}{\sqrt{49.6}}\ (= -2.144)\)M1 Allow no or wrong cc & no \(\sqrt{}\)
\(\Phi(-2.144) = 1 - \Phi(2.144)\)M1
\(= 0.016(0)\)A1 [5] Correct area consistent with their working
## Question 4:

**(i)**
$\lambda = 8 \times 0.32 + 12 \times 0.45\ (= 7.96)$ | M1 |
$1 - e^{-7.96}(1 + 7.96 + \frac{7.96^2}{2})$ | M1 | $1 - P(X \leq 2)$, any $\lambda$ allow one end error
$= 0.986$ (3 sfs) | A1 [3] |

**(ii)**
$\lambda = 155 \times 0.32 = 49.6$ | B1 |
$N(49.6,\ 49.6)$ | M1 | $N(\lambda,\ \lambda)$ any $\lambda$. May be implied
$\frac{34.5 - 49.6}{\sqrt{49.6}}\ (= -2.144)$ | M1 | Allow no or wrong cc & no $\sqrt{}$
$\Phi(-2.144) = 1 - \Phi(2.144)$ | M1 |
$= 0.016(0)$ | A1 [5] | Correct area consistent with their working

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4 Bacteria of a certain type are randomly distributed in the water in two ponds, $A$ and $B$. The average numbers of bacteria per $\mathrm { cm } ^ { 3 }$ in $A$ and $B$ are 0.32 and 0.45 respectively.\\
(i) Samples of $8 \mathrm {~cm} ^ { 3 }$ of water from $A$ and $12 \mathrm {~cm} ^ { 3 }$ of water from $B$ are taken at random. Find the probability that the total number of bacteria in these samples is at least 3 .\\
(ii) Find the probability that in a random sample of $155 \mathrm {~cm} ^ { 3 }$ of water from $A$, the number of bacteria is less than 35 .

\hfill \mbox{\textit{CAIE S2 2012 Q4 [8]}}