CAIE S2 2011 June — Question 1 4 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2011
SessionJune
Marks4
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Mark schemeDownload PDF ↗
TopicApproximating the Binomial to the Poisson distribution
TypeCalculate single probability using Poisson approximation
DifficultyModerate -0.3 This is a straightforward application of the Poisson approximation to the binomial distribution. Students need to identify n=6000, p=2/10000, calculate λ=np=1.2, then find P(X>2)=1-P(X≤2) using Poisson tables or formula. It's a standard textbook exercise requiring recognition of when to use the approximation and routine calculation, making it slightly easier than average.
Spec2.04d 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
1 On average, 2 people in every 10000 in the UK have a particular gene. A random sample of 6000 people in the UK is chosen. The random variable \(X\) denotes the number of people in the sample who have the gene. Use an approximating distribution to calculate the probability that there will be more than 2 people in the sample who have the gene.
AnswerMarks Guidance
Poisson \(\lambda = 1.2\)B1 1.2 seen
\(1 - e^{-1.2}(1 + 1.2 + \frac{1.2^2}{2})\)M1 1 – Poisson P(0, 1, 2, 3) attempted, any \(\lambda\), allow 1 end error
\(= 0.121\)A1 SC: using Bin, ans 0.120: B1
[4] 
Poisson $\lambda = 1.2$ | B1 | 1.2 seen
$1 - e^{-1.2}(1 + 1.2 + \frac{1.2^2}{2})$ | M1 | 1 – Poisson P(0, 1, 2, 3) attempted, any $\lambda$, allow 1 end error
$= 0.121$ | A1 | SC: using Bin, ans 0.120: B1
| [4] |
1 On average, 2 people in every 10000 in the UK have a particular gene. A random sample of 6000 people in the UK is chosen. The random variable $X$ denotes the number of people in the sample who have the gene. Use an approximating distribution to calculate the probability that there will be more than 2 people in the sample who have the gene.

\hfill \mbox{\textit{CAIE S2 2011 Q1 [4]}}
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