CAIE S2 2012 June — Question 5 10 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2012
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeConditional probability with Poisson
DifficultyStandard +0.3 This is a straightforward application of Poisson distribution properties and the Central Limit Theorem. Part (i) involves standard Poisson probability calculations and conditional probability. Part (ii) requires knowing that sample means are approximately normal with mean λ and variance λ/n, then a routine normal probability calculation. All techniques are standard S2 content 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 probabilities5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance

5 A random variable \(X\) has the distribution \(\operatorname { Po } ( 3.2 )\).
  1. A random value of \(X\) is found.
    1. Find \(\mathrm { P } ( X \geqslant 3 )\).
    2. Find the probability that \(X = 3\) given that \(X \geqslant 3\).
    3. Random samples of 120 values of \(X\) are taken.
      (a) Describe fully the distribution of the sample mean.
      (b) Find the probability that the mean of a random sample of size 120 is less than 3.3.

(i) (a)
M1 \(P(X > 3) = 1 - e^{-3.2}\left(1 + 3.2 + \frac{3.2^2}{2!}\right)\)
A1 \(= 0.620\) (3 sf)
Allow one end error
[2]
(i) (b)
M1 \(P(X = 3) = e^{-3.2}\frac{3.2^3}{3!} = 0.22262\)
M1 \(P(X=3 \cap X \geq 3) = \frac{P(X=3)}{P(X \geq 3)} = \frac{\text{'0.22262'}}{\text{'0.62010'}}\)
A1 \(= 0.359\) (3 sf)
May be implied
[3] Their \(P(X=3)\); Their \(P(X \geq 3)\)
(ii) (a)
B1 (Approx) normal with mean \(3.2\)
B1 variance \(= \frac{3.2}{120}\) or \(\frac{2}{75}\) or \(0.0267\) (3 sfs) oe
[2]
(ii) (b)
M1 \(\frac{3.3-3.2}{\sqrt{3.2/120}} = 0.612\)
M1 \(\Phi(\text{"0.612"})\)
A1 \(= 0.730\) (3 sfs)
Allow with cc attempted
Accept \(0.73\)
[3] Or sd \(= \frac{\sqrt{3.2}}{120}\) or \(0.163\) (3 sfs) oe
**(i) (a)**

**M1** $P(X > 3) = 1 - e^{-3.2}\left(1 + 3.2 + \frac{3.2^2}{2!}\right)$

**A1** $= 0.620$ (3 sf)

Allow one end error

**[2]**

**(i) (b)**

**M1** $P(X = 3) = e^{-3.2}\frac{3.2^3}{3!} = 0.22262$

**M1** $P(X=3 \cap X \geq 3) = \frac{P(X=3)}{P(X \geq 3)} = \frac{\text{'0.22262'}}{\text{'0.62010'}}$

**A1** $= 0.359$ (3 sf)

May be implied

**[3]** Their $P(X=3)$; Their $P(X \geq 3)$

**(ii) (a)**

**B1** (Approx) normal with mean $3.2$

**B1** variance $= \frac{3.2}{120}$ or $\frac{2}{75}$ or $0.0267$ (3 sfs) oe

**[2]**

**(ii) (b)**

**M1** $\frac{3.3-3.2}{\sqrt{3.2/120}} = 0.612$

**M1** $\Phi(\text{"0.612"})$

**A1** $= 0.730$ (3 sfs)

Allow with cc attempted

Accept $0.73$

**[3]** Or sd $= \frac{\sqrt{3.2}}{120}$ or $0.163$ (3 sfs) oe

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5 A random variable $X$ has the distribution $\operatorname { Po } ( 3.2 )$.\\
(i) A random value of $X$ is found.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X \geqslant 3 )$.
\item Find the probability that $X = 3$ given that $X \geqslant 3$.\\
(ii) Random samples of 120 values of $X$ are taken.\\
(a) Describe fully the distribution of the sample mean.\\
(b) Find the probability that the mean of a random sample of size 120 is less than 3.3.
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2012 Q5 [10]}}