CAIE S2 2014 June — Question 8 10 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2014
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypePoisson parameter from given probability
DifficultyStandard +0.3 This question tests basic properties of Poisson distributions and standard applications. Part (i) requires recognizing that Poisson distributions only take non-negative integer values (eliminating V) and don't have gaps in their support (eliminating W). Part (ii) involves straightforward manipulation of Poisson probability formulas. Part (iii) uses normal approximation to Poisson, which is a standard S2 technique requiring inverse normal calculation. All parts are routine applications of syllabus content with no novel problem-solving required.
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
8
  1. The following tables show the probability distributions for the random variables \(V\) and \(W\).
    \(v\)- 101\(> 1\)
    \(\mathrm { P } ( V = v )\)0.3680.3680.1840.080
    \(w\)00.51\(> 1\)
    \(\mathrm { P } ( W = w )\)0.3680.3680.1840.080
    For each of the variables \(V\) and \(W\) state how you can tell from its probability distribution that it does NOT have a Poisson distribution.
  2. The random variable \(X\) has the distribution \(\operatorname { Po } ( \lambda )\). It is given that $$\mathrm { P } ( X = 0 ) = p \quad \text { and } \quad \mathrm { P } ( X = 1 ) = 2.5 p$$ where \(p\) is a constant.
    1. Show that \(\lambda = 2.5\).
    2. Find \(\mathrm { P } ( X \geqslant 3 )\).
    3. The random variable \(Y\) has the distribution \(\operatorname { Po } ( \mu )\), where \(\mu > 30\). Using a suitable approximating distribution, it is found that \(\mathrm { P } ( Y > 40 ) = 0.5793\) correct to 4 decimal places. Find \(\mu\).
Question 8:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(V\): cannot have negative valueB1
\(W\): cannot have non-integer valueB1 [2]
Part (ii)(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(e^{-\lambda} = p\) and \(\lambda e^{-\lambda} = 2.5p\) (Hence \(\lambda = 2.5\) AG)B1 [1] or equiv explanation
Part (ii)(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(1 - e^{-2.5}(1 + 2.5 + \frac{2.5^2}{2})\)M1 Allow one end error
\(= 0.456\) (3 sf)A1 [2]
Part (iii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\Phi^{-1}(0.5793) = -0.2\)B1
\(N(\mu, \mu)\) seen or impliedM1
\(\frac{40.5 - \mu}{\sqrt{\mu}} = "-0.2"\)M1 Allow no cc or incorrect cc
\(\mu + "-0.2"\sqrt{\mu} - 40.5 = 0\)
\(\sqrt{\mu} = \frac{"0.2" \pm \sqrt{"0.2"^2 + 4 \times 40.5}}{2}\) \((= 6.4647...)\)M1 For solving quadratic in \(\sqrt{\mu}\) (or \(\mu\)). Ignore other answer for \(\sqrt{\mu}\), but not for \(\mu\)
\(\mu = 41.8\) (3 sf)A1 [5] 
## Question 8:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $V$: cannot have negative value | B1 | |
| $W$: cannot have non-integer value | B1 [2] | |

### Part (ii)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{-\lambda} = p$ and $\lambda e^{-\lambda} = 2.5p$ (Hence $\lambda = 2.5$ **AG**) | B1 [1] | or equiv explanation |

### Part (ii)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 - e^{-2.5}(1 + 2.5 + \frac{2.5^2}{2})$ | M1 | Allow one end error |
| $= 0.456$ (3 sf) | A1 [2] | |

### Part (iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\Phi^{-1}(0.5793) = -0.2$ | B1 | |
| $N(\mu, \mu)$ seen or implied | M1 | |
| $\frac{40.5 - \mu}{\sqrt{\mu}} = "-0.2"$ | M1 | Allow no cc or incorrect cc |
| $\mu + "-0.2"\sqrt{\mu} - 40.5 = 0$ | | |
| $\sqrt{\mu} = \frac{"0.2" \pm \sqrt{"0.2"^2 + 4 \times 40.5}}{2}$ $(= 6.4647...)$ | M1 | For solving quadratic in $\sqrt{\mu}$ (or $\mu$). Ignore other answer for $\sqrt{\mu}$, but not for $\mu$ |
| $\mu = 41.8$ (3 sf) | A1 [5] | |
8 (i) The following tables show the probability distributions for the random variables $V$ and $W$.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$v$ & - 1 & 0 & 1 & $> 1$ \\
\hline
$\mathrm { P } ( V = v )$ & 0.368 & 0.368 & 0.184 & 0.080 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$w$ & 0 & 0.5 & 1 & $> 1$ \\
\hline
$\mathrm { P } ( W = w )$ & 0.368 & 0.368 & 0.184 & 0.080 \\
\hline
\end{tabular}
\end{center}

For each of the variables $V$ and $W$ state how you can tell from its probability distribution that it does NOT have a Poisson distribution.\\
(ii) The random variable $X$ has the distribution $\operatorname { Po } ( \lambda )$. It is given that

$$\mathrm { P } ( X = 0 ) = p \quad \text { and } \quad \mathrm { P } ( X = 1 ) = 2.5 p$$

where $p$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $\lambda = 2.5$.
\item Find $\mathrm { P } ( X \geqslant 3 )$.\\
(iii) The random variable $Y$ has the distribution $\operatorname { Po } ( \mu )$, where $\mu > 30$. Using a suitable approximating distribution, it is found that $\mathrm { P } ( Y > 40 ) = 0.5793$ correct to 4 decimal places. Find $\mu$.
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2014 Q8 [10]}}
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