CAIE S2 2014 June — Question 4 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating the Binomial to the Poisson distribution
TypeState Poisson approximation with justification
DifficultyModerate -0.8 This is a straightforward application of the standard Poisson approximation to the binomial distribution. Students need to recognize n=3500, p=1/1000, identify B(3500, 1/1000) and Po(3.5), state the justification (large n, small p, np moderate), then perform a routine Poisson probability calculation. This requires only recall of conditions and basic calculator work, making it easier than average.
Spec2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)

4 The proportion of people who have a particular gene, on average, is 1 in 1000. A random sample of 3500 people in a certain country is chosen and the number of people, \(X\), having the gene is found.
  1. State the distribution of \(X\) and state also an appropriate approximating distribution. Give the values of any parameters in each case. Justify your choice of the approximating distribution.
  2. Use the approximating distribution to find \(\mathrm { P } ( X \leqslant 3 )\).

Question 4:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(B(3500, 0.001)\)B1
Poisson with mean \(= 3.5\)B1 or \(Po(3.5)\)
\(n > 50\) and \(np < 5\)B1 [3] Both. Or \(n > 50\) and \(\lambda < 5\) or \(3.5 < 5\)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(e^{-3.5}(1 + 3.5 + \frac{3.5^2}{2} + \frac{3.5^3}{3!})\)M1 Allow any \(\lambda\)
\(= 0.537\) (3 dp)A1 [2]
## Question 4:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $B(3500, 0.001)$ | B1 | |
| Poisson with mean $= 3.5$ | B1 | or $Po(3.5)$ |
| $n > 50$ and $np < 5$ | B1 [3] | Both. Or $n > 50$ and $\lambda < 5$ or $3.5 < 5$ |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{-3.5}(1 + 3.5 + \frac{3.5^2}{2} + \frac{3.5^3}{3!})$ | M1 | Allow any $\lambda$ |
| $= 0.537$ (3 dp) | A1 [2] | |

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4 The proportion of people who have a particular gene, on average, is 1 in 1000. A random sample of 3500 people in a certain country is chosen and the number of people, $X$, having the gene is found.\\
(i) State the distribution of $X$ and state also an appropriate approximating distribution. Give the values of any parameters in each case. Justify your choice of the approximating distribution.\\
(ii) Use the approximating distribution to find $\mathrm { P } ( X \leqslant 3 )$.

\hfill \mbox{\textit{CAIE S2 2014 Q4 [5]}}