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Date May 2014 Marks available 5 Reference code 14M.3sp.hl.TZ0.1
Level HL only Paper Paper 3 Statistics and probability Time zone TZ0
Command term Find Question number 1 Adapted from N/A

Question

The random variable X has probability distribution Po(8).

(i)     Find P(X=6).

(ii)     Find P(X=6|5.

[5]
a.

\bar X denotes the sample mean of n > 1 independent observations from X.

(i)     Write down {\text{E}}(\bar X) and {\text{Var}}(\bar X).

(ii)     Hence, give a reason why \bar X is not a Poisson distribution.

[3]
b.

A random sample of 40 observations is taken from the distribution for X.

(i)     Find {\text{P}}(7.1 < \bar X < 8.5).

(ii)     Given that {\text{P}}\left( {\left| {\bar X - 8} \right| \leqslant k} \right) = 0.95, find the value of k.

[6]
c.

Markscheme

(i)     {\text{P}}(X = 6) = 0.122     (M1)A1

(ii)     {\text{P}}(X = 6|5 \leqslant X \leqslant 8) = \frac{{{\text{P}}(X = 6)}}{{{\text{P}}(5 \leqslant X \leqslant 8)}} = \frac{{0.122 \ldots }}{{0.592 \ldots  - 0.0996 \ldots }}     (M1)(A1)

= 0.248     A1

[5 marks]

a.

(i)     {\text{E}}(\bar X) = 8     A1

{\text{Var}}(\bar X) = \frac{8}{n}     A1

(ii)     {\text{E}}(\bar X) \ne {\text{Var}}(\bar X)   {\text{(for }}n > 1)     R1

 

Note:     Only award the R1 if the two expressions in (b)(i) are different.

 

[3 marks]

b.

(i)     EITHER

\bar X \sim {\text{N(8, 0.2)}}     (M1)A1

 

Note:     M1 for normality, A1 for parameters.

 

{\text{P}}(7.1 < \bar X < 8.5) = 0.846     A1

OR

The expression is equivalent to

{\text{P}}(283 \leqslant \sum {X \leqslant 339)} where \sum X is {\text{Po(320)}}     M1A1

= 0.840     A1

 

Note:     Accept 284, 340 instead of 283, 339

     Accept any answer that rounds correctly to 0.84 or 0.85.

 

(ii)     EITHER

k = 1.96\frac{\sigma }{{\sqrt n }} or 1.96{\text{ std}}(\bar X)     (M1)(A1)

k = 0.877 or 1.96\sqrt {0.2}     A1

OR

The expression is equivalent to

P(320 - 40k \leqslant \sum {X \leqslant 320 + 40k) = 0.95}     (M1)

k = 0.875     A2

 

Note:     Accept any answer that rounds to 0.87 or 0.88.

     Award M1A0 if modulus sign ignored and answer obtained rounds to 0.74 or 0.75

 

[6 marks]

c.

Examiners report

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Syllabus sections

Topic 5 - Core: Statistics and probability » 5.4 » Conditional probability; the definition P\left( {\left. A \right|P} \right) = \frac{{P\left( {A\mathop \cap \nolimits B} \right)}}{{P\left( B \right)}} .

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