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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|5X8).

[5]
a.

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

(i)     Write down E(ˉX) and Var(ˉX).

(ii)     Hence, give a reason why ˉX is not a Poisson distribution.

[3]
b.

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

(i)     Find P(7.1<ˉX<8.5).

(ii)     Given that P(|ˉX8|k)=0.95, find the value of k.

[6]
c.

Markscheme

(i)     P(X=6)=0.122     (M1)A1

(ii)     P(X=6|5X8)=P(X=6)P(5X8)=0.1220.5920.0996     (M1)(A1)

=0.248     A1

[5 marks]

a.

(i)     E(ˉX)=8     A1

Var(ˉX)=8n     A1

(ii)     E(ˉX)Var(ˉX)   (for n>1)     R1

 

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

 

[3 marks]

b.

(i)     EITHER

ˉXN(8, 0.2)     (M1)A1

 

Note:     M1 for normality, A1 for parameters.

 

P(7.1<ˉX<8.5)=0.846     A1

OR

The expression is equivalent to

P(283X339) where X is 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=1.96σn or 1.96 std(ˉX)     (M1)(A1)

k=0.877 or 1.960.2     A1

OR

The expression is equivalent to

P(32040kX320+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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a.
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b.
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c.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.4 » Conditional probability; the definition P(A|P)=P(AB)P(B) .

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