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Date May 2009 Marks available 20 Reference code 09M.2.hl.TZ2.11
Level HL only Paper 2 Time zone TZ2
Command term Find and Show that Question number 11 Adapted from N/A

Question

Testing has shown that the volume of drink in a bottle of mineral water filled by Machine A at a bottling plant is normally distributed with a mean of 998 ml and a standard deviation of 2.5 ml.

(a)     Show that the probability that a randomly selected bottle filled by Machine A contains more than 1000 ml of mineral water is 0.212.

(b)     A random sample of 5 bottles is taken from Machine A. Find the probability that exactly 3 of them each contain more than 1000 ml of mineral water.

(c)     Find the minimum number of bottles that would need to be sampled to ensure that the probability of getting at least one bottle filled by Machine A containing more than 1000 ml of mineral water, is greater than 0.99.

(d)     It has been found that for Machine B the probability of a bottle containing less than 996 ml of mineral water is 0.1151. The probability of a bottle containing more than 1000 ml is 0.3446. Find the mean and standard deviation for the volume of mineral water contained in bottles filled by Machine B.

(e)     The company that makes the mineral water receives, on average, m phone calls every 10 minutes. The number of phone calls, X , follows a Poisson distribution such that P(X=2)=P(X=3)+P(X=4) .

  (i)     Find the value of m .

  (ii)     Find the probability that the company receives more than two telephone calls in a randomly selected 10 minute period.

Markscheme

(a)     XN(998, 2.52 )     M1

P(X>1000)=0.212     AG

[1 mark]

 

(b)     XB(5, 0.2119...)

evidence of binomial     (M1)

P(X=3)=(53)(0.2119...)3(0.7881...)2=0.0591   (accept 0.0592)     (M1)A1

[3 marks]

 

(c)     P(X     (M1)

1 - {\left( {0.7881...} \right)^n} > 0.99

{\left( {0.7881...} \right)^n} < 0.01     A1

Note: Award A1 for line 2 or line 3 or equivalent.

 

n > 19.3     (A1)
minimum number of bottles required is 20     A1N2

[4 marks]

 

(d)     \frac{{996 - \mu }}{\sigma } =  - 1.1998   (accept 1.2)     M1A1

\frac{{1000 - \mu }}{\sigma } = 0.3999   (accept 0.4)     M1A1

\mu  = 999(ml), \sigma  = 2.50(ml)     A1A1

[6 marks]

 

(e)     (i)     \frac{{{{\text{e}}^{ - m}}{m^2}}}{{2!}} = \frac{{{{\text{e}}^{ - m}}{m^3}}}{{3!}} + \frac{{{{\text{e}}^{ - m}}{m^4}}}{{4!}}     M1A1

\frac{{{m^2}}}{2} = \frac{{{m^3}}}{6} + \frac{{{m^4}}}{{24}}

12{m^2} - 4{m^3} - {m^4} = 0     (A1)

m = - 6, 0, 2

\Rightarrow m = 2     A1N2

 

(ii)     {\text{P}}(X > 2) = 1 - {\text{P}}(X \leqslant 2)     (M1)

= 1 - {\text{P}}(X = 0) - {\text{P}}(X = 1) - {\text{P}}(X = 2)

= 1 - {{\text{e}}^{ - 2}} - 2{{\text{e}}^{ - 2}} - \frac{{{2^2}{{\text{e}}^{ - 2}}}}{{2!}}

= 0.323     A1

[6 marks]

 

Total [20 marks]

Examiners report

This was the best done of the section B questions, with the majority of candidates making the correct choice of probability distribution for each part. The main sources of errors: (b) missing out the binomial coefficient in the calculation; (c) failure to rearrange 'at least one bottle' in terms of the probability of obtaining no bottles; (d) using 1.2 rather than - 1.2 in the inverse Normal or not performing an inverse Normal at all; (e)(ii) misinterpreting 'more than two'.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.6 » Binomial distribution, its mean and variance.
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