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Date May 2011 Marks available 6 Reference code 11M.2.hl.TZ2.12
Level HL only Paper 2 Time zone TZ2
Command term Find Question number 12 Adapted from N/A

Question

The number of accidents that occur at a large factory can be modelled by a Poisson distribution with a mean of 0.5 accidents per month.

Find the probability that no accidents occur in a given month.

[1]
a.

Find the probability that no accidents occur in a given 6 month period.

[2]
b.

Find the length of time, in complete months, for which the probability that at least 1 accident occurs is greater than 0.99.

[6]
c.

To encourage safety the factory pays a bonus of $1000 into a fund for workers if no accidents occur in any given month, a bonus of $500 if 1 or 2 accidents occur and no bonus if more than 2 accidents occur in the month.

(i)     Calculate the expected amount that the company will pay in bonuses each month.

(ii)     Find the probability that in a given 3 month period the company pays a total of exactly $2000 in bonuses.

[9]
d.

Markscheme

P(x = 0) = 0.607     A1

[1 mark]

a.

EITHER

Using \(X \sim {\text{Po}}(3)\)     (M1)

OR

Using \({(0.6065…)^6}\)     (M1)

THEN

P(X = 0) = 0.0498     A1

[2 marks]

b.

\(X \sim {\text{Po}}(0.5t)\)     (M1)

\({\text{P}}(x \geqslant 1) = 1 - {\text{P}}(x = 0)\)     (M1)

\({\text{P}}(x = 0) < 0.01\)     A1

\({{\text{e}}^{ - 0.5t}} < 0.01\)     A1

\( - 0.5t < \ln (0.01)\)     (M1)

\(t > 9.21{\text{ months}}\)

therefore 10 months     A1N4

Note: Full marks can be awarded for answers obtained directly from GDC if a systematic method is used and clearly shown.

 

[6 marks]

c.

(i)     P(1 or 2 accidents) = 0.37908…     A1

\({\text{E}}(B) = 1000 \times 0.60653... + 500 \times 0.37908…\)     M1A1

\( = \$ 796\,\,\,\,\,\)(accept $797 or $796.07)     A1

 

(ii)     P(2000) = P(1000, 1000, 0) + P(1000, 0, 1000) + P(0, 1000, 1000) +

P(1000, 500, 500) + P(500, 1000, 500) + P(500, 500, 1000)     (M1)(A1)

Note: Award M1 for noting that 2000 can be written both as \(2 \times 1000 + 1 \times 0\) and \(2 \times 500 + 1 \times 1000\) .

 

\( = 3{(0.6065...)^2}(0.01437...) + 3{(0.3790...)^2}(0.6065…)\)     M1A1

\( = 0.277\,\,\,\,\,\)(accept 0.278)     A1

[9 marks]

d.

Examiners report

Most candidates successfully answered (a) and (b). Although many found the correct answer to (c), communication of their reasoning was weak. This was also true for (d)(i). Answers to (d)(ii) were mostly scrappy and rarely worthy of credit.

a.

Most candidates successfully answered (a) and (b). Although many found the correct answer to (c), communication of their reasoning was weak. This was also true for (d)(i). Answers to (d)(ii) were mostly scrappy and rarely worthy of credit.

b.

Most candidates successfully answered (a) and (b). Although many found the correct answer to (c), communication of their reasoning was weak. This was also true for (d)(i). Answers to (d)(ii) were mostly scrappy and rarely worthy of credit.

c.

Most candidates successfully answered (a) and (b). Although many found the correct answer to (c), communication of their reasoning was weak. This was also true for (d)(i). Answers to (d)(ii) were mostly scrappy and rarely worthy of credit.

d.

Syllabus sections

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