A Level Further Mathematics for AQA Statistics Student Book
P( Z > 4) = 1 − P( Z ø 4) = 0.853 (3 s.f.)
Common error
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b The rate of arrival of messages is unlikely to be constant – there will probably be more at some times of the day than at others. Within each distribution messages are not likely to be independent as they may occur as part of a conversation. The two distributions are also probably not independent of each other, as times when more emails arrive might be similar to times when more texts might arrive.
You need to write the required probability in terms of a cumulative probability to use the calculator function.
WORK IT OUT 2.1
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Sometimes people think that the mean rate in a Poisson distribution has to be a whole number. This is not the case.
The number of errors in a computer code is believed to follow a Poisson distribution with a mean of 2.1 errors per 100 lines of code. Find the probability that there are more than 2 errors in 200 lines of code. Which is the correct solution? Can you identify the errors made in the incorrect solutions? A If X is the number of errors in 200 lines then X ~ Po(4.2).
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P( X > 2) = 1 – P( X ø 2) = 1 – (P( X = 0) + P( X = 1) + P( X = 2)) ≈ 0.790
B If X is the number of errors in 100 lines then X ~ Po(2.1). More than 2 errors in 200 lines is equivalent to more than 1 error in 100 lines, so you need P( X > 1) = 1 – P( X ø 1) = 1 – (P( X = 0) + P( X = 1)) = 0.620
C
X ~ Po(4.2).
P( X > 2) = 1– P( X < 2) = 1– P( X = 1) + P( X = 0) ≈ 0.952
EXERCISE 2A 1
State the distribution of the variable in each of these situations. a Cars pass under a motorway bridge at an average rate of 6 per 10 second period. i The number of cars passing under the bridge in one minute. ii The number of cars passing under the bridge in 15 seconds.
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