Chapter 2: The Poisson distribution
3 For the random variable X , where X ~ B(200, 0.01), use a suitable approximation to find P(6 , X , 10). 4 Past records show that the proportion of faulty resistors manufactured by a company is 0.2%. Robert buys a box of 450 resistors. Using a suitable approximation, work out the probability that less than three resistors are faulty.
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5 A machine is known to produce defective components 0.24% of the time. In a production of 500 components, state a suitable approximating distribution and calculate the probability that, at most, three components will be defective.
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6 A rare reaction from a prescribed medicine occurs in 0.05% of patients.
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a Using a suitable approximation, find the probability that in a random sample of 3000 patients prescribed this medicine, four or more will suffer the rare reaction.
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b In a random group of n patients, the probability that none suffer the rare reaction is 0.001. Work out the value of n. 7 For a certain flower, the probability that a seed carries a mutation is 0.0004. A total of 12 000 seeds germinate. Let X be the number of seeds that germinate and carry the mutation. a Justify using the Poisson distribution as an approximating distribution for X . b Use your approximating distribution to find P( X ø 3). c Calculate P( X ø 3) given that P( X . 1).
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Consider again the example of the manufacturer of plastic pipe at the start of Section 2.3. In that example there were n small pieces of pipe in which there was only zero or one defect 4 in each piece, and hence X ~ B n, . To ensure each piece does not contain more than n one defect, the pieces would need to be very small and we would have a large number of pieces.
1000 1000 4 0 4 1 = 0.018169… Suppose n = 1000, then P(X = 0) = − 1000 0 1000 0
10 000 4 4 1− For n = 10 000, P(X = 0) = 10 000 0 10 000 0
10 000
100 000 4 4 1− For n = 100 000, P(X = 0) = 100 000 100 000 0
= 0.018300… 100 000
= 0.018315…
We can see that for increasing values of n the probabilities tend towards 0.018315… = e −4. This is the value given by the Poisson probability for P( X = 0), where X ~ Po(4) .
4 Let’s move on to explore the probabilities for one defect using the binomial X ~ B n, . n 999 1000 4 1 4 1 = 0.072969… For n = 1000 , P( X = 1) = − 1000 1 1000 1
9999 10 000 4 4 For n = 10 000, P( X = 1) = 1 − = 0.072323… 1000 1 10 000
If we continue to calculate probabilities for one defect, increasing the values of n, the result will tend towards 0.073262…, and this is the value given by the Poisson probability for P(X = 1) for e − λ . Original material © Cambridge University Press 2017
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