Financial Risk Manager Part 1
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A motor vehicle production company based in California is assembling its first batch of fully electric cars. After inspecting about 100 newly assembled units, engineers establish that there are a total of 40 defects. While some units have no defects, others have one, two, or more defects. Assume that the distribution of mechanical defects follows a Poisson distribution. Drawing on the first 100 units produced, how many cars, out of every 10,000 units assembled, would we expect to have at least one defect?
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A
About 330
B
About 0.330
C
About 3,300
D
About 1250
Explanation:
Explanation
Let's use X to denote the number of defects in a car. X ~ Poi(40/100) i.e. λ = 0.4
P(X is at least 1) = 1 - P(X = 0) = 1 - exp(-0.4) = 0.330
This is the probability of at least 1 defect in a car. Therefore, for every 10,000 cars, we would expect 0.330 * 10,000 = 3,300 units to have one or more defects.
Further Explanation
From the question, the most important information is that we have 40 defect vehicles in every 100 vehicles manufactured by the company. This implies that, probability of defect is equal to 40/100.
Now, to solve for the probability that there is at least one defect vehicle, putting in mind that this follows a Poisson distribution, with λ = 40/100 = 0.4,
By having at least one defect, means that we expect either 1 defect or more. So that, we don't expect that there will be zero defects, therefore,
P(at least one defect) = 1 - P(No defect) = 1 - P(x = 0)
Now, recall that the pmf of a Poisson distribution is:
P(X = k) = (λ^k * e^(-λ)) / k!
For k = 0: P(X = 0) = (0.4^0 * e^(-0.4)) / 0! = (1 * e^(-0.4)) / 1 = e^(-0.4) ≈ 0.670
Therefore: P(at least one defect) = 1 - 0.670 = 0.330
For 10,000 cars: 0.330 * 10,000 = 3,300 cars*
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