Evidence shows that the probability that a driver will be involved in a serious automobile accident during a given year is .01. A particular corporation employs
100 full-time traveling sales reps. Based on this evidence, use the Poisson approximation to the binomial distribution to find the probability that exactly two of the sales reps will be involved in a serious automobile accident during the coming year.
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We use the formula:
P(x) = [ e^-(np) * (np)^x ]/x!
where P(x) = probability of "x" successes with parameters n and q.
Here n = sample size, and p = probability of success.
x = number of successes in the sample.
So, for the problem above:
n = 100, p = 0.01 and x = 2.
Plugging that on our formula we will have:
P(x = 2) = [ e^-(100*0.01) * (100*0.01)^2 ]/2!
P(x = 2) = 0.1839397
The expected number of accidents among 100 reps in 1 year is 0.01 * 100 = 1. Hence the number of accidents is approximately Poisson(1).
If X~Poisson(1) then P(X=2) = exp(-1)*1^2/2! = 0.1839397.
Hence there is an 18.4% chance.
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