At a service counter , arrivals occur at an average rate of 20 per hour and follow a Poisson distribution. Let X denote the number of arrivals during a particular hour.
a) Using statistical software , we found that P(X 小於等於 20) = 0.559 . Based on that result, find P (X 大於等於 20)
b) If there were at least 20 arrivals during a particular hour , what is the probability that there were exactly 25 arrivals?
probability (7)
2007-11-06 8:28 am
回答 (2)
2007-11-06 7:23 pm
✔ 最佳答案
(a) P(X ≦ 20) = 0.559∴ P(X > 20) = 1 - 0.559 = 0.441
Now, with λ = 20 for the Poisson distribution,
P(X = 20) = 2020 × e-20/20! = 0.088835
Therefore,
P(X ≧ 20) = P(X > 20) + P(X = 20)
= 0.529835
(b) This is equivalent to the probability P(X = 25 | X ≧ 20).
Now, P(X = 25) = 2025 × e-25/25! = 0.000300
Hence,
P(X = 25 | X ≧ 20) = P(X = 25)/P(X ≧ 20)
= 0.000300/0.5298
= 0.000567
參考: My Mahths knowledge
2007-11-11 6:30 am
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