Probability Distribution
2012-09-09 9:51 pm
The number of calls received per minute by an enquiry hotline follows a Poisson distribution with mean 3. Find the minimum value of n such that the probability of receiving at most n calls in a minute is greater than 0.8.
回答 (2)
2012-09-10 3:16 am
✔ 最佳答案
P(X = x) = 3^x exp(-3)/x!P(X = 0) = exp(-3) = 0.049787
P(X = 1) = 3exp(-3) = 0.14936
P(X = 2) = 9exp(-3)/2 = 0.22404
P(X = 3) = 27exp(-3)/3! = 0.22404
P(X = 4) = 81exp(-3)/4! = 0.16803
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 0.815257
So, the min. value of n is 4
2012-09-09 10:51 pm
P(X=0)=0.049787068
P(X=1)=0.149361205
P(X=2)=0.224041808
P(X=3)=0.224041808
P(X=4)=0.168031356
add them up 0.815263245 > 0.8
so, n=4
P(X=1)=0.149361205
P(X=2)=0.224041808
P(X=3)=0.224041808
P(X=4)=0.168031356
add them up 0.815263245 > 0.8
so, n=4
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120909000051KK00289