f(x)=1/9xe^(-x/3) for x>0
f(x)=0 for elsewhere
What is the expected water consumption of water for any given day?
更新1:
myisland8132: 好嘢喎, 咁都記得, 你講得無錯, 雖然audrey個答法比較正路, 但係你個答案又比較直接同技術性, 點選好呢? 一於交付投票啦!
Expected value for continuous random variable
2008-03-20 4:01 am
In a certain city the daily consumption of water (in millions of liters) is a random variable whose probability density is given by
f(x)=1/9xe^(-x/3) for x>0
f(x)=0 for elsewhere
What is the expected water consumption of water for any given day?
f(x)=1/9xe^(-x/3) for x>0
f(x)=0 for elsewhere
What is the expected water consumption of water for any given day?
回答 (2)
2008-03-20 4:37 am
✔ 最佳答案
As follows~~~As follows~~~
圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/A_Hepburn01Mar192036.jpg?t=1205930228
參考: Myself~~~
2008-03-20 5:54 pm
f(x)=1/9xe^(-x/3) for x>0
the expected water consumption of water for any given day
=∫x[1/9xe^(-x/3)] dx [from 0 to infinity]
==∫1/9x^2e^(-x/3) dx
=(1/9) Gamma[3]/(1/3)^3
=(1/9)(2)(27)
=54/9
=6
我明天再正式報答啦﹐因為無手機認證﹐今日已經用盡quota
2008-03-20 09:54:40 補充：
f(x)=1/9xe^(-x/3) for x>0
the expected water consumption of water for any given day
=∫x[1/9xe^(-x/3)] dx [from 0 to infinity]
==∫1/9x^2e^(-x/3) dx
=(1/9) Gamma[3]/(1/3)^3
=(1/9)(2)(27)
=54/9
=6
2008-03-20 10:00:09 補充：
當然Audrey Hepburn沒做錯﹐不過可惜中六七沒教Gamma distribution﹐所以用了2次integration by parts。不過我覺得朋友你一定識(因為你之前問過些好深的統計題)﹐所以用了些快方法答啦(而且條題目是精算題?)
我說過補答啦﹐所以我一定要「言出必行」啦
the expected water consumption of water for any given day
=∫x[1/9xe^(-x/3)] dx [from 0 to infinity]
==∫1/9x^2e^(-x/3) dx
=(1/9) Gamma[3]/(1/3)^3
=(1/9)(2)(27)
=54/9
=6
我明天再正式報答啦﹐因為無手機認證﹐今日已經用盡quota
2008-03-20 09:54:40 補充：
f(x)=1/9xe^(-x/3) for x>0
the expected water consumption of water for any given day
=∫x[1/9xe^(-x/3)] dx [from 0 to infinity]
==∫1/9x^2e^(-x/3) dx
=(1/9) Gamma[3]/(1/3)^3
=(1/9)(2)(27)
=54/9
=6
2008-03-20 10:00:09 補充：
當然Audrey Hepburn沒做錯﹐不過可惜中六七沒教Gamma distribution﹐所以用了2次integration by parts。不過我覺得朋友你一定識(因為你之前問過些好深的統計題)﹐所以用了些快方法答啦(而且條題目是精算題?)
我說過補答啦﹐所以我一定要「言出必行」啦
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