誰可以幫我?? [Poisson distribution](加了兩題)

lyeng

幫幫我解這題好嗎。。。
A sales manager recieves six telephone calls on average between 930 am and 1030 am on a weekday. Find the probability that during a five-day working week, there will be exactly three days on which she recieves NO calls between 10 am and 10.10 am

[ Last edited by lyeng on 9-6-2004 at 11:14 PM ]

微中子

如果没有错的话

既然1小时(930-1030)平均上有六通电话,因此十分钟会有1通电话.

所以λ = 1

在一天的10-1010,P(0)= e^(-1)

过后,定义p = 一天里,10-1010没有任何一通电话的机率.

过后用binomial来找咯, p = P(0)
应该是5C3(e^(-1))(1-e^(-1))

不懂对吗?
statistics都忘了.

lyeng

原來要用binomial..嘿。。。謝拉~~~

微中子 于 5-6-2004 05:25 PM  说 : 应该是5C3(e^(-1))(1-e^(-1))

這裡是应该是5C3(e^(1))^3 (1-e^(-1))^2

微中子

哈哈.. 对对...
死了..
太多东西忘记

lyeng

The probability that a bolt is defective is 0.2%. Bolts are packed in boxes of 500. Two boxes are picked at random from the production line. Find the probabilty that one has two defective bolts and the other has no defective bolts. (ans=0.271)

我的方法>>
X~Po(2)    P(X=2)=e^-2 (2^2/2!) = 0.271

答案是對了。。。但我覺得方法好像不太對。。。可以告訴我why對/不對嗎?

lyeng

An aircraft has 116 seats.The airline has found, from long experience, that on average 2.5% of people who have bought tickets for a flight do not arrive for that flight. The airline sells 120 tickets for a particular flight.

(a) Calculate, using a suitable approximation, the probability that more than 116 people arrive for the flight(0.647)

(b) calculate also the probability that there are empty seats on the flight(0.1828)

這題我用binomial來找答案。。。
可是我想請教下這題可以用poisson 來solve嗎?

铁蛋

lyeng 于 9-6-2004 11:08 PM  说 :
The probability that a bolt is defective is 0.2%. Bolts are packed in boxes of 500. Two boxes are picked at random from the production line. Find the probabilty that one has two defective bolts and ...

请检查。。。是 一 吗？ 如果要用泊松分布，那么应该是 X ~ Poisson (1). 因为 np = 500*2/1000 = 1.

lyeng

铁蛋 于 10-6-2004 08:21 AM  说 :


请检查。。。是 一 吗？ 如果要用泊松分布，那么应该是 X ~ Poisson (1). 因为 np = 500*2/1000 = 1.

是 two沒錯~~~~~

微中子

我觉得lambda应该是1才对.
不过只能拿到答案的一半.

让X ~ Po(1),

P(X=2) = e^{-1}/2
P(X=0) = e^{-1}

所以P(one has 2 defective bolts & one has none) = 2*P(X=2)P(X=0) = e^(-2)

是答案的一半咯.

不过如果问题是要
P(2 defective bolts from 2 boxes) = e^(-2) + (P(X=1))^2 = 2e^(-2)
就有.271啦..

我做对吗?

lyeng

微中子 于 10-6-2004 04:49 PM  说 :
P(2 defective bolts from 2 boxes) = e^(-2) + (P(X=1))^2 = 2e^(-2)
就有.271啦..

i dun understand this part

and why cant lambda be 2?

微中子

因为平均上,每箱有一个defective的bolt..
500*0.002 = 1,
不是吗?
而且np = 1
npq = .998 差不多 = 1

P(2 defective bolts from 2 boxes) = e^(-2) + (P(X=1))^2 = 2e^(-2)

就是说两个箱子有两个defective的bolt的机率



不过不要太相信我..
我的statistics已经换给老师了.

[ Last edited by 微中子 on 10-6-2004 at 10:36 PM ]

lyeng

微中子 于 10-6-2004 10:32 PM  说 :
因为平均上,每箱有一个defective的bolt..
500*0.002 = 1,
不是吗?
而且np = 1
npq = .998 差不多 = 1

hmm...this one i know...
but 2 boxes are picked...
1x2=2 <= cannot do like this? why? in wat condition can we do like this....confusing...

anyway....thank u very much...u really help me a lot....

微中子

嘿嘿..请尽量用中文发表...谢谢

我想啦..
如果问题是要求两个箱子里有2个defective的bolt,那我觉得可以,因为我们不区别这两个箱子..我们可以讲平均上,两个箱子有两个defective的bolt.

不过如果问题是一个箱子有两个,一个箱子没有,
所以必须区别这两个箱子,
因此每个箱子平均有一个.
然后,再来考虑两个箱子的情况.

我的看法啦..
还请高手指点.

铁蛋

lyeng 于 10-6-2004 10:39 PM  说 :


hmm...this one i know...
but 2 boxes are picked...
1x2=2 <= cannot do like this? why? in wat condition can we do like this....confusing...

anyway....thank u very much...u really help me ...

首先我们先了解一下问题。。。设 X 为一盒箱子里故障螺丝的数量。明显的，X 的分布是二项式分布[ X ～ Binomial (n, p) ]。 可是因为 p 很小所以我们可以用泊松分布 (Poisson distribution) 为一个合理的逼近 (Approximation)。故有 X~Poisson(1). 现在任意选两个盒子，假设这些盒子里故障螺丝的数量分布都是X~Poisson(1). 那么根据问题所要求的，是

X_1 = 第一盒子里故障螺丝的数量 ;
X_2 = 第二盒子里故障螺丝的数量 ;

P(X_1 =2, X_2 = 0) OR P(X_1=0, X_2= 2) = e^(-2) .

有时后参考书的答案会不正确, 重要的是逻辑步骤。 。

[ Last edited by 铁蛋 on 11-6-2004 at 12:37 PM ]

铁蛋

微中子 于 10-6-2004 04:49 PM  说 :
我觉得lambda应该是1才对.
不过只能拿到答案的一半.
...
不过如果问题是要
P(2 defective bolts from 2 boxes) = e^(-2) + (P(X=1))^2 = 2e^(-2)
就有.271啦..
我做对吗?

合理。

flash

lyeng 于 9-6-2004 23:14  说 :
An aircraft has 116 seats.The airline has found, from long experience, that on average 2.5% of people who have bought tickets for a flight do not arrive for that flight. The airline sells 120 ticke ...

这题可用 poisson 来做。得到的答案是

(a) 0.647
(b) 0.185

[ Last edited by flash on 15-6-2004 at 04:54 PM ]

akito

A car insurance company has determined that 5% of all drivers were involved in a car accident last year.Among the 10 drivers living on one particular street,3 were involved in a car accident last year.If 10 drivers are randomly selected,what is the probability of getting 3 or more who were involved in a car accident last year?

那个lambda是拿那个...?
是可以用Binomial 做对吗?

akito

原帖由 akito 于 17-12-2006 02:04 AM 发表
A car insurance company has determined that 5% of all drivers were involved in a car accident last year.Among the 10 drivers living on one particular street,3 were involved in a car accident last y ...

是10C3*0.05^3*0.95^7+10C4*0.05^4*0.95^6+.........10C10*0.05^10*0.95^0=Ans对吗?

dunwan2tellu

原帖由 akito 于 17-12-2006 03:02 AM 发表

是10C3*0.05^3*0.95^7+10C4*0.05^4*0.95^6+.........10C10*0.05^10*0.95^0=Ans对吗?

也可以用 1 - P(X=0) - P(X=1) - P(X=2)

where P(X=r) = 10Cr * (0.05)^r * (0.95)^(10-r)

akito

原帖由 dunwan2tellu 于 18-12-2006 09:25 AM 发表


也可以用 1 - P(X=0) - P(X=1) - P(X=2)

where P(X=r) = 10Cr * (0.05)^r * (0.95)^(10-r)

谢谢你！不要告诉你
你的方法很好
我的statistics烂到。。。。
可以再问多你一题吗...?
The participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men and women.Among the last 11 participants there have been only 2 women.If participants are picked randomly,what is the probability of getting 2 or fewer women when 11 people are picked?

11C2*(0.5)^2(0.5)^9 + 11C1*(0.5)^1(0.5)^10 + 11C0*(0.5)^0(0.5)^11= ans 对吗?