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Libniz
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A production line yields two types of devices:
Type1 devices occur with probability p1 and work for an average time T1,
Type2 devices occur with probability p2 and work for an average time T2,
such that p1+p2=1 and T2>T1.
The time a device works is modeled as a geometrical distribution.
Let X be the time a device chosen uniformly at random works.
Find the distribution of X.
 one year ago
 one year ago
Libniz Group Title
A production line yields two types of devices: Type1 devices occur with probability p1 and work for an average time T1, Type2 devices occur with probability p2 and work for an average time T2, such that p1+p2=1 and T2>T1. The time a device works is modeled as a geometrical distribution. Let X be the time a device chosen uniformly at random works. Find the distribution of X.
 one year ago
 one year ago

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Libniz Group TitleBest ResponseYou've already chosen the best response.0
@satellite73
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
I am having trouble setting it up
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
ok i have to think geometric distribution is \(P(n) = p(1p)^n\) right?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
jesus i have no idea. it tell you average is \(T_1\) and i think that means \(\frac{1p_1}{p_1}=T_1\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
let me try to write this correctly \[T_1=\frac{p_2}{p_1}\] and \[T_2=\frac{p_1}{p_2}\] but i am not sure that helps solve the problem do you have an example to work off of ?
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
no, this is a practice exam
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
that's ok, thanks for your help
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
ok forget that last noise i wrote, it is ridiculous and i am embarrassed maybe it is much simpler we pick device \(T_1\) with probability \(p_1\) and device \(T_2\) with probability \(p_2\) could it be as simple as \(p_1T_1+p_2T_2\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
no that isn't right either, at least i don't think so what text are you using?
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
I am actually having troubling understanding what is being asked
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i think you are being asked the following: pick a device at random, put \(X\) = amount of time it runs. Find the distribution of \(X\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
in other words, find a formula for \(P(X=k)\) the probability the device runs for time \(k\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
but really i am stuck so i should shut up. but perhaps i have a text with something similar, which is why i asked what text you were using
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
probability and schocastic process by yates
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
nope, sorry
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
do you know how to combine two geometric distributions?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
that is what they are asking
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i should delete my embarrassing answer now we will get a real one i hope
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
can you show me how to combine distributions
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
No answer from me, I never learned this stuff. But it might be a negative binomial distribution.
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
alright, thanks anyway
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
if you get the answer, please post it.
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
I will ask zarkon, when he comes on; hopefully before my quiz though
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
is that the full problem?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
I'm a little confused ... Type1 devices occur with probability p1 ... Type2 devices occur with probability p2 vs "a device chosen uniformly" are they chosen uniformly or with probabilities p1,p2
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
screen shot in case I mistyped something
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
I don't think the problem is worded very well especally near the end. here is how I see the problem let \(d1,d2\) be device 1 and device 2 then \[P(X\le x)=P((X\le x,d1)\text{ or }(X\le x,d2))\] \[=P(X\le xd1)P(d1)+P(X\le xd2)P(d2)\]
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
\[=P(X\le xd1)p1+P(X\le xd2)p2\]
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
though I could be misinterpreting the problem (since I don't think it is totally clear)
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
yeah, I had hard time even understanding what they were asking for
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
I would ask your prof...problem isn't clear to me and I have taught courses in mathematical statistics.
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
so this is conditional probability problem?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
based on what read it looks like you need to use conditional prob for at least part of the problem
 one year ago
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