GR0177 #16


Problem


This problem is still being typed. 
Lab Methods}Sample
The mean of the ten number is . Thus, the standard deviation of the sample is . (Search for Poisson Distribution on the site for another problem similar to this.)
If the student wants to obtain an uncertainty of 1 percent, then
,
where one assumes the average scales uniformly and C is the time to count. (Note: a good approximation of the uncertainty is given by the ratio of the standard deviation to the average, since that represents the deviation.)
Thus, one has . Thus the student should count C=5000 s.
A Sneak Peek of The NotSoBoring Review of Undergrad Physics
(To be published in the tobeposted library section of http://GREPhysics.NET in Feb 2006.)
The Poisson Distribution is intimately related to the raising and lowering operators of the (quantum mechanical) simple harmonic oscillator (SHO). When you hear the phrase ``simple harmonic oscillator," you should immediately recall the number operator , as well as the characteristic relations for the raising and lowering and . And, don\'t forget the commutation relations that you should know by heart by now, . (That\'s all part of the collective consciousness of being a physics major.)
Now, here\'s some quasiquantum magic applied to the Poisson Distribution. I\'m going to show you how to arrive at the result for standard deviation, i.e., from using the SHO operators.
Let\'s start with something easy to help jog your memory: The mean or average number in the distribution is just the expectation value of the Number operator,
Okay! So, on with the fun stuff: the standard deviation is given by the usual definition, .
The second term is already determined from the above expression for the mean, .
The first term can be calculated from . Now, the commutation relation gives, . Replacing the middle two of the four a's with that result, the expression becomes .
Plugging the above results into the standard deviation, I present to you, this: .
It\'s no coincidence that the above works. The secret lies in the energy eigenfunction that you might not remember...
\subsection{The Poisson Distribution Function is just The Poisson Distribution for a parameter is given by . But, wait, doesn\'t that look a wee bit too familiar? Indeed, the Poisson Distribution is merely the probability of obtaining photons at position : .\footnote{Note that since , each time n increases, it is like we've created an extra photon, since the energy of a photon is . Thus, represents the quanta.}
Why? So you ask. Well...
The energy eigenfunctions of the SHO is given by , where .
This result can be arrived at in the position basis as follows: . The in the exponent can be reexpressed by the relation . Thus, , where the other terms like and we\'ve used the result for from above with the implicit assumption that its complex conjugate is the same, as the average photon number, , is an observable. Thus, the exponent becomes .
The probability in the basis is thus, , where using the definition , we\'ve recovered exactly the Poisson Distribution for a parameter .
Making the following associations, and , you carve the first etchings in the Rosetta Stone between probability and photon statistics...


Alternate Solutions 
NervousWreck 20170328 09:45:40  This is a very complicated problem due to the lack of time. However the fasters solution as I see it is the following\r\n. Here rate is 2, which is multiplied by 1%. On the LHS is an error of poissson distribution with 2 measurements per second taken into account. The solution gives N = 5000, which now corresponds to seconds.   redmomatt 20111004 11:02:41  Much easier way to think of this.
Approximate the variance, .
Thus, the standard deviation is .
Now, as N increases the error decreases by .
Therefore, .  

Comments 
Suman05eee 20171012 08:25:50  If we consider same mean for all different sized samples for this process, then :\r\n\r\nAccording Central Limit Theorem:\r\n\r\nSample SD=Population SD/\r\n\r\nThus we can think of a sample of size N with the same SD (Calculated from the given sample of size 10=) and get the Population SD=. And Mean = (Population SD)^2.= 2N.\r\nAnd at last use the uncertainty = Population Mean/ Population SD= 1/.   NervousWreck 20170328 09:45:40  This is a very complicated problem due to the lack of time. However the fasters solution as I see it is the following\r\n. Here rate is 2, which is multiplied by 1%. On the LHS is an error of poissson distribution with 2 measurements per second taken into account. The solution gives N = 5000, which now corresponds to seconds.   ewcikewqikd 20140705 14:52:10  This problem is just statistics.
Suppose the standard deviation of the true population is .
The standard deviation of the mean of a sample of size 10 is expected to be .
The standard deviation of the mean of a sample of size n is expected to be .
Because the mean of the given sample is 2 and the question ask for uncertainty of 1%, we set
The standard deviation of the given sample is
We can divide the two equations above to get
Solving for n, we get
60000 measurements are needed!   redmomatt 20111004 11:02:41  Much easier way to think of this.
Approximate the variance, .
Thus, the standard deviation is .
Now, as N increases the error decreases by .
Therefore, .
rizkibizniz 20111107 00:38:10 
question: how have you come to approximate the variance as 5?

Rhabdovirus 20121028 18:13:58 
Riz: Variance is squared deviation so since the mean = 2, variance goes like which gives you which is about 5.

luwei0917 20140329 11:42:34 
where come from?

  phoxdie 20101112 21:40:21  I have a quick question. When I looked at this the first time I arrived at the correct solution by the following method. First there are 10 measurements made, this is given. Next out of the ten the maximum spread is from 0 to 5 so I made the uncertainty in their measurement 5. Looking at the answers and what they are asking for, ie that the uncertainty be 1% I simply said ie 1%. This turns out to be the correct answer (5000 seconds) but I am not sure if my naive method is legitimate or not. Does anyone think this is absolutely wrong, and I just got lucky or that there is something behind this? Thanks!   wittensdog 20091008 21:45:38  The language of this problem is indeed pretty vague, so, here is one solution based on the way I interpreted it. I know there has already been a lot of talk on this, I hope maybe I can help sort things out a little...
First, if you just average all of those values, you get 2. So now I guess we just postulate that that should be close enough to the true average for us to get an idea of how long we should count for.
Now, in a Poisson distribution (which describes radioactive phenomenon, or most of it), we know that the standard deviation is the square root of the average. So we can take the standard deviation of this distribution to be sqrt(2). I don't know what the SD is if you actually calculate it for those numbers, but anyone who goes trying to calculate standard deviations from data on the GRE is completely insane.
Now, for reasons that can be seen if you take a course in statistics, the error on the mean is generally taken to be the standard deviation of the measurements divided by the square root of the number of measurements (this stems from the central limit theorem). I believe this is what is meant by uncertainty here. They state an uncertainty of one percent. I don't know exactly what it is that we want one percent of, but I'm guessing they mean 1% of the mean value, aka, the error on the mean should be plus or minus 1 percent of the mean. I don't know what else they would be referencing.
Since we are taking the mean as 2, or at least assuming it should be something in that ballpark, one percent of that would be 0.02. So if we know the standard deviation, the uncertainty we want, and the formula for the uncertainty on the mean, then we get,
uncert = SD / sqrt (n) ==>
0.02 = sqrt(2) / sqrt(n) ==>
4e4 = 2/n ==>
n = 0.5 e +4 ==>
n = 5,000
So we want to make 5,000 measurements, and since each measurement is one second long, this corresponds to 5,000 seconds.
I hope this manages to help someone (and that I'm actually doing it right!).
Prologue 20091105 10:13:11 
Thank you!

kiselev 20110318 11:29:03 
Well done!

timmy 20110502 20:51:54 
this is correct.
also, yosuns solution, as usual is absolutely terrible and makes no sense at all. Honestly who is yosun, and why is all of yosuns solutions so bad???
I mean this site is great, but some of these solutions are just grossly wrong in terms of methodology, even if they are technically correct.

timmy 20110502 20:55:15 
let me elaborate: when I say Yosun's solutions are bad, I don't mean they are wrong, far from it.
The problem is that they go into to much detail and theory to be almost useless to the test taker.
The testtaker needs to UNDERSTAND the basics of what they need to know and QUICKLY SOLVE the problem. Yosun's solution don't EXPLAIN the problem well at all. The solution given above is terrible in terms of explaining what is going on in the problem with the rate and the 1% etc.

Quark 20111005 14:32:36 
yosun created this website so you should actually be thankful for all of his solutions.

rizkibizniz 20111107 00:44:57 
@Quark you mean her solutions. Yosun is a she.

  tensorwhat 20090319 20:39:42  This is way more simple....
/N = 1% = 1E2
Solve for N
N = 1/1E4 = 10,000 s
There are 2 counts per second, so 10,000 s/2 = 5,000 s
Done.
AER 20090402 16:14:24 
Where 2 is still the average of the ten measurements, and each measurement was 1 sec long.

ajkp2557 20091107 05:03:37 
Small typo: 10,000 should be number of counts (unitless), not measured in seconds. Your units will come from the fact that you're dividing by 2 counts / sec.

  eshaghoulian 20070929 19:34:12  I think if you piece together everyone's comments, you'll have a final solution. Here is my thought progression:
The number of counts N is 20 for a time T of 10 seconds, giving a rate R of N/T = 2. Here we invoke the rule without justification, which allows us to say that the uncertainty of an Ncount distribution is . We use the formula for fractional uncertainty which motivates rampancy's form of the uncertainty.
So, for X seconds, we have a total number of counts 2X, and we use the equation above to get
Notice that the fractional uncertainty of the rate is just the fractional uncertainty of the total number of counts. I am not sure about the language here; I want to say that ETS's use of the term "uncertainty" in the question is at best vague, but I am not familiar with this type of experiment (reminiscent of the Q factor, which has as many definitions as it has occurences in physics).
See section "Counting Statistics" in link below for a little more detail
http://www.colorado.edu/physics/phys1140/phys1140_sp05/Experiments/O1Fall04.pdf
ericimo 20071027 14:33:55 
Correct, except there IS justification.
Since all we know from the problem is that it involves radiation detection, the vague nature allows us to assume that the distribution will follow the most common distribution in radiation detection.
And for most radiation measurements, the distribution is a Poisson distribution (hence Yosun's inclusion of the Poission discussion) which is where the employed rule for uncertainty comes into play.

wystra 20161017 04:52:20 
Best answer

  michealmas 20061227 19:09:12  Sorry for the formatting screwup:
Trying to clear up Yosun's solution  Yosun's formula for uncertainty is wrong. He claims it's:
/AverageCounts
when actually it's
/TotalCounts
you can correct Yosun's equation by multiplying the denominator by the total seconds, or C as Yosun calls it. That is what he does, though without explanation.   michealmas 20061227 19:06:57  Trying to clear up Yosun's solution  Yosun's formula for uncertainty is wrong. He claims it's:
\sigma/TotalCounts
you can correct Yosun's equation by multiplying the denominator by the total seconds, or C as Yosun calls it. That is what he does, though without explanation.   simpsoxe 20061130 21:43:30  if you claim that is the average and is the standard deviation, then how do you go from there to get that is the answer? I'm confused as to how the C's get in there   rampancy 20061102 00:50:14  That explanation makes no sense, and seems needlessly complicated.
The total number of counts in 10 seconds is 20. The error in that is Sqrt(20).
Counts = 20 +/ sqrt(20).
The average number of counts is 2, so in N seconds, we should see,
2N +/ sqrt(2N) counts.
We want the fractional error to be .01, so,
sqrt(2N)/(2N) = .01
So N = 5000.
nitin 20061113 14:23:09 
I agree with rampancy. Yosun, your solution is nonsense, and it seems you don't even know what you're talking about.

mr_eggs 20090816 18:21:53 
You don't have to be a jerk, nitin. This is an open community site to help those trying to get into grad school. If you don't have anything intelligent to add, then fuck off.rnrnA little late.. haha..

mr_eggs 20090816 18:31:54 
You don't have to be a jerk, nitin. This is an open community site to help those trying to get into grad school. If you don't have anything intelligent to add, then fuck off.
A little late.. haha..

FutureDrSteve 20111107 14:23:11 
Also late, but totally agreed. There are still no decent books and few good resources to prepare for the PGRE. This site is an absolute godsend. Yosun's solutions are not always the best solutions for me to have that "Ah ha!" moment, but I imagine that's why she was brilliant enough to make this a community site. And aside from the HATERS, it's a good community. This site is the sole reason I will do well on test day. Thanks, Yosun!

wystra 20161017 04:51:25 
Best answer

  yosun 20051127 01:50:12  Poisson Distribution, the way it was meant to be.
Blake7 20070919 06:02:24 
It's beautiful, Yosun! How can I get a copy of your wonderful book?

Ge Yang 20101005 12:46:46 
Right, Yosun, where can we get your wonderful book?
This website definitely has its own memory...

 

Post A Comment! 
You are replying to:
I think if you piece together everyone's comments, you'll have a final solution. Here is my thought progression:
The number of counts N is 20 for a time T of 10 seconds, giving a rate R of N/T = 2. Here we invoke the rule without justification, which allows us to say that the uncertainty of an Ncount distribution is . We use the formula for fractional uncertainty which motivates rampancy's form of the uncertainty.
So, for X seconds, we have a total number of counts 2X, and we use the equation above to get
Notice that the fractional uncertainty of the rate is just the fractional uncertainty of the total number of counts. I am not sure about the language here; I want to say that ETS's use of the term "uncertainty" in the question is at best vague, but I am not familiar with this type of experiment (reminiscent of the Q factor, which has as many definitions as it has occurences in physics).
See section "Counting Statistics" in link below for a little more detail
http://www.colorado.edu/physics/phys1140/phys1140_sp05/Experiments/O1Fall04.pdf

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