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Error in counting detector

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phoenix95
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Homework Statement


A detector is used to count the number of gamma photons emitted by a radioactive source. If the number of counts recorded in exactly 20 seconds is 10000, the error in counting rate per second is
(a) 5.0
(b) 22.4
(c) 44.7
(d) 220.0

Homework Equations




The Attempt at a Solution


I don't know anything about this problem. I thought the counting rate would be 500 per second. I went through other threads, and in some Poisson distribution is mentioned. I don't know if it is relevant here... Can someone help me?

Thank you for your time...
 
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Answers and Replies

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Buzz Bloom
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Hi phoenix:

I am unsure I understand the vocabulary of the question, but I think the assumption of a Poisson distribution is important. Take a look at the Wikipedia article your broswr can find using
poisson distribution !wiki​
Look at the definition section of the article. The key idea is that the variance equals the average. I think the question about "the error" means the expected error, which is the expected value with respect to the distribution of possible error values. Do you understand how this value relates to the variance?

Hope this helps.

Regards,
Buzz
 
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phoenix95
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I am unsure I understand the vocabulary of the question, but I think the assumption of a Poisson distribution is important. Take a look at the Wikipedia article your broswr can find using
poisson distribution !wiki
Thanks for that buzz bloom :smile:.
Look at the definition section of the article. The key idea is that the variance equals the average. I think the question about "the error" means the expected error, which is the expected value with respect to the distribution of possible error values. Do you understand how this value relates to the variance?
But they haven't given a set of data right? I mean to calculate the mean and variance I would need a set of measurements, instead of just one...:oops:
 
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Buzz Bloom
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But they haven't given a set of data right? I mean to calculate the mean and variance I would need a set of measurements, instead of just one
Hi pheonix:

You have an estimate for the mean count rate of 500/s. 500 is also related to the expected error for 1 count. Do you have a formula that calculates from that the expected error given 10000 counts? Take a look at Wikipedia articles for variance and for standard error.

Regards,
Buzz
 
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  • #5
Ray Vickson
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Homework Statement


A detector is used to count the number of gamma photons emitted by a radioactive source. If the number of counts recorded in exactly 20 seconds is 10000, the error in counting rate per second is
(a) 5.0
(b) 22.4
(c) 44.7
(d) 220.0

Homework Equations




The Attempt at a Solution


I don't know anything about this problem. I thought the counting rate would be 500 per second. I went through other threads, and in some Poisson distribution is mentioned. I don't if it is relevant here... Can someone help me?

Thank you for your time...
I will explain the issue to you, then let you deal with the rest of the analysis. You have a Poisson decay process with some unknown "rate" parameter ##r## (usually denoted by ##\lambda##, but ##r## is the first letter of the word "rate" and is also a lot easier to type). Anyway, the probability mass function for ##X_T## = number of decays in ##T## intervals of time is
[tex] \text{Probability that} \; \{X_T = n \} = P(X = n) = \frac{(rT)^n}{n!} e^{-rT}, \; n = 0,1, 2,3, \ldots [/tex]
The expected value of ##X_T## (= the mean of ##X##) is just ##rT##, as is the variance. It is a fundamental property of the Poisson distribution that its individual sub-parts are also Poisson-distributed, so (for example), the number of decays in second 1 is Poisson with mean ##r##, the number in second 2 is also Poisson with mean ##r##, the number in the first two seconds is Poisson with mean ##2r##, etc.

So, the number of decays in 20 seconds is Poisson with mean ##20r##. However, we do not know the value of ##r## or of ##rT = 20r##. All we have is a single observation of ##X_{20}##, and from that we want to say something about the value of ##r## or ##20r##. If you were asked to pick a single value of ##m = 20r## to bet on, what value would you choose? Personally, I would choose the observed value, 10,000. Basically, I have no reason at all to pick any other value than the one observed, so I would pick ##\hat{r} =10,000/20 = 500## (per second) as my estimate of ##r##. However, my estimate is bound to be at least a bit wrong because it is based on a finite sample of a random process. So, we want to estimate just how badly wrong our estimate could plausibly be, and that amounts to asking how large a variation in ##m = 20r## could we have while still making an observation of ##10,000## believable? The problem is that the observation itself is improbable, so we are really asking "how much less probable could our observation be before it becomes not-credible"?

Perhaps a coin-tossing example would be easier to follow, so let's imagine tossing a coin 100 independent times, with ##\Pr(H) = p## and ##\Pr(T) = 1-p## per toss. If ##X## is the observed number of heads in 100 tosses, the mean of ##X## is ##m = 100p##, and the "mode" (= most probable single value) is close to ##m## as well. So, if asked to bet on a value of ##m## based on a single observation ##X = n## I would choose the estimate (##\hat{m}##) of ##m## as ## \hat{m} = n##. For those reasons, if I observed exactly 50 heads, I would estimate that ##m = 50##, but I could very well be wrong! The value ##m = 50## would imply that ##p = 1/2## (a fair coin). For ##p = 1/2## exactly, ##\Pr{X = 50} \doteq 0.07959##, which is small. In other words, even if our estimate is exactly right, the observation is still not very likely---it is just more likely than any other single observation! In other words, for ##p \neq 1/2## the probability of ##\{X = 50\}## would be even smaller. For example, for ##p = .0.4## we would have ##P(X = 50) \doteq 0.01004##, which is still, perhaps believable. However, if ##p = 0.3## then ##P(X = 50) \doteq 0.1303 \: 10^{-4}##, and that seems too small to be believable: it is about 10,000 times less likely than the value we obtain when ##p = 1/2##.

Basically, we want to devise an interval (a so-called "confidence interval") ##[\underline{p},\overline{p}]## so that we would feel "comfortable" making the claim that ##\underline{p} \leq p \leq \overline{p}##, but would feel very uncomfortable about asserting that either ##p < \underline{p}## or ##p > \overline{p}##. The remaining issues are what we mean be being comfortable or uncomfortable, and how to actually perform the computations of ##\underline{p}## and ##\overline{p}##. Statisticians have developed many tools and methods for dealing with these aspects of the problem.

The issues are similar for the Poisson-distribution case: you want to determine an interval ##[\underline{r}, \overline{r}]## for the unknown value of ##r##. It will certainly contain your single estimated value ##\hat{r} = 500##. You can Google "Poisson confidence interval" to find example and formulas, etc.
 
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