Standard deviation and count rate

[Zuzana]

Hello,

I watched MIT course on Nuclear physics (13. Practical Radiation Counting Experiments on ytb) and I do not understand why 2*sigma (standard deviation) = 0.05* countRate. As far as I know, integral of normal distribution from -2sigma to 2 sigma gives 95 % probability, but how can 2*sigma equals 100%-95% of count rate?

Thank you for the answer.

 

[BvU]

Hello @Zuzana,

$\qquad$!
​

As you may know, a counting rate obeys a Poisson distribution. ( e.g. sheet 17 here ). For reasonable counting rates, such a Poisson distribution is very close to a Gaussian distribution ( ibid sheet 20 ). Hence the 95%.

$\$

 

[Hornbein]

u = measured mean
s = measured standard deviation

There is a 95% chance that the true mean lies in the interval
-1.96s+u to u+1.96s

 

[Zuzana]

u = measured mean
s = measured standard deviation

There is a 95% chance that the true mean lies in the interval
-1.96s+u to u+1.96s

yes, I understand this, but I do not understand why should 2*sigma = 0.05*countRate.

 

[gleem]

I don't know where you got that the standard deviation of count rate is count rate but the standard deviation for count rate r is r^1/2 / t^1/2 or (r / t)^1/2 where t is the counting time.

and I do not understand why 2*sigma (standard deviation) = 0.05* countRate.

I don't understand this statement either. Perhaps you misinterpreted something in the video.

95% are between ±2σ meaning 5% is outside this interval or 2.5% above and 2.5% below.

 

[Hornbein]

yes, I understand this, but I do not understand why should 2*sigma = 0.05*countRate.

I don't understand it either. I'd say you should disregard this confused concept.