Calculations based on a Poisson distribution (2) In the study of radioactive dec
ID: 3173580 • Letter: C
Question
Calculations based on a Poisson distribution (2) In the study of radioactive decay during a one-minute period 270 counts are recorded. Calculate: the mean count rate; the error in the mean count rate; and the fractional error in the mean count rate. Were the experiment to be repeated with a 15-minute counting interval, what is the expected count; and the probability of obtaining exactly 270 times 15 counts? the results of that exercise to predict the mean and standard deviation expected for this distribution. How do these compare with the results for the year 2000, with a mean of 25.4, and standard deviation 14.3? The lottery has six numbers selected, with the mean readily calculated. Based on the central limit theorem we expect the distribution of the means to follow a Gaussian distribution with a standard deviation which is Squareroot 6 smaller than the original distribution. What do you predict for the mean and standard deviation of the means? How do these compare with the results for the year 2000, with a mean of 25.4, and standard deviation 5.7?Explanation / Answer
Since radioactive decay is described by a Poisson distribution, we use the estimators for this distribution to find
mean = 270 counts per minute
If the number of counts recorded in a time T is N, then the uncertainty in the number of counts is N . That is, if you repeat the experiment many times, you will get a gaussian distribution of N’s, centered on N , the average value of N, , with a standard deviation of N .
Uncertainity in the number of counts here would be sqrt(270) would be 16.431
The fractional uncertainity of the count is equal to dR/R which comes to (dN/T)/(N/T)
That makes dN/N = dR/R = SQRT(N)/N = 1/SQRT(N) where R is the average number of counts N in time T OR r = n/t
Hence fractional error here is 1/sqrt(270) = 0.0608
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