Probability of counts from the counting rate of a radioactive sample?

[carnivalcougar]

Homework Statement

The average counting rate of a radioactive sample is 486 cpm (counts per minute). Find the probability that in any given 10s interval one gets less than 72 counts. Is this the same as the probability of getting less than 72 X 6 = 432 counts in 60s?

Homework Equations

The Attempt at a Solution

I'm not really sure where to start. I do not think that the probability of less than 72 counts in 10s is the same as getting less than 72 x 6 counts in 60s though.

 

[Curious3141]

Radioactive decay can be modeled as a Poisson process. You're given the average event rate (486cpm = 81 counts per 10 sec interval). The variance equals the mean (here that's the average event count for the respective time interval).

Remember that for a sufficiently large mean, the Poisson can be well-approximated by the normal distribution, with a continuity correction applied. That should simplify the calculations.

Does that help?

 

[carnivalcougar]

We were not told what a Poisson process is, just that the standard deviation will equal to \sqrt{navg} where navg is the average number of counts, n. If there are 468 counts per minute then the average count rate /10s will be 81 and the standard deviation of that will be 9. So you would get less than 72 counts 15.8% of the time if you use a normal distribution (15.8% of values will lie below one standard deviation).

As for the probability of getting less than 72 X 6 = 432 counts in 60S, this is not the same because one standard deviation from 486 is 464.

Is this correct?

 

[Curious3141]

We were not told what a Poisson process is, just that the standard deviation will equal to \sqrt{navg} where navg is the average number of counts, n. If there are 468 counts per minute then the average count rate /10s will be 81 and the standard deviation of that will be 9. So you would get less than 72 counts 15.8% of the time if you use a normal distribution (15.8% of values will lie below one standard deviation).

This is OK *if* you're not supposed to know about a Poisson process and if you're expected to apply the normal distribution without a continuity correction.

However, applying a continuity correction means you have to find $P(X \leq 71.5)$ in a normal distribution with mean 81 and s.d. 9, which is 0.145 or 14.5%.

Doing it by hand for a Poisson distribution is almost impossible, but a Poisson calculator will give a very similar value.

As for the probability of getting less than 72 X 6 = 432 counts in 60S, this is not the same because one standard deviation from 486 is 464.

Correct.