# A counting question

dear reader,
here is a quick counting question:
A counter near a long-lived radioactive source measures an average of 100 counts per minute. The probabilty that more than 110 counts will be recorded in a given one-minute interval is most nearly
(A) zero
(B) .001
(C) .025
(D) .15
(E) .5
I kinda guess that it is D, .15, but I am not able to explain it accurately, other than it is within one standard deviation. ## Answers and Replies

Your questions says nothing about the distribution of the counts/minute so technically, the answer could be anything.

ehild
Homework Helper
quantumworld said:
dear reader,
here is a quick counting question:
A counter near a long-lived radioactive source measures an average of 100 counts per minute. The probabilty that more than 110 counts will be recorded in a given one-minute interval is most nearly
(A) zero
(B) .001
(C) .025
(D) .15
(E) .5
I kinda guess that it is D, .15, but I am not able to explain it accurately, other than it is within one standard deviation. This should be Poisson distribution and Poisson distribution can be approximated with Gaussian one of the same mean and standard deviation if the number of counts is high. P(n>110) = 1-F(110), where F is the probability distribution function. To calculate with the normalized Gaussian distribution, you transform the variable n (number of counts) to u=(110-100)/10=1,

$$F(110)=\Phi(1)$$,

From a table for normalized Gaussian distribution $$\Phi (1) = 0.8413$$, so the probability of getting a count number greater than 110 is 1-0.8413=0.1587. So your answer seems to be all right.

ehild