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A counting question

  • #1
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. :confused:
 

Answers and Replies

  • #2
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Your questions says nothing about the distribution of the counts/minute so technically, the answer could be anything.
 
  • #3
ehild
Homework Helper
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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. :confused:


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,

[tex]F(110)=\Phi(1)[/tex],

From a table for normalized Gaussian distribution [tex]\Phi (1) = 0.8413 [/tex], 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
 

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