1. 3/2003 Rev 1 I.2.13 – slide 1 of 48 Part IReview of Fundamentals Module 2Basic Physics and Mathematics Used in Radiation Protection Session 13Statistics - Distributions Session I.2.13 IAEA Post Graduate Educational Course Radiation Protection and Safety of Radiation Sources

2. 3/2003 Rev 1 I.2.13 – slide 2 of 48  We will discuss several distributions including  Binomial  Poisson  Gaussian  Log Normal Overview

3. 3/2003 Rev 1 I.2.13 – slide 3 of 48  The Central Limit Theorem assumes that random samples of n number of measurements are taken from a population with a mean, , and a standard deviation of .  If n is large enough, the sample means will have a distribution that is approximately normal with a mean = , and a standard deviation of  nnnn Statistics Central Limit Theorem

4. 3/2003 Rev 1 I.2.13 – slide 4 of 48  For a large number of samples, a population will have a mean of  A standardized variable is a transformation brought about by subtracting a variable’s mean from the variable, and dividing the result by the variable’s standard deviation Statistics Central Limit Theorem (Y -  )  nnnn Zn =Zn =Zn =Zn = yiyiyiyin Y =

5. 3/2003 Rev 1 I.2.13 – slide 5 of 48  By conducting the transformation, the standardized variable is approximately normally distributed with a mean = 0, and a standard deviation = 1.  The advantage of conducting the transformation is that it enables us to compare populations. Statistics Central Limit Theorem

6. 3/2003 Rev 1 I.2.13 – slide 6 of 48 Confidence Levels  If, for a statistic, S, the mean of a population is  and the standard deviation is , we can be confident that S will be in the interval between (  s +  s ) and (  s -  s ), 68% of the time.  We are 95% confident that S will be in the interval (  s + 2  s ) and (  s - 2  s ). The percentage confidence is called the confidence level.

7. 3/2003 Rev 1 I.2.13 – slide 7 of 48 Confidence Levels  The confidence level can be expressed for S as well as for the mean  as  68% of the time (S +  s ) and (S -  s )  we are 95% confident that S will be in the interval  s + 2  s and  s - 2  s. The percentage confidence is called the confidence level.

8. 3/2003 Rev 1 I.2.13 – slide 8 of 48   - 1   + 1   + 2   - 2    2  2  2  2   1  1  1  1 Confidence Levels

9. 3/2003 Rev 1 I.2.13 – slide 9 of 48 Confidence Level Number of Standard Deviations 50%0.68 68%1.0 90%1.645 95%1.96 96%2.0 99%2.575 Standard Deviations Corresponding to Various Confidence Levels

10. 3/2003 Rev 1 I.2.13 – slide 10 of 48 Confidence Levels  A “2-tailed” confidence level of 95% (1.96  ) means that the 5% uncertainty is distributed on both sides of the mean value, or 2.5% on both sides.

11. 3/2003 Rev 1 I.2.13 – slide 11 of 48  -1.96  +1.96  2 Tailed Distribution at 95% Confidence Limit

12. 3/2003 Rev 1 I.2.13 – slide 12 of 48 Binomial Distribution  The binomial distribution is used to predict the probability of a certain outcome based on the conditions that  there are only two possible outcomes  the probability for each outcome is constant  there are “n” independent trials where “n” is a finite sample.

13. 3/2003 Rev 1 I.2.13 – slide 13 of 48 Binomial Distribution

14. 3/2003 Rev 1 I.2.13 – slide 14 of 48 Binomial Distribution  If P B (x,n,p) is the probability of observing “x” successes in “n” trials  P B (x,n,p) = (p x )(q n-x )  Where “p” is the probability of an event occurring for one trial, and  Where “q” is the probability of an event not occurring for one trial n!x!(n-x)!

15. 3/2003 Rev 1 I.2.13 – slide 15 of 48 Binomial Distribution  The binomial distribution is used to calculate the probability of obtaining “x” decay events out of “n” radioactive atoms  All other distributions to be presented are approximations of the binomial distribution.

16. 3/2003 Rev 1 I.2.13 – slide 16 of 48 Binomial Distribution  The mean,, of a binomial distribution is the product of n, the number of observations and p, the probability: = np = np  The standard deviation, s, of a binomial distribution is: S =  npq =  [ (1 – p)] _x _x _x

17. 3/2003 Rev 1 I.2.13 – slide 17 of 48 Binomial Distribution  The binomial distribution may be applied to counting a single radionuclide  The number of radioactive atoms may be thought of as the population to be analyzed  The probability of observing a count from a counting system is the binomial probability.

18. 3/2003 Rev 1 I.2.13 – slide 18 of 48 Binomial Distribution If C N is the actual net count and is the expected net count, then = np =  C N, and = np =  C N, and S 2 = np(1-p), or S 2 = npq  C N q _x

19. 3/2003 Rev 1 I.2.13 – slide 19 of 48 In cases where the probability of an event is extremely low, the binomial distribution approaches a Poisson distribution, p(n): P(n) = Poisson Statistics (N n x e -n ) n!

20. 3/2003 Rev 1 I.2.13 – slide 20 of 48  The Poisson statistic is applicable to describe the decay of a radioactive atom.  The advantage of a Poisson distribution is that it can be represented by a single parameter, the mean, Poisson Statistics _x

21. 3/2003 Rev 1 I.2.13 – slide 21 of 48 Poisson Statistics

22. 3/2003 Rev 1 I.2.13 – slide 22 of 48 If you have 37 Bq of activity, the mean number of disintegrations per second is 37. The probability of observing exactly 37 disintegrations in 1 second is: P(37) = = 0.066 Sample Poisson Statistics (37 37 x e -37 ) 37!

23. 3/2003 Rev 1 I.2.13 – slide 23 of 48 In counting statistics for radioactivity, the standard error of the mean for a count rate is given as:  (poisson) =  n Poisson Statistics

24. 3/2003 Rev 1 I.2.13 – slide 24 of 48 So the count rate and error is then R   R =  R   R =   R = = =  R = = = R   R = R  R   R = R  Poisson Statistics nt nnnnt nnnnt nt 1 t x  Rt  Rt 

25. 3/2003 Rev 1 I.2.13 – slide 25 of 48 If we observe 10,000 counts in 10 minutes, the standard deviation is then  (poisson) =  n =  10,000 = 100 counts in 10 min =  10,000 = 100 counts in 10 min Poisson Statistics

26. 3/2003 Rev 1 I.2.13 – slide 26 of 48 Accounting for background, the background count rate:  net =  (  2 g +  2 bkg )  net = Poisson Statistics rgrgrgrg tgtgtgtg t bkg r bkg + 

27. 3/2003 Rev 1 I.2.13 – slide 27 of 48 You have counted a sample and measured 100 counts in 1 minute. The standard deviation for this count is  100 = 10 cpm. A 1 minute background count reveals 10 counts, so the standard deviation of the background is  10 = 3.2 cpm. What is the true activity and uncertainty of the sample? Sample Problem #1

28. 3/2003 Rev 1 I.2.13 – slide 28 of 48 Net Rate = (sample + background) – background = (100 –10) = 90 counts/minute (cpm) Uncertainty of Net Rate =  (10 2 + 3.2 2 ) =  (100 + 10) = 10.5 cpm At the 68.3% confidence level (1  ), the activity is then 90  10.5 cpm Solution to Sample Problem #1

29. 3/2003 Rev 1 I.2.13 – slide 29 of 48 A 5 minute sample count resulted in 510 counts while a 60 minute background count resulted in 2,400 counts. What is the net rate of the sample count and standard deviation? Sample Problem #2

30. 3/2003 Rev 1 I.2.13 – slide 30 of 48 R net = - R net = 102 cpm – 40 cpm = 62 cpm  net = = = 4.6 62  4.6 cpm Solution to Sample Problem #2 510 counts 5 min 2,400 counts 60 min rgrgrgrg tgtgtgtg t bkg r bkg + 10256040+ 

31. 3/2003 Rev 1 I.2.13 – slide 31 of 48 Count Time The relationship between sample count time and rate (T s+b, R s+b ) and background count time and rate (T b, R b ) is: T total = T s+b + T b TbTbTbTb T s+b RbRbRbRb R s+b ½ =

32. 3/2003 Rev 1 I.2.13 – slide 32 of 48 Sample Problem #3 If an environmental sample produces a count rate of 30 cpm and is counted for one hour, what is the optimum counting interval to determine the background rate if the background counting rate is 10 cpm?

33. 3/2003 Rev 1 I.2.13 – slide 33 of 48 T b = 34.6 minutes If T total = 60 minutes T s+b = 60 – 34.6 = 25.4 minutes Solution to Sample Problem #3 TbTbTbTb T s+b RbRbRbRb R s+b ½ = TbTbTbTb 60 min 10 cpm 30 cpm ½ =

34. 3/2003 Rev 1 I.2.13 – slide 34 of 48 The Lower Limit of Detection (LLD) is the lowest amount of activity that will result in a net count The LLD at the 95% confidence level is Counting Statistics Lower Limit of Detection 2.71 T s+b + 3.29 TbTbTbTb RbRbRbRb T s +b TbTbTbTb½ 1 +

35. 3/2003 Rev 1 I.2.13 – slide 35 of 48 If the background count time and sample count time are equal, (T b = T s+b = t) then Counting Statistics Lower Limit of Detection 2.71 T s+b + (1.645)(2  2)  bkg LLD = where  bkg is the standard deviation of the background, or 2.71t + 4.65  bkg LLD =

36. 3/2003 Rev 1 I.2.13 – slide 36 of 48 What is the LLD of a counting system with a background counting rate of 120 cpm? Assume the background counting interval is 60 minutes and the sample counting time is 10 minutes. Sample Problem #4

37. 3/2003 Rev 1 I.2.13 – slide 37 of 48 Solution to Sample Problem #4 2.71 T s+b + 3.29 TbTbTbTb RbRbRbRb T s +b TbTbTbTb½ 1 + 2.71 10 + 3.29 60 120 10 60½ 1 + 0.271 + 3.29 (2 (1 + 6)) = 0.271 + 3.29 (3.74) = 12.6 cpm ½

38. 3/2003 Rev 1 I.2.13 – slide 38 of 48 The Minimum Detectable Activity (MDA) is used to determine if the sample count rate is statistically different from the background count rate. The MDA at the 95% confidence level is Counting Statistics Minimum Detectable Activity If T b = T s+b, 1.645 TbTbTbTb RbRbRbRb T s +b TbTbTbTb½ 1 + MDA = T 2R b ½ MDA = 1.645

39. 3/2003 Rev 1 I.2.13 – slide 39 of 48 What is the MDA at the 95% confidence level for a system that results in 10 counts per minute in a 10 minute counting interval. Assume the sample counting interval is also 10 minutes. Sample Problem #5

40. 3/2003 Rev 1 I.2.13 – slide 40 of 48 T 2R b ½ MDA = 1.645 10 2x10½ = 1.645 = 2.3 cpm Solution to Sample Problem #5

41. 3/2003 Rev 1 I.2.13 – slide 41 of 48 The Minimum Detectable Concentration (MDC), has been defined as Counting Statistics Minimum Detectable Activity 3 + 4.65  C B (  I )(  s )(T) probe area 100 cm 2 MDC = C B is the background count in time T for paired observations of the sample and blank,  I is the instrument efficiency,  s is the surface efficiency (typical values of  s are 0.25)

42. 3/2003 Rev 1 I.2.13 – slide 42 of 48  If the mean is less than 20, the Poisson distribution is the appropriate statistical model for error analysis and the probability distribution  For a mean of greater than 20, and if the probability of observation is small, both the Poisson and binomial distribution can be approximated by the Gaussian or “normal” distribution. Gaussian Statistics

43. 3/2003 Rev 1 I.2.13 – slide 43 of 48  Two important characteristics of the Gaussian distribution are:  It is symmetric about the mean  Probabilities of adjacent values of x observed do not differ greatly which results in a large value for the statistical mean of the distribution Gaussian Statistics e [s  (2  )] 2 (x- ) 2 (x- ) 2- P G (x,, s) = _x _x _x

44. 3/2003 Rev 1 I.2.13 – slide 44 of 48 Gaussian Distribution

45. 3/2003 Rev 1 I.2.13 – slide 45 of 48 Log-Normal Distribution A log-normal distribution is one that is normally distributed after taking the logarithm of the data. That is, if a variable, x, is lognormally distributed, then Y = ln(x) is normally distributed.

46. 3/2003 Rev 1 I.2.13 – slide 46 of 48 Log-Normal Distribution

47. 3/2003 Rev 1 I.2.13 – slide 47 of 48  Dan Lurie and Roger H. Moore, “Applying Statistics,” NUREG-1475, U.S. Nuclear Regulatory Commission, (1994).  “Statistics Manual,” U.S. Naval Ordnance Test Station, NAVORD Report 3369, Dover Publication, Inc., (1960). Statistics – References

48. 3/2003 Rev 1 I.2.13 – slide 48 of 48  Cember, H., Johnson, T. E., Introduction to Health Physics, 4 th Edition, McGraw-Hill, New York (2008).  Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6 th Edition, Hodder Arnold, London (2012).  C. H. Wang, D. L. Willis, and W. D. Loveland, “Radiotracer Methodology in the Biological, Environmental, and Physical Sciences,” Prentice-Hall (1975). Where to Get More Information