Transcript Session I213 Distributions

Session I.2.13
Part I Review of Fundamentals
Module 2 Basic Physics and Mathematics
Used in Radiation Protection
Session 13 Statistics - Distributions
3/2003 Rev 1
IAEA Post Graduate Educational Course
Radiation Protection and Safe Use of Radiation Sources
I.2.13 – slide 1 of 48
Overview
 We will discuss several distributions
including




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Binomial
Poisson
Gaussian
Log Normal
I.2.13 – slide 2 of 48
Statistics
Central Limit Theorem
 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
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I.2.13 – slide 3 of 48
Statistics
Central Limit Theorem
 For a large number of samples, a population
will have a mean of
yi
Y= n
 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
(Y - )
Zn =

n
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I.2.13 – slide 4 of 48
Statistics
Central Limit Theorem
 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.
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I.2.13 – slide 5 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.
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I.2.13 – slide 6 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.
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I.2.13 – slide 7 of 48
Confidence Levels
  2
  1
 - 2
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 - 1

 + 1
 + 2
I.2.13 – slide 8 of 48
Standard Deviations Corresponding
of Various Confidence Levels
Confidence Level
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Number of Standard
Deviations
50%
0.68
68%
1.0
90%
1.645
95%
1.96
96%
2.0
99%
2.575
I.2.13 – slide 9 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.
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2 Tailed Distribution
at 95% Confidence Limit
-1.96 
+1.96 

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I.2.13 – slide 11 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.
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I.2.13 – slide 12 of 48
Binomial Distribution
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Binomial Distribution
 If PB(x,n,p) is the probability of observing
“x” successes in “n” trials
n!
 PB(x,n,p) =
(px)(qn-x)
x!(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
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I.2.13 – slide 14 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.
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Binomial Distribution
_
 The mean, x , of a binomial distribution is
the product of n, the number of
observations and p, the probability:
_
x = np
 The standard deviation, s, of a binomial
distribution is:
_
S = npq =  [ x (1 – p)]
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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.
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I.2.13 – slide 17 of 48
Binomial Distribution
If CN is the actual net count and <CN> is the
expected net count, then
_
x = np = <CN>  CN , and
S2 = np(1-p), or
S2 = npq  CNq
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I.2.13 – slide 18 of 48
Poisson Statistics
In cases where the probability of an event is
extremely low, the binomial distribution
approaches a Poisson distribution, p(n):
(Nn x e-n)
P(n) =
n!
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I.2.13 – slide 19 of 48
Poisson Statistics
 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, x
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I.2.13 – slide 20 of 48
Poisson Statistics
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I.2.13 – slide 21 of 48
Sample Poisson Statistics
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:
(3737 x e-37)
P(37) =
= 0.066
37!
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Poisson Statistics
In counting statistics for radioactivity, the
standard error of the mean for a count rate is
given as:
 (poisson) = n
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I.2.13 – slide 23 of 48
Poisson Statistics
So the count rate and error is then
n

t
n
R  r =
t
r = n =
t

R  r = R 
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
n x 1
t
t
=

r
t
r
t
I.2.13 – slide 24 of 48
Poisson Statistics
If we observe 10,000 counts in 10 minutes, the
standard deviation is then
 (poisson) = n
= 10,000 = 100 counts in 10 min
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Poisson Statistics
Accounting for background the background
count rate:
net = (2g + 2bkg)
net =
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
rg
rbkg
tg + tbkg
I.2.13 – slide 26 of 48
Sample Problem #1
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?
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Solution to Sample Problem #1
Net Rate = (sample + background) – background
= (100 –10) = 90 counts/minute (cpm)
Uncertainty of Net Rate = (102 + 3.22)
= (100 + 10) = 10.5 cpm
At the 68.3% confidence level (1), the activity is
then 90  10.5 cpm
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Sample Problem #2
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?
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Solution to Sample Problem #2
Rnet = 510 counts - 2,400 counts
5 min
60 min
Rnet = 102 cpm – 40 cpm = 62 cpm
net =

rg
rbkg
tg + tbkg
=

102 40
5 + 60
= 4.6
62  4.6 cpm
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Count Time
The relationship between sample count time
and rate (Ts+b, Rs+b) and background count
time and rate (Tb, Rb) is:
Ts+b
Tb
=
Rs+b
Rb
½
Ttotal = Ts+b + Tb
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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?
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Solution to Sample Problem #3
Ts+b
Tb
=
60 min
=
Tb
Rs+b
Rb
½
30 cpm
10 cpm
½
Tb = 34.6 minutes
If Ttotal = 60 minutes
Ts+b = 60 – 34.6 = 25.4 minutes
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Counting Statistics
Lower Limit of Detection
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 as
2.71
+ 3.29
Ts+b
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Rb
Tb
Tb
1+
Ts +b
½
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Counting Statistics
Lower Limit of Detection
If the background count time and sample
count time are equal, (Tb = Ts+b = t) then
LLD =
2.71 + (1.645)(22) 
bkg
Ts+b
where bkg is the standard deviation of the
background, or
LLD =
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2.71
+ 4.65 bkg
t
I.2.13 – slide 35 of 48
Sample Problem #4
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.
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Solution to Sample Problem #4
2.71
+ 3.29
Ts+b
2.71
+ 3.29
10
Rb
Tb
120
60
Tb
1+
Ts +b
½
60
1+
10
½
½
0.271 + 3.29 (2 (1 + 6)) =
0.271 + 3.29 (3.74) = 12.6 cpm
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Counting Statistics
Minimum Detectable Activity
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
½
Rb
Tb
MDA = 1.645
1+
Tb
Ts +b
If Tb = T s+b,
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2Rb
MDA = 1.645
T
½
I.2.13 – slide 38 of 48
Sample Problem #5
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.
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Solution to Sample Problem #5
2Rb
MDA = 1.645
T
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½
2x10
= 1.645
10
½
= 2.3 cpm
I.2.13 – slide 40 of 48
Counting Statistics
Minimum Detectable Activity
The Minimum Detectable Concentration
(MDC), has been defined as
MDC =
3 + 4.65 CB
(I)(s)(T) probe area
100 cm2
CB 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)
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Gaussian Statistics
 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.
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Gaussian Statistics
_
PG(x, x , s) =
_
(x- x_ )2
2x
e
[s(2)]
 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
result in a large value for the statistical
mean of the distribution
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Gaussian Distribution
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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.
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Log-Normal Distribution
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Statistics – References
 Dan Lurie and Roger Moore, “Applying
Statistics,” NUREG-1475, U.S. Nuclear
Regulatory Commission, February 1994.
 “Statistics Manual,” U.S. Naval Ordnance
Test Station, NAVORD Report 3369, Dover
Publication, Inc.; 1960.
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Statistics – References
 C. H. Wang, D. L. Willis, and W. D.
Loveland, “Radiotracer Methodology in the
Biological, Environmental, and Physical
Sciences,” Prentice-Hall, 1975.
 Herman Cember, “Introduction to Health
Physics,” Pergammon Press, 1976.
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