Transcript Ludlum Measurements, Inc.

Ludlum Measurements, Inc.
User Group Meeting
June 22-23, 2009
San Antonio, TX
Counting Statistics
James K. Hesch
Santa Fe, NM
Binary Processes


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




Success vs. Failure
Go or No Go
Hot or Not
Yes or No
Win vs. Lose
1 or 0
Disintegrate or not
Count a nuclear event or not
Uncertainty
 Shades of gray – neither black nor white
 How gray is gray?
 More black than white, or more white than
black?
Some Familiar Real World Applications
What is the probability of drawing a
Royal Flush in five cards drawn
randomly from a deck of 52 cards?
The first card must be a member of
the set [10, J, Q, K, A] in any of the
four suites. Thus it can be any one
of 20 cards.
20
p
 0.3846
52
The set of valid cards diminishes to
four for the second card out of the
remaining 51 cards, etc.
20 4 3 2 1
p
  

52 51 50 49 48
Probability 1 : 649740
20(4!)( 47!)
p
 0.000001359
52!
Plato’s Real vs. Ideal Worlds

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


Observed vs. Expected
Predicting with uncertainty
Science is inexact
Stating the precision
“+/- 2% at the 95% confidence level”
Toss of One Die
Single Die Results Distribution
18%
16%
Frequency
14%
12%
10%
8%
6%
4%
2%
0%
1
2
3
4
Value
5
6
Toss of Two Dice
Two Dice Results Distribution
18.00%
16.00%
14.00%
Frequency
12.00%
10.00%
8.00%
6.00%
4.00%
2.00%
0.00%
2
3
4
5
6
7
Value (Sum)
8
9
10
11
12
Four Tosses of a Pair of Dice







3
10
5
2
Total = 20
Average (Mean) = 20/4 = 5
Compute the average value by which each
toss in this sample VARIES from the mean.
Variance = σ²
(x  X )

2
 
(
x

X
)
i
n
1
2
i 1
 
n 1
n
2
Toss of Three Dice
Three Dice Results Distribution
30
25
Frequency
20
15
10
5
0
3
4
5
6
7
8
9
10
11
Value (Sum)
12
13
14
15
16
17
18
Toss of Four Dice
Four Dice Results Distribution
12%
10%
Frequency
8%
6%
4%
2%
0%
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Value (Sum)
Probability Distribution Functions
 Binomial
 Poisson
 Gaussian or Normal (the famous bell curve)
Binomial Distribution Function
N
N!
k
N k
Pp   
p (1  p)
 k  k!( N  k )!
Poisson Distribution Function
p ( x) 
x 
e
x!
Sample Exercise
In a counting exercise where the average
number of counts expected from background
is 3, what should the minimum alarm set point
be to produce a false alarm probability of
0.001 or less?
Lambda = 3
Discrete
Cumulative
x
p(x)
∑p(x)
0
0.04979
0.04979
1
0.14936
0.19915
2
0.22404
0.42319
3
0.22404
0.64723
4
0.16803
0.81526
5
0.10082
0.91608
6
0.05041
0.96649
7
0.02160
0.98810
8
0.00810
0.99620
9
0.00270
0.99890
10
0.00081
0.99971
11
0.00022
0.99993
12
0.00006
0.99998
Poisson Distribution, Lambda = 3
Poisson Distribution, Lambda = 3
25%
Probability
20%
15%
10%
5%
0%
0
1
2
3
4
5
6
7
Discrete Value, (x)
8
9
10
11
12
Poisson Distribution, Lambda = 1.25
Poisson Distribution, Lambda = 1.25
40%
35%
30%
25%
20%
15%
10%
5%
0%
0
1
2
3
4
5
6
7
Discrete values of x
8
9
10
11
12
Gaussian Distribution Function
P( x) 
1

2
x
2
e


2
dx
Gaussian Distribution Function
 Is a Density Function, or cumulative
probability (as opposed to discreet).
 Can use look-up table or Excel functions to
apply
 Scale to data by use of Mean and Standard
Deviation
 Single-sided confidence – but can be used
to determine two-sided confidence function
“Erf(x)”.
Gaussian Distribution Function
Excel Function
F(2) = NORMDIST(2, 0, 1, TRUE) = 0.97725
2 StdDev
Mean = 0
StdDev of Data = 1
Cumulative = True
If NORMDIST() set to FALSE…
Controlling False Alarm Probability
 Determine expected number of background
counts that would occur in a single count
cycle.
 Determine the StdDev of that value
 Set the alarm setpoint a sufficient number of
Standard Deviations above average
background counts for an acceptable false
alarm probability.
False Alarm Probability

PFA  1 F ( K B )
N

How Many Sigmas?
1 N
K B  F ( 1  PFA )
In Excel…
KB = NORMINV((1-PFA)^(1/N),0,1)
False Alarm Probability
Mean
StdDev
Computing Alarm Setpoint
N A  NB  KB NB  NB
Simplify and Divide by Time
NA
 KB
T
NB
T
…almost!
RA( MIN )  K B
RB
T
Final Form:
RA(min)  K B
RB RB

T TB
Slight detour … 2-sided distribution
±σ
±1 StdDev = 68%
±2 StdDev = 95%
±3 StdDev = 99.7%
In Excel…
 Two sided distribution…
 …=2*(NORMDIST(x, 0, 1, TRUE) – 0.5)
Getting Back to Alarm Setpoint…
R A( MAX )  MDA Eff  K S  B
MDA Eff  RB RB

T
TB
MDA-Driven Alarm Setpoint
Maximum Alarm Set Point
“Minimum” Count Time
 Solve for T using the simplified equation below, and round
up to a full no. of seconds:
 Compute a new value for MDA (see next slide) using the
resulting “T” as
 As needed, iteratively, add 1 second to the T and
recompute MDA until the result is < the desired MDA
 K B RB  K S  B MDA Eff  RB 
T 

MDA Eff


²
Computing MDA
 Start with MDA=1 for the right side of the following
equation, and compute a new value for MDA
 Substitute the new value on the right hand side and repeat.
 Continue with the substitution/computation until the value
for MDA is sufficiently close to the previous value.
R
R
MDA
MDAEff
EffRRBB  RRBB
K BK B BRB  BRB KK

S SB B
T T TBTB
TT
TTBB
MDA

MDA
Eff
Eff
Activity Other than MDA
Alarm Probability vs. Activity Level
1
0.9
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Percent MDA
95
%
10
0%
90
%
85
%
80
%
75
%
70
%
65
%
60
%
55
%
50
%
45
%
40
%
35
%
30
%
25
%
20
%
15
%
10
%
5%
0
0%
Alarm Probability
0.8
Approximation of Nuisance Alarms
Nuisance Alarms
100%
90%
70%
60%
50%
40%
30%
20%
10%
Percent of MDA
10
0%
95
%
90
%
85
%
80
%
75
%
70
%
65
%
60
%
55
%
50
%
45
%
40
%
35
%
30
%
25
%
20
%
15
%
10
%
5%
0%
0%
Alarm Probability
80%
With Extended Count Time
Nuisance Alarms
100%
90%
70%
60%
50%
40%
30%
20%
10%
Percent of MDA
10
0%
95
%
90
%
85
%
80
%
75
%
70
%
65
%
60
%
55
%
50
%
45
%
40
%
35
%
30
%
25
%
20
%
15
%
10
%
5%
0%
0%
Alarm Probability
80%
A Look at Q-PASS
1000 cps Background
1600
1400
1200
1000
800
600
400
200
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
Elapsed Time
Clean
High Alarm