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Chap. 3 Statistics & Measurement
Fact: Randomness of decay leads to statistical fluctuations.
• Given statistical fluctuations, how accurate is the measurement?
• Statistical analysis can tell if equipment is working properly.
• Random events in time should follow the “interval” distribution.
Goals:
(I) 
Characterization of fluctuations in data: The problem is: a set of
measurements of one quantity gives a range of answers. How should
we describe the set ?
Experimental Mean:
1 N
xe = ∑ xi
N 1
N
Residual:
di = x i − x e
note
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i
=0
1
1 N
σ = ε = ∑ xi − x
N 1
Deviation from (the) meanεi = x i − x
And its variance:
Sample Variance:
∑d
2
1 N
s =
xi − xe
∑
N −1 1
2
(
(
2
)
2
)
2
( )
practical expresssion s2 ≈ x i2 − x e
2
Statistics & Measurement – Models –1–
(II)  Fluctuation Models:
A) The binomial distribution, three parameters in general (a,b,n)
(a+b)n = an / 0! + (nan-1 b1)/1! + [n(n-1) an-2 b2 ]/ 2! + … [n! an-x bx ]/ (n-x)!x! + bn / 0!
Special case of a fractional probability, p, where one variable is removed,
b = p, a = 1-p so that a+b = 1 then the normalized probability function is:
(a+b)n = Σ Pn (x) = (1-p)n + … [n! (1-p)n-x px ]/ (n-x)!x! … + pn
Thus, if “p” is the probability of success and “n” is the number of tries, then x is the
number of successes.
Example 1: Coin flip: p = ½ (1-p) =1/2
Example 2: Toss die: p = 1/6 (1-p) = 5/6
Example 3: Probability of radioactive decay:
p = (N0 – N)/N0 = (1 – e –λt)
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(1-p) = e –λt
Statistics & Measurement – Models –2–
(II)  Fluctuation Models:
A) Binomial distribution, special case: n tries, fractional probability p
Pn(x) =
[n! / (n-x)!x!] (1-p)n-x px
Mean = p*n,
σ2 = p n (1-p),
σ2/mean = (1-p)
Toss die: p = 1/6 , (1-p) = 5/6 …
Toss three dice all at once, chance to get a “5” ?
(the same as three throws of one die to get a “5”)
n=3, x is number of 5’s
Mean = p*n = 1/6 * 3 = 1/2
σ2 = p n (1-p) = 1/6*3*(5/6) = 5/12
x=0
P3 (0) = [3! / (3-0)! 0!] (1/6)0 (5/6)3
0.567
1
P3 (1) = [3! / (3-1)! 1!] (1/6)1 (5/6)2
0.347
2
P3 (2) = [3! / (3-2)! 2!] (1/6)2 (5/6)1
0.0694
3
P3 (3) = [3! / (3-3)! 3!] (1/6)3 (5/6)0
0.00463
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Σ Pn (x) = 1
Statistics & Measurement – Models –3–
(II)  Fluctuation Models:
B) The Poisson distribution (p→0) and large values of n
Pn (x) = [n! (1-p)n-x px ]/ (n-x)!x! → Pn (x) = [nx (1-p)n-x px ]/x! Since n-x ~ n
Pn (x) = [nx px (1-p)n-x]/x!
()
n x p x = (np) x = x
x)
(
P (x) =
n
€
x
&
p = x /n
x
 x
1− 
x!  n 
when n >> x
∑ PP (x) = 1
n−x
x)
(
P (x) =
P
x = pn
x!
x
e−x
Only one parameter, the mean
σ 2 = x (1− p) = x
Asymmetric distribution with a tail on high side
€
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Statistics & Measurement – Models –3a–
(II)  Fluctuation Models:
B) Example of Poisson Distribution: the birthday problem, how many
people in this room have the same birthday?
x)
(
P (x) =
P
x = pn
x!
x
e−x
1
p=
365.25
x)
(
P (x = 0) =
P
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e−x
0!
x)
(
P (x = 1) =
P
0
1!
1
e−x
Statistics & Measurement – Models –4–
(II)  Fluctuation Models:
C) The Gaussian approximation, p→0, n>>1 & mean > 20
PG (x) =
1
2π x
e
−(x−x )2 / 2x
1
−( x−x )2 / 2σ 2
or G(x) =
e
σ 2π
Symmetric Continuous distribution (one parameter)
2 −χ 2 / 2
G( χ ) =
e
2π
€
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(
also chi-squared distribution:
)
where χ = x − x /σ
Statistics & Measurement –Analysis –1–
(III) Analysis of data
A)  Are fluctuations in N-values consistent with statistical accuracy?
Measure: xe and s2
Model: mean gives σ2 … thus the test: Is s2 = σ2 ?
Define a statistic χ in terms of measurables that will provide a simple test.
N
1
2
χ = ∑ xe − x
x e i=1
(
N
χ2 =
∑( x
e
−x
i=1
xe
)
)
2
N
=∑
i=1
(
xe − x
xe
)
2
N
=∑
(
xe − x
i=1
2
s2 (N −1)
=
xe
→
χ2
s2
=
(N −1) x e
But in a “good” data set with large N: s2 = σ2 = mean
χ2
∴
= 1 or
(N −1)
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s2
χ 2 = (N −1)
)
2
 x − x
= ∑ e

s 
i=1 
N
2
Statistics & Measurement – Analysis –2–
(III) Analysis of data
B) Only one measurement, estimate uncertainty?
There is only one measurement, strictly speaking you are out of luck.
However, one can posit that it must be the mean, and σ2 = mean
One further assumes that the distribution is symmetric: x +/- σ
Fractional error: σ /x = √x / x = 1 / √x
(IV)  Measurements in the presence of Background (itself measureable)
(1) measure background (B); (2) measure source + background (S+B).
S = (S+B) – B … nothing new here but there are two contributions to error.
How should the time be allocated to give minimum uncertainty in S ie s S ?
Simplest case, constant rates the minimum in s S comes when:
B diminishes in importance for large S
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Statistics & Measurement – time distribution
(V)  Time between random events, the Interval distribution:
The product of the probability that nothing happens for some length of
time, t, times the probability that the next event occurs in the small
time
interval, dt.
*
*
I1 (t)dt = [ PP (x = 0)] [ r dt ] where r is average rate
(
)
I (t)dt =
x
1
0
e−x [ r dt ] where
0!
I1 (t)dt = e−r t r dt
r
x = r*t
I(t)
rt
Similar arguments give the distribution of the j-th event as:
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which has a peak near the value (rt) = j-1
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*
**
*
time