Week 4: Chap. 3 Statistics of Radioactivity

Transcription

1 Week 4: Chap. 3 Statistics of Radioactivity Vacuum Technology General use of Statistical Distributions in Radiation Measurements -- Fluctuations in Number --- distribution function models -- Fluctuations in Time General Detector Properties Without deviation from the norm, progress is not possible. Frank Zappa, 986

2 Chap. 3 Statistics & Measurement Fact: Randomness of decay leads to statistical fluctuations. Goals: Given statistical fluctuations, how precise is the measurement? Statistical analysis can tell if equipment is working properly. Random events in time should follow the interval distribution. (I) Characterization of fluctuations in data: The problem is: A set of measurements of one quantity contains a range of answers. How should we describe the set? [could enumerate a small set: {7,2,5,},.25 +/- 3.3 ] Experimental Mean: x e = N N x i Experimental Residuals: d i = x i x e note d i = N Deviation from (the) mean And (the) variance: ε i = x i x σ 2 = ε 2 = N N ( x i x) 2 Expt. Sample Variance: ( ) 2 s 2 = N x e x i practical expresssion s 2 x 2 i x e N ( ) 2

3 Statistics & Measurement Models (II) Fluctuation Models: A) The binomial distribution, a theory with three general parameters (a,b,n) (a+b) n = a n /! + (na n- b )/! + [n(n-) a n-2 b 2 ]/ 2! + [n! a n-x b x ]/ (n-x)!x! + b n /! Special case of a fractional probability, p, which allows one variable to be removed, b = p, a = -p so that a+b = and automatically normalizes the distribution: (a+b) n = S P n (x) = (-p) n + [n! (-p) n-x p x ]/ (n-x)!x! + p n Thus, if p is the probability of success and n is the number of tries, then x can be seen to be the number of successes. Example : Coin flip: p = ½ (-p) =/2 Example 2: Toss die: p = /6 (-p) = 5/6 Example 3: (fractional) Probability of radioactive decay: p = (# decay) / (# sample) = (N N) / N = ( e lt ) ; (-p) = e lt

4 Statistics & Measurement Models 2 (II) Fluctuation Models: A) Binomial distribution function (theory), special case: n tries with p fractional probability, gives x successes P n (x) = [n! / (n-x)!x!] (-p) n-x p x Mean = p*n, s 2 = p n (-p), s 2 /mean = (-p) Need to know p & n to evaluate each element of this distribution. Question: Toss three dice all at once, what is chance of getting a 5? (the same as three throws of one die to get a 5 an experiment beckons) n=3, x is number of 5 s Toss die: p = /6, (-p) = 5/6 Mean = p*n = /6 * 3 = /2 s 2 = p n (-p) = /6*3*(5/6) = 5/2 x = P 3 () = [3! / (3-)!!] (/6) (5/6) P 3 () = [3! / (3-)!!] (/6) (5/6) P 3 (2) = [3! / (3-2)! 2!] (/6) 2 (5/6) P 3 (3) = [3! / (3-3)! 3!] (/6) 3 (5/6).463 Note: S P n (x) =

5 Statistics & Measurement Models 2a Test random dice-roller app ( ) Mean = p*n = /6 * 3 = /2 s 2 = p n (-p) = /6*3*(5/6) = 5/2 x = P 3 () = [3! / (3-)!!] (/6) (5/6) P 3 () = [3! / (3-)!!] (/6) (5/6) P 3 (2) = [3! / (3-2)! 2!] (/6) 2 (5/6) P 3 (3) = [3! / (3-3)! 3!] (/6) 3 (5/6).463 Note: S P n (x) =

6 Statistics & Measurement Models 2a Test random dice-roller app throws of 3 dice Mean = p*n = /6 * 3 = /2 s 2 = p n (-p) = /6*3*(5/6) = 5/2 x = P 3 () = [3! / (3-)!!] (/6) (5/6) P 3 () = [3! / (3-)!!] (/6) (5/6) P 3 (2) = [3! / (3-2)! 2!] (/6) 2 (5/6) P 3 (3) = [3! / (3-3)! 3!] (/6) 3 (5/6).463 Note: S P n (x) =

7 Statistics & Measurement Models 2a Test random dice-roller app throws of 3 dice Mean = p*n = /6 * 3 = /2 s 2 = p n (-p) = /6*3*(5/6) = 5/2 x = P 3 () = [3! / (3-)!!] (/6) (5/6) P 3 () = [3! / (3-)!!] (/6) (5/6) P 3 (2) = [3! / (3-2)! 2!] (/6) 2 (5/6) P 3 (3) = [3! / (3-3)! 3!] (/6) 3 (5/6).463 Note: S P n (x) =

8 (II) Statistics & Measurement Models 3 Fluctuation Models: B) The Poisson distribution (p ) and large n (many tries before success) P n (x) = [n! (-p) n-x p x ]/ (n-x)!x! P n (x) = [n x (-p) n-x p x ]/x! Since n!/n-x! ~ n x n x p x = (np) x = ( x) x also p = x / n P n (x) = [n x p x (-p) n-x ]/x! rearrange This image cannot currently be displayed. P n ( ) x ( x) = x x! " x % $ ' # n & n x lim n # + & % ( $ n ' n # e lim n + a & % ( $ n ' n e a when n >> x P Poisson x ( ) x ( ) = x x! e x P p (x) = x = pn σ 2 = x( p) = x Only one parameter, the mean Asymmetric (theoretical) distribution with a tail on high side

9 Statistics & Measurement Models 3a (II) Fluctuation Models: C) Example of Poisson Distribution: Easy variant of the birthday problem: what is the probability that at least one person in this room has the same birthday as me? ( P Poisson (x) = x ) x x! e x n >> x p = [ / ] x = pn p = n= 6 <x> = pn = ( P (x = ) = x ) n! e x P (x > ) = P (x) = P (x = ) x= x P(x) E E E E-2

10 Statistics & Measurement Models 3b (II) Fluctuation Models: D) Example of Poisson Distribution: Low counting rate experiments can be nerve racking an experiment to produce 4 Mg ran for 7.6 days and only 3 events were observed. What is the probability that any more counts would be observed if the experiment was extended by 48 hours (i.e., 2 days)? P ( x) = x P = pn ( x) x! p x = e -x 3 7.6d =.395d - p =.395 n = 2 P P P P ( x ( x = > ) = ( x)! ) = - P e P -x ( x = ) =.546 Remember, distribution is normalized <x> = pn =.79 x P(x)

11 Statistics & Measurement Models 4 (II) Fluctuation Models: C) The Gaussian approximation, p, n>> & mean = p*n > 2 P G (x) = 2π x e (x x ) 2 / 2x or G(x) = Symmetric Continuous distribution (one parameter) G(χ) = 2 2π e χ 2 / 2 σ where χ = ( x x) /σ 2 2π e (x x) / 2σ 2 AKA chi-squared distribution: p*n = /6 p*n = /6

12 Statistics & Measurement Analysis (III) Analysis of data A) Are fluctuations in a set of N-values consistent with statistical accuracy? Measure: x e and s 2 Model: mean gives s 2 thus the test: Is s 2 = s 2? Define a statistic c in terms of measurable values that will provide a simple test. N " χ 2 x = e x % N x $ ' 2 e x = = # s & s 2 χ 2 = i= N i= ( x e x) 2 i= ( ) 2 x e = s2 (N ) x e But in a good data set with large N: s 2 = s 2 = mean N i= ( x e x) 2 x e (Bottom of Slide today) χ 2 (N ) = s2 x e χ 2 (N ) = or χ 2 = (N )

13 Statistics & Measurement Analysis 2 (III) Analysis of data B) Only one measurement, estimate uncertainty? There is only one measurement of an integer quantity, strictly speaking, you are out of luck. However, one can posit that it must be the mean, and s 2 = mean One further assumes that the distribution is symmetric: x +/- s Fractional error: s /x = Öx / x = / Öx Caution: this refers to the error in the number of counts and not derived values (see text p.85). (IV) Measurements in the presence of Background (itself measureable) () 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: time æ + S + B S B S + B S B 2 = = + & s = ç S + timeb B B T B è B B diminishes in importance for large S ö ø T (S+B) / T(B) S/B

14 Statistics & Measurement time distribution (V) Time between random events, the Interval distribution: A) 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 interval, dt. I (t)dt = [ P P (x = ) ][ r dt] where r is average rate ( I (t)dt = x )! I (t)dt = e r t r dt Note the most-probable time between events occurs at rt =, but r > The average time between events e x [ r dt] where x = r * t t = t I (t)dt I (t)dt = I(t) tr e r t dt * * r = r t time (r > )

15 Statistics & Measurement time distribution (V) Time between random events, the Interval distribution: B) Some people like to scale-down high counting rates by a factor, N. Similar arguments give the distribution of the N-th event as: I N (t)dt = (rt)n (N )! r e r t dt * * * * time which is a curve that has a peak near the value (rt) = N- [find most-probable time from di N /dt = t mp = (N-)/r ] and an average time between scaled events: t N = t I N (t)dt I N (t)dt = N r (r > )

16 Statistics & Measurement time distribution (V) Time between random events, the Interval distribution C)Examples of Interval Distribution from nuclear physics: ) One of the early results from the Sudbury Neutrino Observatory (SNO) was that 968 charged-current neutrino events were observed in 36.4 days of operation. One concern on any given day in low counting rate experiments is that the detector has failed. What is the probability that no events were observed on any given day during this run? I( t) = e r r = = -rt rdt t ò I( t) = t ò e -rt rdt = - e -rt Time passes until evt occurs 968evts -r* = 6.423/ d - e =.9984 Probability of event in day 36.4d -r* e =.6 Probability of not in day 2) The confirmation in 23 of Element 5 claimed to have measured 3 atoms in a three week run. Similarly, what is the probability that this experiment went day without observing an event? 3evts -r* =.429 / d e = d