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STATISTICS AGEN-689 ADVANCES IN FOOD ENGINEERING Statistics of atoms and radiation • The mathematical descriptions used to make quantitative predictions about nature on the atomic scale do so on a statistical basis • How much energy will a 1 MeV proton lose in its next collision with an atomic electron? • Will a 400-keV photon penetrate a 2-mm lead shield without interacting? • How many disintegration will occur during the next minute with a given radioactive source? Radioactive disintegration – Exponential decay • To determine the activity of a long-lived radionuclide in a sample, the sample can be counted for a specific length of time • Knowledge of the counter efficiency (# of counts/# of disintegrations) then yields the sample activity Radioactive disintegration – Exponential decay • If the counting experiment is repeated many times, the numbers of counts observed in a fixed length of time will be found to be distributed about their mean value ( the best estimate of the true activity) • The spread of the distribution about the mean is a measure of the uncertainty of the determination Radioactive disintegration – Exponential decay • We will calculate the probability for a given number of disintegration to occur during time t when N atoms of a radionuclide with decay constant λ are initially present • The half-life of the nuclide need not be long compared with the observation time Radioactive disintegration – Exponential decay • We consider the probability of decay for each atom, assuming that: – all atoms are identical and independent and – that the decay process is spontaneous and random. Radioactive disintegration – Exponential decay • The relative number of atoms that have not decayed in a sample in time t is: e − λt • Exponential decay can be interpreted as the probability that an atom survives at time t without disintegrating: q = probability of survival = e − λt • For decay during time t: p = probability of decay = 1 − q = 1 − e − λt Note that there are only two alternatives for a given atom in time t, since p+q=1 Radioactive disintegration – A Bernoulli process • Consider a set of N identical atoms • Characterized by a decay constant λ at time t =0 • What is the probability that n atoms will decay between t = 0 and a specific later time t ? • 0<n<N • During t, each time is ‘trying’ to decay (success) or not decay (failure) Radioactive disintegration – A Bernoulli process • Observations of a set of N atoms from 0 to t can be seen as: – It consists of N trials (N atoms each having a chance to decay) – Each trial has a binary outcome success or failure (decay or not) – The probability of success (decay) is constant from trial to trial (all atoms have an equal chance to decay) – The trial are independent • These 4 conditions characterize a Bernoulli process • The number of success, n, from N trials is a binomial random variable Example • A sample of N=10 atoms of K-42 (half-life=12.4 h) is prepared and observed for time t = 3 h a) What is the probability that atoms 1, 3, and 8 will decay during this time? b) What is the probability that atoms 1, 3, and 8 will decay while none of the others decay? c) What is the probability that exactly three atoms (any three) decay during the 3 hours? d) What is the probability that exactly 6 atoms will decay in 3 hours? e) What is the chance that no atoms will decay in 3 hours? f) What is the general formula for the probability that exactly n atoms will decay, where 0 < n < 10? g) What is the sum of possible probabilities from (f)? h) If the original sample consisted of N = 100 atoms, what would be the chance that no atoms decay in 3 hours? Solution –(a) • The decay constant for K-42 is λ = 0.693/12.4 h = 0.0559 1/h • The probability that a given atom survives at time t = 3 hours without decay is: q = e − λt = e −0.0559×3 = e −0.168 = 0.846 • The probability that a given atom will decay is: p = 1 − q = 0.154 • The probability that atoms 1, 3 and 8 decay in this time is: p 3 = (0.154)3 = 0.00365 Solution –(b) • The answer to (a) is independent of the fate of the other seven atoms • The probability that none of the others decay in the 3 hours is: q7 = 0.8467 = 0.310 • The probability that only atoms 1, 3, and 8 decay while others survive is therefore: p 3q7 = (0.00365)(0.310) = 0.00113 Solution –(c) • • • • • • • The answer to (b) gives the probability that a particular, designated 3 atoms decay – and only those 3 – in the specific time. The probability that exactly 3 atoms (any 3) decay is p3q7 times the number of ways that a group of 3 can be chosen from among the N = 10 atoms To make such group, there are 10 choices for the first number, 9 choices for the second member, and 8 choices for the third member Thus there are N(N-1)(N-2) = 10 x 9 x 8 = 720 ways of selecting the 3 atoms for decay However, these groups are not all distinct In a counting experiment, the decay of atoms 1, 3, and 8 in that order is not distinguished from decay in the order 1, 8, and 3 The number of ways of ordering the 3 atoms that decay is 3 x 2 x 1: 3!=n!=6 Solution –(c) • Therefore, the number of ways that any n = 3 atoms can be chosen from among N = 10 is given by the binominal coefficient N 10 10! 10 × 9 × 8 720 = = = 120 = ≡ 3! 6 n 3 3!7! • The probability that exactly 3 atoms decay is: 10 3 7 P3 = p q = 120 × 0.00113 = 0.136 3 Solution –(d) • The probability that exactly 6 atoms decay in the 3 hours is: 10 6 4 10! (0.154)6 (0.846)4 = 0.00143 P6 = p q = 6!4! 6 Solution –(e) • The probability that no atoms decay is: 10 0 10 10! (0.846)10 = 0.188 P0 = p q = 0!10! 0 • Thus there is a probability of 0.188 that no disintegration occur in the 3-hour period which is approximately one-fourth of the halflife Solution –(f) • We can see from (c) and (d) that the general expression for the probability that exactly n of the 10 atoms decay is: 10 n 10−n P0 = p q n Solution –(g) • The sum of probabilities for all possible numbers of disintegrations, n=0 to n=10, should be unit, so: 10 n 10−n 10 P p q = ( p + q ) = ∑ ∑ n n =0 n =0 n 10 10 • Since p + q = 1, the total probability is unity Solution (h) • With N = 100 atoms in the sample, the probability that none would decay in 3 h is: q100 = (0.846)100 = 5.46 × 10−8 • This is a much smaller probability than in (e), where there are only 10 atoms in the initial sample • Seeing no atoms decay in a sample of size 100 is a rare event The binomial distribution (or Bernoulli) • The probability that exactly n will disintegrate in time t is: N n N −n Pn = p q n • Since Pn are just the terms in the binomial expansion and since p + q =1, the probability distribution is normalized: N N P = ( p + q ) =1 ∑n n =0 The binomial distribution (or Bernoulli) • The expected, or mean, number of disintegration in time t is given by the mean value µ of the binomial distribution: N n N −n µ = ∑ nPn = ∑ n p q n =0 n =0 n N • And N µ = Np • Thus, the mean is just the product of the total number of trials and the probability of the success of a single trial The binomial distribution (or Bernoulli) • The scatter, or spread, of the distribution of n is characterized quantitatively by its variance σ2 or standard deviation σ: N σ = ∑ (n − µ )2 Pn 2 n =0 • The standard deviation is given by σ = Npq Poisson distribution – discrete • Consider conditions N>>1, N >> n and p << 1 • Small p implies that q is near unity and therefore λt <<1 • So the Poisson distribution is: Pn = µ ne− µ n! • The mean and standard deviation are: µ = Np ; σ = µ Normal Distribution - continuous • The normal distribution can be obtained from the Poisson under specific conditions • It can be written as the probability density for the continuous random variable x f ( x) = 1 −( x − µ ) 2 / 2σ 2 e 2π σ Direction of increasing variance -3 -2 -1 0 Normal Distribution 1 2 3 Normal Distribution - continuous • • • • The probability that x lies between x and x+dx is f(x)dx The function is normalized to unit area when integrated over all values of x, from - ∞ < x < ∞ In place of the single Poisson parameter m, the normal distribution is characterized by 2 independent parameters, is mean µ and standard deviation σ It has inflection points at x = µ + σ where the changes from concave downward to concave upward in going away from mean Direction of increasing variance -3 -2 -1 0 Normal Distribution 1 2 3 Normal Distribution • The probability that x has a value between x1 and x2 is equal to the area under the curve f(x): 1 P( x1 ≤ x ≤ x2 ) = 2π σ x2 ∫e x1 −( x − µ ) 2 / 2σ 2 dx Normal Distribution – the Standard • Mean = 0 and standard deviation = 1 1 P( z1 ≤ z ≤ z2 ) = 2π • where z= x−µ σ z2 ∫e z1 −z2 / 2 dx