Download STATISTICS

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