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population
complete set of objects relevent to study
sample
subset of population
-use to make infrences about the whole
Methods to collect data
retrospective
observational
designed experiments
Retrospective study
historical data
avantages:
existing data
low cost
doesnt interupt production
drawbacks:
limited key factor varation
missing data
possible recording errors
unmeasured variables
hard to isolates effects/variables
observational study
monitoring process/pop. with minimal interference, recording varibales of interest over period of time
advantages:
minimal process distrubance
potential to measure additonal variables
improved data accuracy and reliability
drawbacks:
limited to ability to manipulate factors
difficulty isolating individual effects
designed experiments
changing controllable variables, observing results, and using statistical principles to identify which factors cause observed effects
advantages:
cause-effect relationships
enables process improvement and validation of engineering theroies
drawbacks:
time consuming
Models
describe how variables in a process are related
two types:
mechanistic
empirical
Mechanistic models
based on known physical laws or mechanisms describing a process
ohm’s law (voltage=current x resistance)
well understood underlying mechanism
limitation: real data often deviate due to uncontrolled factors
adding variability:
real processes rarely behave as theory predicts
error term ( ε ) : accounts for environmental fluctuations, operator differences, and measurement noise
include error term to account for variability
empirical models
developed from observed data when no simple theory exists (data relationships)
mechanisms are unclear
Product Quality = β₀ + β₁(Conveyor Speed) + β₂(Operator Interventions) + β₃(Machine Temp) + ε
Coefficients (β’s) are estimated using regression analysis
include error term to account for variability
help understand and improve real processes
production line
probability model
help quantify risk and uncertainty in decisions
examples:
sample size selection
process monitoring
quality control
discrete model - countable events (defects)
continuous model - measurable data (concentrations)
Why probability models are needed?
sampling always involves chance → results may not represent reality
measure uncertainty and control the risk of wrong decisions
examples:
detecting a defect
acceptable risk
probability
study of randomness and uncertainty
methods to quantify likelihood
random experiment
any activity whose outcome is subject to uncertainty
sample space (S)
set of possible outcomes from the experiment
represented as tree diagram
event (E)
any subset of sample space (S)
simple - one outcome of interest
E_1={(0,0)}, … E_n = {(6,6)}
compound - mult outcomes of interest
A={(0,0), (0,1)}
complement - A’
outcomes in sample space S that are not in A
Let 𝑨 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒 , 𝑩 = 𝟑, 𝟒, 𝟓, 𝟔 , 𝑪 = 𝟏, 𝟑, 𝟓
𝑨′ = 𝟓, 𝟔
Union - AUB
outcomes in A or B or both
Let 𝑨 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒 , 𝑩 = 𝟑, 𝟒, 𝟓, 𝟔 , 𝑪 = 𝟏, 𝟑, 𝟓
𝑨 ∪ 𝑩 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔 = 𝑺
intersection - A∩B
outcomes in both A and B
Let 𝑨 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒 , 𝑩 = 𝟑, 𝟒, 𝟓, 𝟔 , 𝑪 = 𝟏, 𝟑, 𝟓
𝑨 ∩ 𝑩 = 𝟑, 𝟒
P(A)
prob of A in S
must follow axioms of probabiliy
For any event 𝑨, 𝑷 𝐀 + 𝑷 𝐀′ = 𝟏, from which 𝑷 𝐀 = 𝟏 − 𝑷 𝐀′
For any event 𝑨, 𝑷 𝑨 ≤ 𝟏
When events 𝑨 and 𝑩 are not mutually exclusive, simply adding 𝑷(𝑨) and 𝑷(𝑩) double-counts their intersection
(addition rule): For any two events 𝑨 and 𝑩 not mutually exclusive, 𝑷 𝑨 ∪ 𝑩 = 𝑷 𝑨 + 𝑷 𝑩 − 𝑷(𝑨 ∩ 𝑩)
For any three events 𝑨, 𝑩, and 𝑪, 𝑷 𝑨 ∪ 𝑩 ∪ 𝑪 = 𝑷 𝑨 + 𝑷 𝑩 + 𝑷 𝑪 − 𝑷 𝑨 ∩ 𝑩 − 𝑷 𝑨 ∩ 𝑪 − 𝑷 𝑩 ∩ 𝑪 + 𝑷(𝑨 ∩ 𝑩 ∩ 𝑪)
Axioms of prob
axiom 1
P(A) >= 0
Axiom 2
P(S)=1
Axiom 3
If 𝑨𝟏, 𝑨𝟐, … is an infinite collection of disjoint events, then
𝑷 (𝑨𝟏 ∪ 𝑨𝟐 ∪ ⋯ )= Sum from i=1 to infinity of P(A_i)
relative frequencies
probability
as experiemnt is repeated, rel frequency of an event stablizes
long run rel frequency defines the P(A) probability
counting techniques
find number of outcomes in sample space or events
permutation
ordered arrangement of all elements in set
The number of permutations of 𝒏 different elements is 𝒏! where: 𝒏! = 𝒏 × 𝒏 − 𝟏 × 𝒏 − 𝟐 × ⋯ × 𝟐 × 𝟏
The number of permutations of subsets of 𝒓 elements selected from a
set of 𝒏 different elements is 𝑷𝒏,r = 𝒏 × (𝒏 − 𝟏) × (𝒏 − 𝟐) × ⋯ × 𝒏 − 𝒓 + 𝟏 =
𝒏!/(𝒏 − 𝒓 )!
permuation of similar objects (not all different)
The # of permutations of 𝒏 = 𝒏𝟏 + 𝒏𝟐 + ⋯ + 𝒏𝒓 objects of which 𝒏𝟏 are
of one type up to 𝒏𝒓 of a 𝒓𝒕𝒉 type is 𝒏! /(𝒏𝟏! 𝒏𝟐! … 𝒏𝒓!)
combination
selection of r elements when order does not matter
Cn,r = (n r) = n!/ r!(n-r)!
if random experiemnt where may or may not be replaced → combine combination with multication rule
mult rule (counting techniques)
squence of k steps, number of ways to complete step 1 are n1, nuber of ways to complete step 2 are n2
total number of ways to complete operation = n1xn2x..nk
conditional prob
prob depends on what is known
new info may change prob
P(A|B)
prob of A given B
B is conditioning event
𝑷 (𝑨 )= 𝑵 (A) / N
For any two events 𝑨 and 𝑩 with 𝑷(𝑩) > 𝟎
P(A|B) = 𝑷(𝑨∩𝑩) / 𝑷(𝑩)
P(A|B) is proportional to 𝑷(𝑨 ∩ 𝑩)
multiplication rule for 𝑷(𝑨 ∩ 𝑩) : 𝑷 (𝑨 ∩ 𝑩 )= P(A|B)× 𝑷(𝑩)
Bayes theorem - law of total prob
with enough conditional prob, overall event prob can be determined
For any event 𝑩, we can write 𝑩 as the union of the part of 𝑩 in 𝑨 and the part of 𝑩 in 𝑨′. That is: 𝑩 = (𝑩 ∩ 𝑨) ∪ (𝑩 ∩ 𝑨′)
Because 𝑨 and 𝑨′ are mutually exclusive events, so are 𝑩 ∩ 𝑨 and 𝑩 ∩ 𝑨′, leading to the total probability rule for two events: P(B) = P(𝑩 ∩ 𝑨) + P(𝑩 ∩ 𝑨’) = P(B|A)P(A) + P(B|A’)P(A’)
total prb rule for mult events: P(B) = P(𝑩 ∩ E1) + … + P(𝑩 ∩ Ek) = P(A|E1)P(E1) +…+P(A|Ek)P(Ek)
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴 | 𝐵)𝑃(𝐵) = 𝑃(𝐵 ∩ 𝐴) = 𝑃(𝐵 | 𝐴)𝑃(𝐴)
𝑃(𝐴 | 𝐵) = 𝑃(𝐵|𝐴)𝑃(𝐴) / 𝑃(𝐵)
𝑃(E1|B)= 𝑷 (𝑩 |𝑬𝟏 )𝑷(𝑬𝟏) / 𝑷 (𝑩| 𝑬𝟏) 𝑷 (𝑬𝟏) + ⋯ + 𝑷 (𝑩| 𝑬𝒌) 𝑷(𝑬𝒌) for P(B)>0
independence
Two events A and B are independent when learning that one event occurred does not change the probability of the other
Two events 𝑨 and 𝑩 are independent if 𝑷( 𝑨| 𝑩) = 𝑷(𝑨) and are dependent otherwise
𝑨 and 𝑩 are independent if 𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑨) × 𝑷(𝑩)
Two events A and B are mutually exclusive if they
cannot occur at the same time, i.e. 𝑷 𝑨 ∩ 𝑩 = 𝟎
random variable
for given subspace (S), RV is any rule that associates a number with each outcome in S
RV is function whose domain is the sample space and range is the set of real numbers
denoted with upercase letters (X)
lowercase letter (x) id given outcome of X
2 outcomes → S={S,F}
RV in this sample space → X(S)=1, X(F)=0
bernoulli RV - RV whose only possible outcomes are 0 and 1
discrete - possible values form finite set or countable infinite list
continous - possible values lie in one interval or union of intervals
no possible vlaue of te varibale has positive prob → P(X=c)=0
prob distr for discrete RV
how total prob of mass 1 is distr among possible values of X
f(x)
prob mass function (PMF) of discrete rv: 𝒇 (𝒙 )= 𝑷 (𝑿 = 𝒙)
f(x)>=0, and sum of all x f(x) = 1
PMF - prob mass function
distr total prob mass (=1)
prob histogram shows how prob mass allocated across values
all rectangles share same base width
rectangle above each value of x with f(x)>0
height is proportional to f(x)
parameter of prob distr
fixed numerical value that determines how prob distr assigns prob to outcomes
NOT random
changing parameter changes shape of distr
parameter (p)
if p inc → success is more likely
𝑷( 𝑿 = 𝟏) = 𝒑
𝑷 (𝑿 = 𝟎) = 𝟏 − 𝒑
f(x) = (1-p)^x-1 p for x=1,2…
cumulative dist function CDF
how prob accumulate up to x
The CDF 𝑭(𝒙) of a discrete random variable 𝑿 with PMF 𝒇(𝒙) is defined for every number 𝒙 by: 𝑭 𝒙 = 𝑷 (𝑿 ≤ 𝒙) = SUM_y:y<=x for f(y)
For any 𝒙, 𝑭(𝒙) is the probability that the observed value of 𝑿 will be at most 𝒙
CDF
non-decreasing
right cont : each horz step included its left endpoint and extend t the right
limits: lim_x→-inf F(x)=0
lim_x→+inf F(x)=1
For any numbers 𝒂 ≤ 𝒃: 𝑷 (𝒂 ≤ 𝑿 ≤ 𝒃) = 𝑭 (𝒃 )− 𝑭(𝒂−)
a- is largest possible value of X less than a
If 𝑿 can only take integer values, 𝒂 and 𝒃 are integers; then: 𝑷( 𝒂 ≤ 𝑿 ≤ 𝒃) = 𝑭 (𝒃) − 𝑭 (𝒂 − 𝟏)
Special case: 𝒂 = 𝒃 (probability at a single value) :
𝑷( 𝑿 = 𝒂) = 𝑭( 𝒂 )− 𝑭 (𝒂 − 𝟏)
mean
center or middle of prob distr
-Two different distributions can have the same mean and variance
μ= E(X)= SUM_x for 𝑥𝑓(𝑥)
E(X) is weighted average
the point at which a bar balances
variance
measure of dispersion in distr
-Two different distributions can have the same mean and variance
σ² = V(X) = E(X-μ )² = SUM_x for (X-μ )² f(x) = SUM_x for x²f(x) - μ²
standard deviation
σ = square root of σ²
prefered over variance
σ is in same units as the data (variance is in unites²)
X units away from the mean
easier to compare across data sets
binomial distr
series of repeated random independent trials
rv X i= number of sucesses in n trials
parameters: n trials, p=prob of success
symmetric shape skewked pepending on n and p
counts number of sicesses in a fixed number of tri
each trial is success or failure (2 outcomes)
success meets criterion
prob of success is constant
number of trials n is fixed
only 2 possible outcomes → bernouilli trial
general prob, 0<p<1
P(X=x) = (n x)(p^x)(1-p)^n-x
no closed form for CDF
CDF: F(X<=x)= SUM_i=0 to x for (n i)(p^i)(1-p)^n-i
mean: E(X)=np
variance: V(X): np(1-p)
general prob formula
(n x) = n! / x!(n-x)!
general prob
P(X=x) = (n x)(p^x)(1-p)^n-x
geometric distr
series of repeated random independent trials
counts number of trials until 1st sucess
rv Y=trial number of 1st success
parameter: p = success prob
shape: always skewed right
in bernoulli trials → f(x) = (1-p)^x-1 (p) for x=1,2,…
CDF: F(x)= 𝑷 (𝑿 ≤ 𝒙) = 𝟏 − (𝟏 − 𝒑)^x
mean: E(X)= 1/p
variance: V(X)= (1-p)/p²
lack of memory
success can be started at any trial w/o changing prob dist of rv
system does not wear out and lacks memory
prob of finding defect in next n chips is always the same
distr forgets the past: 100 defect free chips doesnt chage prob of next 100 chips beign dfect free
each inspection is ind with constant defect prob
system doesnt improve or degreafe during operation
negative binomial distr
X= number of trials required to obtain r successes
Total # of trials to obtain 𝒓 successes is 𝑿𝟏 + 𝑿𝟐 + ⋯ + 𝑿𝒓
lack of memory property, each rv follows a geometric distribution with same value of 𝒑
shape: right skewed, more skew as p dec
bernoulli trials → f(x) = (x-1 r-1) (1-p)^x-r (p^r) , x=r,r+1,….
CDF: P(X<=x)= SUM, k=r to x for (k-1 r-1) (1-p)^k-r (p^r)
mean: E(X): r/p
variance: V(X): r(1-p)/p²
Poission distr
number of events occuring in fixed interval (time, space, volume)
independent trials
at known average rate
randomly through interval
defect per batch
shape: right skewed or symmetric, as λ inc → gets more symmetic
for λ>0 → f(x)= (λ^e e^-λ) / x! for x=0,1…
CDF: P(X<=x)= SUM, k=r to x for (λ^e e^-λ) / k!
mean: E(X): λ
variance: V(X): λ
mean and variance are the same
when to use: need all 4
count discrete events
random in time or space
events are rare relative to opportunity
events dont influence each other
per unit quantities
rare event in cont domain
rate based process
prob density function
f(x)>=)
∫−∞ to +∞ 𝒇 (𝒙 )𝒅𝒙 = 𝟏
𝑷 (𝒂 ≤ 𝑿 ≤ 𝒃 )= ∫ 𝒂 to 𝒃 𝒇( 𝒙) 𝒅𝒙 = area under 𝒇(𝒙) from 𝒂 to 𝒃 for any 𝒂 and 𝒃
histograms - approx of pdf
area under bar = rel frequency = est of the prob that measurement falls in the interval
area under f(x) over interval = true prob meas falls in interval
density est at xf(x) = 0
assigns mass only over interval → prob at point =0
𝑷 (𝒙𝟏 ≤ 𝑿 ≤ 𝒙𝟐) = 𝑷 (𝒙𝟏 < 𝑿 ≤ 𝒙𝟐) = 𝑷 (𝒙𝟏 ≤ 𝑿 < 𝒙𝟐) = 𝑷(𝒙𝟏 < 𝑿 < 𝒙𝟐)
cumulative distr function
F(x)= ∫−∞ to x f(u)du for −∞ <x<+∞
density: d/dx (F(x)= ∫−∞ to x f(u)du) = f(x)
as long as derv exists
mean of cont rv
E(X)=∫−∞ to +∞ xf(x)dx
variance of cont rv
V(X)=∫−∞ to +∞ (X-μ )² f(x)dx = ∫−∞ to +∞ x²f(x)dx - μ²
standard deviation does not change for cont rv
normal (gaussian) distr
model for cont measurements
quality control, measurement sys, central limit theorem (many small ind effects add up to normal distr)
parameters: − ∞ < μ < ∞ , and 𝝈>0
density: 𝒇( 𝒙) = 𝟏/ ( 𝝈 square root (𝟐𝝅)) • 𝒆^(−(𝒙−𝝁)² / 𝟐𝝈² ). for − ∞ < 𝒙 < ∞
mean: E(X): μ
variance: V(X): 𝝈²
notation: N(μ,𝝈²)
from symmetry: 𝑷 (𝑿 < 𝝁) = 𝑷( 𝑿 > 𝝁 )
shape: symmestric about 𝝁 and bell shaped
center of bell is mean and mediam of distr
𝝁 is location parameter (changing 𝝁 shifts side to side)
𝝈 is scale parameter (changing 𝝈 strenches or compreses curve w/o chanigng shape)
lower 𝝈 → taller bell
standard normal distr
Computing 𝑷(𝒂 ≤ 𝑿 ≤ 𝒃) requires integrating the pdf which cannot be done by standard integration techniques
The probability has been calculated for 𝝁 = 𝟎 and 𝝈 = 𝟏 and tabulated for certain values of 𝒂 and 𝒃 - standard normal rv
use table for other mu and sigma probs
CDF (Z): 𝚽( 𝒛) = 𝑷(𝒁 ≤ 𝒛)
If 𝑿 is a Normal rv with 𝑬 𝑿 = 𝝁 and 𝑽 𝑿 = 𝝈², the rv: 𝒁 = (𝑿 − 𝝁)/𝝈
Normal approx of binomial distr
if X is binimal distr with parameters n and p
Z=(X-np)/ (square root(np(1-p)))
np=E(X)
NP(1-p) = V(X)
good approx of E(X) and V(X) >5
continuity correction:
P(X<=x) = 𝑷(𝑿 ≤ 𝒙 + 𝟎. 𝟓) = P(Z<= (x + 0.5 -np)/ (square root(np(1-p)))
P(X>=x) = 𝑷(𝑿 >= 𝒙 - 𝟎. 𝟓) =P(Z<= (x - 0.5 -np)/ (square root(np(1-p)))
normal approx to poisson distr
If 𝑿 is a Poisson rv with 𝑬 𝑿 = 𝝀 and 𝑽 𝑿 = 𝝀 → Z= (X-𝝀)/ (sqaure root(𝝀))
same continutiy correct ad binomial
approc is good if 𝝀>5
uniform distr
notation: U(A,b)
application: reandom number generation in simulations, and arrival time withon known interval
density: f(x) = 1/(b-a). for a<=x<=b
mean: E(X): (a+b)/2
variance: V(X)= (a+b)² / 12
exponential distr
The rv 𝑿 that equals the distance between successive events from a Poisson process with mean number of events 𝝀 > 𝟎 per unit interval is anexponential rv with parameter 𝝀 and pdf
density: f(x)= 𝝀𝒆^(−𝝀𝒙), for 𝟎 ≤ 𝒙 < inf
Mean: E(X): 1/𝝀
variance: V(X): 1/𝝀²
lacks memory
For an Exponential rv 𝑿: 𝑷(𝑿 < 𝒕𝟏 + 𝒕𝟐 | 𝑿 > 𝒕𝟏) = 𝑷(𝑿 < 𝒕𝟐)
weibull distr
model lifetimes and failure behavoirs (durability and preformance)
shape parameter allows to represent inc, dec, constant failure rates
parameters:
scale: 𝜹 > 𝟎
shape: 𝜷 > 𝟎
density: f(x)= (𝜷 / 𝜹) (𝒙 / 𝜹)^(𝜷−𝟏) e𝒙𝒑 [− (𝒙 / 𝜹)^𝜷 ]. for 𝒙 > 𝟎
CDF: 𝑭 (𝒙) = 𝟏 − 𝒆^[-(x / 𝜹)^𝜷 ]
mean: E(X): 𝜹𝚪( 𝟏 + (1/𝜷))
variance: V(X)= 𝜹²𝚪 (𝟏 + (2/𝜷))- 𝜹² [ 𝚪 (𝟏 + (1/𝜷))]²
bathtub curve (failure rates)
early life failure (decreasing failure rate, 𝜷 < 𝟏)
Cause: Manufacturing defects, poor quality components, assembly errors
Useful Life (constant failure rate, 𝜷 = 𝟏)
Cause: Random events, external shocks, and human error
failure is not related to the age of the component
Wear-Out Failures (increasing failure rate, 𝜷 > 𝟏)
Cause: Material fatigue, corrosion, and natural aging
not included: lognormal
lognormal