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EML 4550: Engineering Design Methods
Probability and Statistics
in Engineering Design
Class Notes
Hyman: Chapter 5
EML4550 2007
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Basic Concepts
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Probability
Functions of a probabilistic variable
Probability distributions
Statistics
Reliability vs. Availability
MTBF and MTTR
FTA
FMECA
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Probability
 Dictionary definition
 The quality or state of being probable; the chance that a given event will
occur
 Mathematical definition
 The ratio of the number of outcomes in an exhaustive set of equally likely
outcomes that produce a given event to the total number of possible
outcomes
n # of outcomes that an even A will occur

N
N possible outcomes
# of outcomes that an even A will not occur
P( A ) 
 1  P( A)
N possible outcomes
P( A) 
 Examples: Tossing coins? Rolling dice?
Pcoin(head)=1/2;
Pdice(6)=1/6
Pcoin(tail)=1-P(head)=1/2, Pdice(<6)=1-1/6=5/6
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Probability
 Independent events: either A & B has no effect on the other
Example: event A: die 16, event B: die 24, P(A)=1/6, P(B)=1/6, two
independent events, P(A and B)= P(A)P(B)=(1/6)(1/6)=1/36
 Mutual exclusive: outcome of A will exclude outcome of B and vice versa
 Joint probability  both A & B occur:
P(A and B) = P(AB) = P(A)P(B) = P(BA) for independent events A & B;
if A & B are mutually exclusive P(AB)=0
Example: event A: die 16, event B: 1+2<6, two mutually exclusive events,
P(A and B)=0, it is not possible to have both events
 The probability of A or B (non-mutually exclusive):
P(A or B) = P(A+B) = P(A) + P(B) - P(AB)
Example: Event A: die 16, event B: 1+210, P(A)=1/6, P(B)=1/6
P(A or B)=P(A+B)=P(A)+P(B)-P(AB)=1/6+1/6-1/12=9/36=3/12
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Examples
 Conditional probability: P(A|B)=probability that A will
occur given B has occurred
Example: event A: die 16, event B: 1+210, P(A)=1/6, P(AB)=1/12,
P(B|A)=(1/12)/(1/6)=1/2
Since once die 1 is 6. There are three (4, 5, 6) out of six (1,2,3,4,5,6)
tries that die 2 will be greater than or equal to 4 to satisfy event B:
1+210
P( AB)
P ( AB )
P( A B ) 
, P( B A) 
P( B )
P( A)
Show that (Homework 4 - 3)
P(B)  P(AB)  P(AB)
P(B)  P(A)P(B A)  P(A)P(B A)
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Probability
Example: given the following data table
Vendor-1 (V1)
Vendor-2 (V2)
Total
Defective (D)
500
300
800
Acceptable (A)
11500
8700
20200
total
12000
9000
21000
P(V1)=12000/21000=0.57, P(V2)=9000/21000=0.43=1-P(V1)
P(D|V1)=500/12000=0.042, P(D|V2)=0.033
P(A|V1)=0.958=1-P(D|V1), P(D)=800/21000=0.038
P(DV1)=P(D|V1)P(V1)=(0.042)(0.57)=0.024=500/21000
P(DV2)= P(D|V2)P(V2)=(0.033)(0.43)=0.014=300/21000
P(D)=P(DV1)+P(DV2)=0.038
What is the probability that a defective part comes from vendor-1?
P(V1 D) 
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P( D V1 ) P(V1 ) (0.042)( 0.57)

 0.63
P( D )
0.038
Probability in Measurements: The Histogram
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Probability Density Distribution
 If the measurement variable is continuum (ranges tend to
zero, number of ranges tend to infinity), the resulting
histogram is a probability distribution (experimental set
tends to infinity)
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Normal distribution (Gaussian)
1
( y   )2 / 2 s 2y 
f ( y) 
e
s y 2
 Mean:

 Standard deviation: sy
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

 f ( y )dy  1

y*
1
Pr( y  y )   f ( y )dy 
s y 2

*
Pr( y  y * )  1  Pr( y  y * )
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y*
e


 ( y   ) 2 / 2 s 2y

dy
Standard normal distribution (Z-distribution)
 y

z
 s 
 y 
z 0
sz  1
1
Pr( z  z ) 
2
*
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z*

e

 z2 / 2
dz
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Z-Table,
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Example
 A group of steel rods (population) with an average yield strength of
50,000psi with standard deviation of 6,000psi
 Probability of drawing a bar with yield strength less than 40,000psi?
 Z*=(40000-50000/6000)=-1.67, Pr(z<-1.67)~0.0475
 Probability of drawing a bar with yield strength greater than
55,000psi?
 Z*=(55000-50000/6000)=0.83
 Use symmetry, for -0.83, Pr(z<-0.83)=0.2033
 Pr(z>0.83)=0.2033
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Implications to design
 The mean is an important indicator of quality
 The standard deviation is just as important
 Quality control to ensure minimum spread in properties
 Economic penalty of a ‘broad’ distribution
 “derating” to ‘guarantee’ a value
 Predictability for design purposes
 More on this when we cover reliability
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Probability distributions
General definitions
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Probability in Measurements: The Histogram
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First Moment: The Average (Mean)
 First moment of data is zero (by definition of average, or
mean)
1 n
y   yi
n i 1
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1 n
1
1    yi  y 
n i 1
Second Moment: The Variance
 S2 is the sample variance, S is the sample standard
deviation
Note: n-1 is used instead of the obvious
choice of n (give an unbiased estimate of s
1 n
2
2  S 
  yi  y 
n  1 i 1
2
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S  S2
Third Moment: The Skew
 Positive skew: data biased to the right
1 n
3
3    yi  y 
n i 1
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3
Skew  3
S
Fourth Moment: The Kurtosis
 Zero-based kurtosis is defined as 3
 Kurtosis = 3 (mesokurtic distribution)
 Kurtosis < 3 (platykurtic distribution)
 Kurtosis > 3 (leptokurtic distribution)
1 n
4
 4    yi  y 
n i 1
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4
Kurtosis  4
S
Kurtosis - Examples
Leptokurtic
Platykurtic
Mesokurtic
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Probability of failure: Example
 Slender rod of cross section A subject to load P
 Stress, s=P/A
 Rod will fail if s>y (yield strength)
 Suppose P=20,000lb, A=0.5sq.in., s=40,000psi
 Stock of rods has mean yield strength y=50,000psi with
standard deviation=6000psi (same as last example)
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Design to safety factor
S=40,000psi, y=50,000psi
FS=y/s=50,000/40,000=1.25
What if we were given the safety factor? FS=1.25
A = P/s, s = y/FS ---> A = FS*P/y = 1.25*20,000/50,000=0.5
sq.in.
 Is this a safety factor high? Low?
 The safety factor approach to design is NEVER a direct measure
of safety because it completely ignores the standard deviation
 It is a simple approach (simplistic), but inappropriate
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Design to probability of failure
 From the previous example (last lecture) we know the
probability of the yield strength to be less than 40,000psi is
4.75%
 The safety factor of 1.25 would lead to a probability of
failure of nearly 5%
 Suppose we want the probability of failure to be 0.5% when
we pick a rod at random from the stock and subject it to a
20,000lb load.
 Pr(y<y*), or Pr(y<s)=0.005, s=P/A, y*=P/A
 Standard distribution, Pr(z<z*)=0.005
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Example (Cont’d)
From the table, z* = -2.575
-2.575 = (s-50,000/6000)
s = 34,550psi, P=20,000lb ---> A = 0.58 sq. in.
So, the area needs to be 0.58 sq.in. to have a
probability of failure of 0.5%
 (if rods come in 0.1 sq. in. increments, design should
call for 0.6 sq.in. rod, s=33,300psi, z*=-2.78,
Pr=0.27%<0.5%, ‘safety factor’ would have been
FS=50,000/33,300=1.5, but note that safety factor is
dependent on standard deviation)
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Implications
 Probabilistic considerations to design
 Fully specifying a variable requires knowing not only a
mean value, but also a standard deviation
 Many systems follow a normal distribution, but this is
not always the case. It is important to fully characterize
systems statistics before design
EML4550 -- 2007