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EML 4550: Engineering Design Methods Probability and Statistics in Engineering Design Class Notes Hyman: Chapter 5 EML4550 2007 1 Basic Concepts Probability Functions of a probabilistic variable Probability distributions Statistics Reliability vs. Availability MTBF and MTTR FTA FMECA EML4550 -- 2007 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 EML4550 -- 2007 Probability Independent events: either A & B has no effect on the other Example: event A: die 16, event B: die 24, 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 16, 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 16, event B: 1+210, 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 EML4550 -- 2007 Examples Conditional probability: P(A|B)=probability that A will occur given B has occurred Example: event A: die 16, event B: 1+210, 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+210 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) EML4550 -- 2007 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) EML4550 -- 2007 P( D V1 ) P(V1 ) (0.042)( 0.57) 0.63 P( D ) 0.038 Probability in Measurements: The Histogram EML4550 -- 2007 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) EML4550 -- 2007 Normal distribution (Gaussian) 1 ( y )2 / 2 s 2y f ( y) e s y 2 Mean: Standard deviation: sy EML4550 -- 2007 f ( y )dy 1 y* 1 Pr( y y ) f ( y )dy s y 2 * Pr( y y * ) 1 Pr( y y * ) EML4550 -- 2007 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 * EML4550 -- 2007 z* e z2 / 2 dz EML4550 -- 2007 Z-Table, EML4550 -- 2007 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 EML4550 -- 2007 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 EML4550 -- 2007 Probability distributions General definitions EML4550 2007 16 Probability in Measurements: The Histogram EML4550 -- 2007 First Moment: The Average (Mean) First moment of data is zero (by definition of average, or mean) 1 n y yi n i 1 EML4550 -- 2007 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 EML4550 -- 2007 S S2 Third Moment: The Skew Positive skew: data biased to the right 1 n 3 3 yi y n i 1 EML4550 -- 2007 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 EML4550 -- 2007 4 Kurtosis 4 S Kurtosis - Examples Leptokurtic Platykurtic Mesokurtic EML4550 -- 2007 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) EML4550 -- 2007 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 EML4550 -- 2007 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 EML4550 -- 2007 EML4550 -- 2007 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) EML4550 -- 2007 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