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# Prob Meth In Engr CIVL 7012

University of Memphis

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This 33 page Class Notes was uploaded by Dana Yundt on Friday October 23, 2015. The Class Notes belongs to CIVL 7012 at University of Memphis taught by Stephanie Ivey in Fall. Since its upload, it has received 16 views. For similar materials see /class/228414/civl-7012-university-of-memphis in Civil Engineering at University of Memphis.

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Date Created: 10/23/15

Continuous Distributions Probability Density Function 7 De nition 1 fx 2 0 2 Tfxdxl 3 Pa s X s b jfxdx Palt X lt b fX You use fX to calculate an area that represents the probability that X takes on a value in ab CI VL 70128012 Probabilistic Meihadsfar Engineers Example Determine the probability that X assumes a value between 06 and 12 x for 0 lt x lt 1 Where fx 2 xforlelt2 0 elsewhere fX 05 For a continuous random variable PX X 0 Cumulative distribution function FxPXSxjfudu for ooltxltoo Pa s X s b Fa Fa CI VL 70128012 Probabilistic Methadsfar Engineers Example Suppose the cumulative distribution function of the random variable X is 0 xlt0 Fx 02x 0Sxlt5 1 x25 Determine The probability density function of X and PX lt 28 Mean and Variance of a quot Random Variable u mean Ex fxfx variance 0392 i x u2 fxdx foo CI VL 70128012 Probabilistic Methadsfar Engineers Uniform Distribution fX 1ba fx aSXSb b a Fx Ifudub1ajduxa anSb Ex jmxmxjx biajdxba Z 1 Z 039 b a 12 Example A bomb is to be dropped along a mile long line that stretches across a practice target The target center is at the midpoint of the line The target will be destroyed if the bomb falls within 01 miles to either side of the center Find the probability that the target is destroyed if the bomb falls randomly along the line CI VL 70128012 Probabilistic Methadsfar Engineers Exponential Distribution f x le where l average rate of occurrence Fxl e Ex i I l 0392 Events are Poisson distributed time between events will be exponentially distributed thus the derivation of this model Example At a stop sign location on a cross street vehicles require headways of 6 seconds or more in the main street traffic before being able to cross If the total ow rate of the main street traffic is 1200 vph what is the probability that any given headway will be greater than 6 seconds Example An electronic component is known to have a useful life represented by an exponential density with failure rate of 10395 failures per hour What fraction of the components will fail before the mean life CI VL 70128012 Probabilistic Methadsfar Engineers The exponential distribution is unique in that it can be said to be memoryless This means that the probability of a success in a certain time period does not change if the start time of the interval changes Example The lifetime of a particular integrated circuit has an exponential distribution with mean 2 years Find the probability that the circuit lasts longer than three years Now suppose the circuit is now four years old and is still functioning Find the probability that it functions for more than three additional years Compare this with the preVious probability a new circuit functions for more than three years So we are interested in a conditional probability 39 After 4 years the probability of the circuit lasting more than three years is the same as if it is new thus the lack of memory property Gamma Distribution The exponential distribution is a special case of the Gamma distribution where X is the interval length until r counts occur in a Poisson process CI VL 70128012 Probabilistic Methadsfar Engineers 4 m N W9 we no GammaFlmcnan 17 FxPTSX n j elxewhere nx Fxgure 4736 Montgoma39y text CM 70123012 Probabtltmc Mmodsfov EMgtmzth The Gamma Function Fr Ixr39le39xdx 0 Properties 1 F11 Forrgt1 2 Frr1 Fr1 3 r12fn 4 Fr1r for large values of r r may be approximated using Stirling s approximation Stirling s Approximation to n 71 V2727 Z n gt 00 e Accurate to within 1 for ngt10 and within 01 for ngt100 Example Suppose the survival time in weeks of a randomly selected male mouse exposed to 240 rads of gamma radiation has a gamma distribution with r 8 and 7t 115 Find a The expected survival time b variance c the probability that a mouse survives between 60 and 120 weeks d the probability that a mouse survives at least 30 weeks CI VL 70128012 Probabilistic Methadsfar Engineers GAMMA FUNCTION Vulnu 011 n c lxIldz 1 01 1 nl39n 0 n 1quot n n 139 n n 1 n n 1 n 1 25 90640 150 58623 115 919 1 26 909 151 58659 176 92137 1 27 90250 152 8871 177 92376 1 28 90072 153 88751 119 92623 1 8990 154 9381 179 92871 1 30 891quot 155 958 130 93133 1 31 99600 1 181 931103 1 32 8946 157 99049 182 93085 1 33 9933 15 39142 183 93969 1 31 99222 159 19243 184 94251 1 35 M115 1 60 5935 135 94561 1 36 8901 1 01 39463 196 91369 1 37 88931 1 92 quot592 187 9518 1 8895 1 63 59721 188 95507 1 88785 1 64 all 189 95838 115 1 8726 1 65 M12 190 96177 110 1 98670 1 06 90167 191 95523 117 1 81136 1 67 90330 191 96578 113 1 8860 193 90500 193 97240 119 1 38580 169 79 194 97610 120 91911 1 95 88565 170 90964 195 97985 121 9155 1 46 885 171 9105 196 99174 122 91311 1 17 88563 172 91258 197 98768 123 91075 1 48 33575 173 91466 198 99171 L2 90952 1 49 83595 174 91933 99 99551 For large punitive vlluu n 1 Hz unproximua Burlint39l nymmnuc maria sz 1 m an quot39 39 l1 12 as sxuw annoy CIVL 70128012 Probabilirtic Methods for Engineers The Incomplete Gamma Function The incomplete gamma function is often used for ease of application It is a transformation of the gamma function We will rewrite the Gamma function ra and 1 Where 0i B are parameters which determine the shape of the curve 1 1x 139le for xgt0 FOO So fx Now EX a B and VX a 52 The incomplete gamma function may then be defined as 3 y FxaJx dy xgt0 0 mile Fa If X has a gamma distribution with parameters 0i and B then for any Xgt0 PX SxFa Where F 0i is the incomplete gamma function CI VL 70128012 Probabilistic Methadsfar Engineers MI III 9H comma Function Hz a EnidVP dy 39 I 1 z a 4 5 0 7 9 9 lo 1 a rm 0110 019 004 001 000 000 000 000 a 1594 325 143 05 017 005 001 000 000 3quot w 190 577 355 195 094 034 11239 004 001 4 992 5911 909 702 507 571 215 111 051 021 000 5 993 37571960 375 735 560 384 233 133 031 0 998 5190995 938 949 715 554 399 250 155 094 7 0997 970 919 927 099 550 401 271 170 5 1000 M2997 990 959 900 097 547 7 203 9 999 994 979 945 894 799 676 544 419 10 1000 997 990 971 953 970 750 007 542 11 999 995 995 902 921 957 7011 059 12 1000 999 992 990 954 911 945 751 15 999 990 999 974 940 900 1154 14 1000 999 994 9110 909 959 591 15 999 997 992 9112 90 930 Example Suppose the survival time in weeks of a randomly selected male mouse exposed to 240 n Find rads ofgamma radiation has a gamma distribution with a 8 a d B a The expected survival time b variance 0 the probability that a mouse survives between 60 and 120 weeks d the probability that a mouse survives at least 30 weeks CIVL 70128012 Probabilirtic Methods for Engineers Wexbull Distxubuu on Used to model ume unal fmlure ofvanous physical systems H 1 s v gj em for x gt 0 where 5 seaze parameter and p shape parameter Fx17277 E 5r1 x 2 V0 03 5 r13e 1quotHlj 9 9 Figure 4727 Montgomery text Cm 70123012 onbnbmmc Memasar EMgmea Example Navidi text Weibull 39 39 39 L A 39 bake step in L e r L L L step for a randorn1y chosen 1ot Ifoollows a Weibull distribution having a 03 and 10 39 39 39 L 39 r nranronr uqu whatis the probabihty that it takes between two and seven hours The Normal Distribution Most widely used model for distribution ofrandom variables May also be called Gaussian distribution You aria in 39 39 39 distribution as long as you choose appropriate center and width of curve mean and standard deviation x 2 03971 021 62 1115 115 x CIVL 70128012 Probabilistic MELhadsfar Engineers Probabilig Densig Function 1 3H2 fx 2 2quot icoltxltuo G427 Ex Vx r2 The distribution is symmetrical PXgtp PXltp 05 For any normal random Variable P0176 lt X ltya 06827 P u726 lt X lty2 r 09545 P0173a ltXlty3 r09973 fax g39i sin 6 M730 L172u p70 p n0 p21r 30 x L39 LI 997 CIVL 70128012 Probabilistic Methodsquota Engineers rehabilin Dmitv Fumam 1M3 M22 inE v exp z p 2 sea Eq 1012 Cuer shown use A O gt Flam 1061 A normal random variable is called a standard normal random variable z if u 0 and Cummulative Distribution Function Fz FZ E z Tables are used to determine the value of Fz Table II in Appendix A EX from book FZ gt 126 1 PZ12617 089616 7 010384 FZ lt 086 7 019490 FZ gt 137 FZ lt 137 7 091465 by symmetry P125 lt z lt 037 7 FZ lt 037 7 FZ lt 125 7 0664317 010565 7 053866 CIVL 70128012 Probabilirtic Methods for Engineers A transformation must be applied for normal distributions which do not correspond to the standard normal distribution in order to make use of the normal tables for Fz Z transformation If X is a normal random variable with EX u and VX 62 then where the transformed variable Z is a standard normal random variable Example In diaphragms of rats tissue respiration rate under standard temperature conditions is normally distributed with u 203 and 6 044 a What is the probability that a randomly selected rat has rate Xgt25 b What is the probability that X falls outside the interval 159 247 CI VL 70128012 Probabilistic Methadsfar Engineers Example In an industrial process the diameter of a ball bearing is an important component part The buyer sets specifications on the diameter to be 30 i 001 cm The implication is that no part falling outside these speci cations will be accepted It is known that in the process the diameter of a ball bearing has a normal distribution with mean 30 and standard deViation 0005 On the average how many manufactured ball bearings will be scrapped Example Gauges are used to reject all components where a certain dimension is not within the speci cation 15 i d It is known that this measurement is normally distributed with mean 15 and standard deViation 02 Determine the value d such the speci cations cover 95 of the measurements CI VL 70128012 Probabilistic Methadsfar Engineers The Normal 39 quot to the Binomial Valid only for cases where n is large relative to p np gt 5 If X is a binomial random variable X X Z np quotp 1s approx1mately a standard normal random variable an p Vnpq Example In 6000 tosses ofa die what is the probability of obtaining a 3 between 980 and 1030 times inclusive Example The probability that a patient recovers from a rare blood disease is 04 If 100 people are known to have contracted this disease what is the probability that 30 or less survive CI VL 70128012 Probabilistic Methadsfar Engineers Continuity correction for the binomial gQQrOXIrnaIIO a a isdismtey we have to do a continuity correction for x in our approximation ofthe binomial with the nonnal distn39bution CIVL 70128012 Prababtttsm Me iodsfar Engmeerx ANOVA 7 Analysis of Variance Used for comparison of multiple samples All differences in sample means are judged statistically signi cant or not by comparing them to the variation within samples We performed a comparison of means previously on two samples using hypothesis testing Analysis of variance is used when comparing more than 2 population means Example Sclnple 1 Sclnple 2 Sclnple 3 49 451 401 492 450 400 491 450 399 489 449 398 488 450 402 J71490 J71450 J71400 Variability among samples is large but within samples is small So we would be inclined to believe there is truly a difference in means Sclnple 1 Sclnple 2 Sclnple 3 49 531 352 342 254 593 651 373 348 689 620 455 278 472 252 J71490 J71450 J71400 Now the within sample variability is large in comparison to between sample variation so we are less inclined to believe there is a true difference in population means here CI VL 70128012 Probabilistic Methadsfar Engineers OneWay Classification Each observation is classified according to the population from which it came Notation yij jth observation from population i m observations from population i n total sample size n Z n Ti sum total of sample observations from population i T Z average of n observations from population 1 T x n x G grand sum of all observations from all populations G 2 Ti 7 average of all sample observations 7 n The variability of the 71 sample measurements about their mean 7 can be measured using the sum of the squared deviations This quantity TSS 2204 Z is called the total sum of squares of the measurements about their mean It is possible to partition the total sum of squares as follows 22o m2 ZZo f2 Znf 7Z I f 1 j I Since the rst quantity on the right hand side of the equation measures the variability of an observation yij about its sample mean f we call ssw 22o f2 I J the withinsample sum of squares and we will use it to compute swz The second expression on the right hand of the equation measures the variability of the sample means f about the overall mean 7 This term measures the variability between the sample means and we will refer to SSB ZINE W As the sum of squares between samples and use it to compute sBz CI VL 70128012 Probabilistic Methadsfar Engineers It is much easier to make AOV calculations using the shortcut formulas given below Shortcut formulas for a One Way Classi cation G2 Tss 22x37 1 J 2 2 SSB 21 6 1 n n SSW TSS 7 SSB An analysis of variance for a oneway classi cation of t populations is performed using the following null and alternative hypotheses Ho ul u ug ut that is the tpopulation means are equal Ha At least one of the tpopulation means differs from the rest We can use the shortcut formulas to get 2 SSB SB t l 2 SSW sW n t where tl 9 the degrees offreedom for s n 7 t 9 the degrees offreedom for sVZV Again s is called the mean square between samples and sVZV is called the mean square within samples The null hypothesis is rejected if F wk gt F mug that is if Ira CUN F ml 4 gm exceeds the tabular value ofF for level oftest a v1 t l and v nt The results ofthe study are summarized in an analysis of variance table as shown below CI VL 70128012 Probabilistic Methadsfar Engineers 3 AOV table for one wax classi cation Source Sum of Squares Degrees of Mean Square F Test Freedom Between SSB tl 2 SS3 32 samples SE E 3 5 W Within samples SSW nt 2 SSW SW rt t Totals TSS n l CI VL 70128012 Probabilistic Meihadsfar Engineers Example One way Classi cation A clinical psychologist wished to compare three method of reducing hostility levels in university students A certain test HLT was used to measure the degree of hostility A high score on the test indicated great hostility Eleven students obtaining high and nearly equal scores were used in the experiment Four were selected at random from the eleven problem cases and treated with Method 1 Four of the remaining 7 students were selected at random and treated with Method 3 All treatments were continued for a one semester period Each student was given the HLT test at the end of the semester with the results shown in the table below Method Test Scores Totals 1 80 92 87 83 342 2 70 81 78 74 303 3 63 76 70 209 Total 854 Using 0i 005 perform an analysis of variance on these data to determine if there are differences among mean scores for the three methods CI VL 70128012 Probabilistic Methadsfar Engineers ANOVA for a Randomized Block Design A randomized block design is an experimental design for comparing t treatments in b blocks Treatments are randomly assigned to experimental units within a block with each treatment appearing exactly once in every block Notation for a Randomized Block Design yij observation for treatment i in block j n number of sample measurements t number of treatments b number ofblocks T total for all observations receiving treatment i B j total for all observations in block j G total for all sample measurements G y overall sample mean y n The total sum of squares of the measurements about their mean y is de ned as before TSS 22 if I J Partitioning TSS The total sum of squares can be partitioned into three sources of variability The rst is due to treatments the second is due to blocks and the third is due to random error 22o 72 bZf W MEG 2 22 47 1 7 1 j 1 j I J Variability of the treatment means SST 122039 W SST is called the betweentreatment sum of squares CIVL 70128012 Probabilistic Methadsfar Engineers Variability between blocks SST t2 Ej W J This term measures the variability between the block means Bj and the overall mean and is called the betweenblock sum of squares Shortcut Formulas for Randomized Block Design G2 Tss 22x37 1 J 2 SST 217 6 2 n B2 G2 SSB Znt n SSWTssisSTisSB The Model yij 04 81 where yij the response on treatment i in row j u an overall mean a constant 0i an effect due to treatment i a constant B an effect due to block j a constant sij A random error associated with yij It is assumed that the sij s are independent and normally distributed with mean 0 and variance 0 2 These assumptions imply that yij is normally distributed with mean Eyijl1 Wit 51 2 And variance 0 CI VL 70128012 Probabilistic Methadsfar Engineers 7 Test of Hypotheses Treatments Ho 0i 1 0i 2 0i 1 0 the treatment means are identical Ha At least one of the 0i i s differs from the others Test statistic F MSTMSE Test for Blocks Ho B1 B 2 B1 0 ie no effect due to blocks Ha At least one ofthe B S differs from zero Test statistic F MSBMSE AOV Table for Randomized Block Design Source Sum of Squares Degrees of Mean Square F Test Freedom Treatments SST tl MST SSTtl MSTMSE Blocks SSB bl MSB SSBbl MSBMSE Error SSE bltl MSE SSEbltl Totals TSS btl CI VL 70128012 Probabilistic Methadsfar Engineers 8 Example Randomized Block An experiment was conducted to compare the effect of 3 insecticides on a particular variety of string beans Four different plots were prepared with each plot subdivided into 3 rows A suitable distance was maintained between the rows within a plot Each row was planted with 100 seeds and then maintained under the insecticide assigned to the row The insecticides were randomly assigned to the rows within a plot so that each insecticide appeared in one row in all 4 plots The response of interest was the number of seedlings that emerged per row The data are given below Plot 39 quot 39J l 2 3 4 l 56 49 65 60 2 84 78 94 93 3 80 72 83 85 a Set up an appropriate statistical model for this experimental situation b Run an analysis of variance to compare the three insecticides c Summarize the results in an AOV table Use 0i 005 CI VL 70128012 Probabilistic Methadsfar Engineers 9 ANOVA for a Latin Square Design A t X t Latin square design contains trows and tcolumns The t treatments are randomly assigned to experimental units within the rows and columns so that each treatment appears in every row and every column Notation for a t X t Latin Square Design yij response on treatment i in row j and column k n number of sample measurements n t2 t number of treatments same as the number of rows and columns T total for all observations receiving treatment i R j total for all observations in row j C k total for all observations in column k G total for all sample measurements G y overall sample mean y n Partitioning TSS The total sum of squares can be partitioned into four components The rst three measure variability among treatments rows and columns and the fourth source is due to random error 22 Z rZCT W HZUE 2 r2 2 Zyk 47 13 2W 1 1 k We Variability of the treatment means SST if if SST is called the betweentreatment sum of squares Variability between rows CI VL 70128012 Probabilistic Meihadsfar Engineers 10 SSR t2 Ff W J Variability between columns ssc tZQ W k Shortcut Formulas for Latin Square Design G2 I J T2 G2 SST t n R2 G2 SSR Z t n C2 G2 SSC k t n SSETSS788T7SSRSSC TheModel yxjk luax 7kgyk where yijk the response on treatment i in row j and column k u an overall mean a constant 0i an effect due to treatment i a constant B an effect due to row j a constant y an effect due to column k a constant sijk A random error associated with yijk It is assumed that the sijk s are independent and normally distributed with mean 0 and variance 0 2 CI VL 70128012 Probabilistic Methadsfar Engineers 11 These assumptions imply that yij is normally distributed with mean 2 EyijkHaiBjyk And variance 0 Test of Hypotheses Treatments Ho 0i 1 0i 2 0i 1 0 the treatment means are identical Ha At least one of the 0i 1 s differs from the others Test statistic F MSTMSE Test for Rows Ho B1 B 2 B1 0 ie no effect due to rows Ha At least one ofthe B 1 s differs from zero Test statistic F MSRMSE Test for Columns Ho 31 3 z 31 0 ie no effect due to columns Ha At least one ofthe y 1 s differs from zero Test statistic F MSCMSE AOV Table for t x t Latin Sguare Design Source Sum of Degrees of Mean Square F Test S ares Freedom Treatments SST tl MST SSTtl MSTMSE Rows SSR tl MSR SSRtl MSRMSE Columns SSC tl MSC SSCtl MSCMSE Error SSE t2 7 3r 2 MSE SSE ti 3r 2 Totals TSS tZl CI VL 70128012 Probabilistic Methadsfar Engineers 12 Example Latin Square A traffic engineer wished to compare the total unused green time the total amount of time the signal is green but with no vehicles progressing through the intersection for four different signal timing plans at four intersections of a city It was assumed that the intersections were sufficiently far apart that each functioned as an isolated intersection In addition to comparing the four timing plans at the four intersections the engineer wished to compare the timing plans at different time periods during the day night time AM peak offpeak and PM peak The 4 X 4 Latin square design below was chosen for the experimental design where Roman numerals are used to denote the four signal timing plans being tested Time Period Intersection AM Peak O Peak PM Peak Night 1 IV II III I 2 II III I IV 3 III I IV II 4 I IV II III The timing plans were then assigned to experimental units intersection and traffic peak combinations according to the following scheme On the appointed test day Timing Plan IV was implemented at Intersection 1 during the AM peak period Similarly Timing Plans II III and IV were implemented at Intersections 2 3 and 4 respectively during the same peak During off peak the timing plans were changed according to the table above The procedure was repeated for each time period Delay measurements were obtained for each of the 16 study periods Use the data given in the following table to perform an analysis of variance using 0t 005 TimePeriod Intersection AM Peak O Peak PM Peak Night 1 IV 155 II 339 III 132 I 291 2 II 163 III 266 I 194 IV 228 3 III 108 I 311 IV 171 II 303 4 I 147 IV 340 II 197 III 216 CI VL 70128012 Probabilistic Methods for Engineers 13 Fisher s Least Signi cant Difference LSD Test We are interested in determining which population means differ after we have rejected the hypothesis of equality of t population means in an analysis of variance Fisher 1949 developed a procedure for making pairwise comparisons among a set of t population means Fisher s procedure is summarized below Procedure 1 Perform an analysis of variance to test Ho ul u ut against the alternative hypothesis that at least on of the t population means differs from the rest If there is insufficient evidence to reject Ho using E 015 2 s f stop There 1s no SW need to perform the test Assuming Ho is rejected define the least significant difference LSD to be the difference between two sample means necessary to declare the corresponding population means different 4 For a given value of 0i the least significant difference for comparing in to W is l l m 2 LSD LIAM SW quot1 n J Where a number of treatments For the special case of equal sample sizes 5 0 CIVL 23 LSD iwm All pairs of sample means are then compared If El 2 LSD the population it means in and W are declared different For each pairwise comparison of population means the probability of a type I error is fixed at a specified value of 0i For this method sometimes referred to as Fisher s protected LSD Carmer and Swanson 1973 have shown that the error rate is controlled on an experimentwide basis at a level approximately equal to the 0i value for the F test 70128012 Probabilistic Meihadsfar Engineers

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