Final Exam on Statistics for Electrical Engineering with Answers | STAT 4714, Exams of Probability and Statistics

Download Final Exam on Statistics for Electrical Engineering with Answers | STAT 4714 and more Exams Probability and Statistics in PDF only on Docsity! STAT 4714 Final Exam December 12, 2008 Name______________________ 1. Passwords for a system are required to consists of either 6, 7, or 8 characters, where a character can be one or the digits 0,1,2, ,9… , one of the 26 lower case letters, or one of the 26 upper case letters. Characters are allowed to be repeated in the password. Then, for example, a possible 7-character password is g8aNNe3. How many passwords are possible for this system? 2. A box of 10 components is received, and 3 components are selected at random from this box and installed in a system. The system will not operate if one or more of the 3 components are defective. A test of the system will show whether there are any defective components in the system, but this test will not indicate which component or components are defective. Suppose that the box contains 4 defective components (and 6 non-defective components). When 3 randomly selected components from this box were installed in the system, it was found that the system would not operate. Given that the system would not operate, find the conditional probability that there is exactly one defective component in the system. 3. An elevator is being designed to accommodate 10 people, and the nominal maximum capacity is to be set at 1800 lb. It is estimated that the distribution of the weight of people who will be riding on this elevator can be modeled as a normal distribution with mean μ = 160 lb. and standard deviation σ = 40 lb. If 10 people ride on this elevator, then the total weight of these 10 people can be represented as 101i iT X== ∑ , where iX is the weight of the i th person. (a) If 10 people ride on the elevator, and the weights of these 10 people are independent, find the probability that the total weight will exceed the nominal capacity of 1800 lb. (b) Suppose that it is desirable to design the elevator with a nominal capacity such that the probability of the total weight of 10 people exceeding this capacity is only .005. Find this nominal capacity. 4. The proportion of time X during a day that a terminal is in use is assumed to be a continuous random variable with density 1( ) (1 ) , 0 1f x x xδδ −= − ≤ ≤ , where 0δ > is an unknown parameter of the distribution. The mean of this distribution is 1/(1 )δ+ and the variance is 2/((1 ) (2 ))δ δ δ+ + . Suppose that the proportion of time that the terminal is used is measured on n different days. Let 1 2, , , nX X X… represent these n proportions, and consider the problem of estimating δ. (a) Find an expression for the method of moments estimator of δ. (b) Find an expression for the maximum likelihood estimator of δ. 5. Consider a random process consisting of independent binary random variables 0 1 2, , ,X X X … , where ( 1)tP X p= = and ( 0) 1tP X p= = − for each 0,1,2,t = … , where 0 1p< < . Define the random process 1 2 3, , ,Y Y Y … by 1t t tY X X −= − for each 1,2,3,t = … , and define the random process 1 2 3, , ,Z Z Z … by 1t t tZ X X−= + for each 1,2,3,t = … . Find the crosscovariance function , ( , 1)Y ZC t t + for the tY and tZ processes (here we are considering only the case of 1t t= and 2 1t t= + ). 6. Two instruments are used to record process temperatures. One instrument is coupled to a process computer and the other is used for visual control. Ideally, these two instruments should give the same temperature reading. Daily temperature readings were recorded for the two instruments over a seven-day period and the readings are shown in the table below. The question of interest here is whether the two instruments have the same mean. Using α = .05, test the hypothesis that the two instruments have the same mean. Please state the null hypothesis, the alternative hypothesis, and the conclusion. Day 1 2 3 4 5 6 7 Instrument 1 87.7 84.5 88.4 84.6 86.8 87.3 83.8 Instrument 2 88.2 85.1 89.6 85.3 87.2 88.5 83.9 STAT 4714 Final Exam May 8, 2002 Name___________________________ 1. A company is evaluating two software packages, labeled A and B, for use by the company. A random sample of 8 people in the company who would be using this type of softwear was selected. Each of the 8 people was asked to rate A and B on a number of factors relating to company requirements, and the total scores are given below (higher scores are better). Person 1 2 3 4 5 6 7 8 A 85 79 78 72 89 77 83 85 B 64 76 69 74 71 65 78 79 Do these data provide strong evidence that there is a difference between A and B in terms of meeting the requirements of the company? Use α = .05 and state the null hypothesis, the alternative hypothesis, and the conclusion. 2. Components used in an assembly arrive in boxes of 24. Before the components are installed in an assembly, a sample of 3 components from each box is selected and inspected. Each component in the sample is classified as grade A, grade B, or grade C, where grades B and C indicate lower than standard quality. Suppose that a box arrives with 18 grade A components, 4 grade B components, and 2 grade C components. (a) Find the probability that the sample of 3 from this box will contain exactly one grade C component. (b) Find the expected number of grade C components in the sample of 3. (c) If the sample contains exactly one grade C component, find the conditional probability that the other two components in the sample are grade A. 3. The number of defects in a circuit board has a Poisson distribution with density ,2 ,1 ,0 , ! )( == − x x exf xλλ , where λ > 0. (a) If λ = 1 find the probability that a circuit board has more than two defects. (b) Suppose now that λ is unknown and the number of defects nXXX ,,, 21 … in a random sample of n circuit boards are observed. Derive the maximum likelihood estimator of λ. 4. The joint density of the two continuous random variables X and Y is ⎪⎩ ⎪ ⎨ ⎧ ≤≤≤≤+= otherwise. 0 10 and 10 )( 7 12 ),( 2 yxxyxyxf (a) Find )(XE . (b) Find )2/1 | 2/1( =≤ XYP , the conditional probability that Y is no more than ½, given that we know that X = ½ . 5. Parts (a) and (b) of this problem are not related. (a) A manufacturer of coils for portable heaters is interested estimating the mean life length of the coils. If the coils are tested under normal operating conditions then the time required for testing may be so long that the coils are obsolete before the testing is completed. Therefore an accelerated life test is used in which the product is tested under severe conditions so that life length is dramatically reduced. The information from this testing will eventually be used to predict average life at normal operating temperatures. The data below represent the average life (Y, in hours) for sets of coils tested at five different temperatures (X, in degrees Fahrenheit). Temperature (X) 500 600 700 800 900 Average Life (Y) 804 791 658 599 562 For these data ∑ =∑=∑==∑ ,2379706 ,3414 ,000,550,2 ,3500 22 iiii yyxx and ∑ = 200,322,2ii yx . Estimate the linear regression equation that models mean life as a function of temperature, and estimate the mean life when the temperature is 650 degrees. (b) Let ,...2 ,1 , =tXt , represent the signal received by a station at time t. In some situations it may be useful to look at the process ,...3 ,2 ,1 =−= − tXXY ttt , where tY is the difference between the signal at t and the previous signal. A large value of tY represents a large change in the signal received, and this may indicate some problem with transmission or reception. For the current problem, assume that X X1 2, , ... are independent continuous random variables with mean Xμ and variance 2 Xσ . Find the autocovariance function ),( 21 ttCY for the process Yt for the case in which 21 =t and 32 =t . STAT 4TI4 FINAL EXAM ANSWER SHEET May 10,2002 T a 2 [. The ebsarvedvons one paved So We MWe ee B=, Sy=4.709 B-o Wo- Hol Mp=O , Wii Mp #0 & stlishe = Ss ae = 3.302 bore 22305 wth T dh Reject Ho au corclude TWA Thar fs a differrme latwo, A wd B CHC) _ tet an 2 i 2, (a) Plow t) = ay = done 72183 ee) 4 (a E-S) = 2205 | (ey Kena of grade © vtems — P(K=0)= ae = Weer P(K=2)= Sats oton EC%) = O(.1604) + 16.2283) + 2C0109) = 0.15 YR) E(x)2 35,2025 PCT As REY CENCE Le, WSIL (c) PCwe As low c) = Plawe pee i Ed Ih) +2283 ~ i ae -1\e oli =V ye dG) PCRrr) = I> P(KE2) = le (et eres = 0803 Mow FL —nd =X G\ Lays Mesa = eM Reaves ny + BEAN WC = A 7 ree A =X, _s Fits = -ne te 20 ns Sy NEE 4. C) Fyn = 9) Ferg ldy = [Sy rteey, = FRY, cee y Ss EK) = wile = Soe CG ere S wddx = (3974 5) = Teas EK) = SS. 4 Fenaytey = S$, . C5 Gate ai Tee (wy PCr e AL Xe) = ch * Ee olAaydy Sey Git = ae = he AAG ds x ee Tee = § a re = ee sy Th s a Dx (pn estas ee. ~!~O~SsS J bh tage Eee a EEE “VSS0 900 ~ E (ase0)™ aa _ bo = Gr = CRY -G ume) (82) = FUSE Mey = IS@ —.6Te* Gh *= GSO Te tehyat of Th wen AY A MSG ~ 676 @S0)= 71b.G 2 Rye) Y= ged My MG ORVE EB OK = Ke) = ECOW- KI C%-%\) - 0 eS EO - “Xr - XiX3 4X, %.) = Ky > (Sy, +My )- wae a-8y  SpA Py FINAL EXAM ANSWER SKEET Maw @ , 209 | we Gn-1)st 2, myst (24) (198.2) a CHyCIe2y~ 4, @) oe Ee OE ete Ee SAT £08257 ~ lowly — race ? ? (e) destehishe = “apayyes = 73.338. For at af , Boos = 2.787 boo = 3.467 So .001 < p-value < 005 . This pevelae is very small so we reject Hy clove Tat reizeo anh cmelude teat &<1200, COCR feos 2, (a) Plone B ond oc) = machen = ae TNS (oR) 3! P(ABc) = OM a a * : ie iat Bod one c) _ — et ay Way Plow Bowe) = ROS B MESS = REY = ee HT CE BQ Gey Trak Hans ty oom ©) The preb. that ga the othe tuo we A mt 8 we CCC) = aa =: 37 } KLO 3. Gy Fe) = $2 ot de = Le i = oe xed ie Fe) = | ous 0424 | CH £ %2)\ (ry) BCX) = So ven de = (e+ J, = ar 6 ie 6 Vax ix) = ECX™) - Ce@y = CH Bd Carry (Ott _ 6 Tes on Se a © ECO) = BT. rf R=3a Ton (ny k= 8 ud 6 = '-X 4. @ Fal Aa Ee are (KY COCA + YGOTY = 105 Ce) = (0) (fe) + (1S(.06) = Teige . =, 08 &) ECKY) = LE vy Fg) = 07 ; Cov (x¥) = ECKr) -ECRE(Y) = OT ~ (096.08) c) Cy edb tr) = Cou(%a, Yen), Tf bx tes Ke, mk Ye, pectary « xy; . 13 p to ¢ FGeve} “ems which ore thditpesdint so Cov (Ki, Yn.) 20, TF hy thy ten Xe, oh Yer petals = 0628 ty We same VR so Te > the uelen gow com pred wy Cb), Covenraur % .0628 a7 qLB- ‘lo @) PCRSIHL ov K7INB) = PCBS Taye a ov BP tayge) = P(e<-3 or 273) = 0026 (uy PC Ker ov ¥ PITS Wer p= 12) LAUD -B-TLEL = PCa HEGRE w ar ae) = (a e-S ov B71) = 1587 A ces ARL = PGignay = 1seT = G.30 Stat 4714 Final Exam May 5, 2000 Name__________________________ 1. The thickness of the insulation on a component must be at least 1.2 mm to meet specifications. From past measurements of thickness it has been determined that the process applying the insulation produces a distribution of thickness that is normal with mean 1.4 mm and standard deviation 0.16 mm. (a) What proportion of the components from this process will meet the insulation specification? (b) Find the probability that at least one of the next 10 components from the process will not meet specifications. Assume that these 10 components are independent. (c) The proportion of components meeting the specification is not very high, and one approach to improving quality is to increase the mean thickness for the process. What value of the mean thickness would be required to have 99% of the parts meet the specification? (d) Increasing the mean thickness will increase costs due to the extra insulation required. Another approach to improving quality is to keep the mean constant at 1.4 mm, but try to modify the insulation process to reduce the variability. What value of the standard deviation of thickness would be required to have 99% of the parts meet the specification? 2. The number of defects in a circuit board has a Poisson distribution with density L,2 ,1 ,0 , ! )()( == − x x exf xλλ , and moment generating function )1()( −= te X etm λ , where λ > 0. (a) Let nX represent the numbers of defects in a random sample n circuit boards. Use moment generating functions to find the distribution of ∑= = n i iXY 1 . XX ,,, 21 K (b) Derive the method of moments estimator of λ. (c) If λ = 1 and n = 10 find the probability that each of the 10 boards has no more than one defect. 3. Suppose that you play a sequence of games and let X tt , , , . . . = 1 2 , represent the amount that you win in game t. Assume that X X1 2, , ... are independent continuous random variables with mean Xμ and variance 2Xσ . Let tXY ti i represent the average amount that you win in the first t games, ,...2 ,1 =t . t /1∑= = (a) Find the mean function )(tYμ for the process Y . t (b) Find the variance function )(2 tσ for the process Yt . Y (c) Is the process Y wide sense stationary? t (d) Find the autocorrelation function ),( 21 ttRY for the special case in which t1 = 2 and t2 = 4. 4. You win the lottery if your ticket has the same 6 numbers as the 6 balls that will be drawn at random without replacement from 49 balls numbered 1, 2, 3, …,49 (the order of the numbers does not matter). (a) Find the probability that you win if you have one ticket. (b) Find the probability that your ticket has at least 5 of the 6 numbers correct. (c) Given that at least 5 of the 6 numbers are correct, find the conditional probability that all 6 of the numbers are correct. 5. When complex devices such as TV sets are inspected, defects are sometimes classified by categories. Suppose that defects are classified by two broad categories, major defects and minor defects. Let X represent the number of major defects in an item and let Y represent the number of minor defects. From past records of inspections it has been determined that the joint density of X and Y is y=0 1 2 x=0 .69 .05 .03 1 .04 .07 .04 2 .01 .02 .05 (a) Determine whether X and Y are independent. Please justify your answer. (b) Find E(X). (c) Find the conditional expected value of X, given that Y = 1. | STAY 4114 FINAL BXAYy ANSWER SHEET Mei ® 2000 Lo-h4 Jc PCK2 LLY) = PCBS BR) = PCar tas) =. 8944 (Ly) Yee. wm lo vot merctray spec. P(Y21) = I-P(Ys0) = I= Ce \Gtose) Caquat” W2= = C)-4,0) = -B3Ue = Se > Med (O.le)(2.306) = 1872 aa ‘Leh Lael B)-3.0.7 29m = AE > c= TRL = 0800 Loe * (@) Myc = R x, (8) = R en v. ere ty > & Poisson | dyed ACeb=i) ACH1) (tb) oh = vee Ab K=O we get ECK)= he =) TSK eshwtes ECx\=) thy T= =I = Ce) We no, beads WWD with x41 PcKe\) =e s set = 1¢'=.7388 POwate) = CIS) PEN) PERS = C7388)" = ues ca) yy = ECE Ex) = ROEM) = ECKL) = My (ey oe Ce) = Va (Bs) = halk Va leay) = VaR = eT ces No, lecouse TEC) = Cy (aye) dpeds on %, ECE (Kath) F CK, KEK Ky) ) BECKY tk Kot Keka t hey + HK) + KE 4 Kes Key) zu a =B (2G) tome) = FE tomy (ay Ry lty4) = Eee) " ats Ss 4 out i _ (a) Keno, cowed PCK=6) = (49 Cy = ae 4448 4746'S HY = 13903816 43) 43 Use tl 1S4 G@) PCKZS) = POKES) + PKKEG) = ean S aXe a 13403Blb 131836l% 288 TeR) Fae PCK=S) from (4 ee eu Fk 1B. = [g903016 _ _ PCKse wk RES) _ Keo) _ a steibie aa. > Sites (AEE = PeKes) PERS) ~ 259 /j038\6 4 CA) FCOOd= 04» FCO CON= CITC TH) = S098 4.67 X,Y tot taperdech wy ECKy= Trteced =X.77) #CGis) + (23.08) = 0.34 OO COtwet) = SxhuyGetey = Z x HOY = YB) +0054) +A) Fay = 1657  ‘stat 4TI4 FINAL EXAM AWSWER SHEET Mou to 1499 I | x = Stl te a Ss 1@ ECxyet & Le. Ke eS, he galt 2 a ts -(2s is Yen. TK ove month , PCrz1\ = \- PCreo) = \-e ; PCrziy = (-e = 73S Ye ¥ (4) Khe, & 4 moulhs ) PCK>L) = I P(KER) = I Ze os a & -S 5% - PCK7r) = \- Ze Sy = I-.\247 = ,8753 G) discrete (LTE OH) = Von Xe % gay) = Vee (he) + Ven (Kee) = 25 2. Icey Ry Katt Ce ety = EC Xe Yer) = ECXe (Xa4, - X2)) = E (XaXaer) — ECKL L ' = ML ~ COM Y= 8x | B) _ besoee8 _ oe @) _ tcoreust 3, ce) PC visitle dafeets) = = azo =. 4B [oe] “(a) “Tan tet i a 49, 34 38 | MBE § Byx) Pro missle od wo vik og 3G : = = = = 6168 | 1/2) Pra te umeilhen) = Pre vie) “Sey 44GB (2) - 38 31, He SL @ ols )_ i ace OT! BD ee Roa sae ee Be) “CEY =.016 OB Ce 768 It 4G) The cbsmvations ave pared , The diffevtnus me 0.3) 1.15 Lf 57 9-E,03,0.5 04 -;. . Bro _oS-0 © Het Me =P one Ho: Mp0 Restetishe = Sn = ee = 2,842 Hy. Mg #Ma Myt MoO gt oce aus wilh © Af, Reyeck the Conclude thet teow pucctuve has an effeet on the Chmcebmiie i | i (uy) £,oeg 2 2447 Boro 3-43 =P 010% P(t-stehshe 2 %.B42) 07S | | H => 010 < prvelre 6.050 Pt ay (eee 344 Ca (SBE sy i903 d > 2087 © PETS i! ah SOG) 1h -BY _ Ca (34) (06) @) ¢ = oy JEP =m y= [eer |. se FN Port ae oe = 538.6, iI Luspect 939-50 = 464 cddvtiondl components, Ths answev used 9 =.34 a The vbur fo p. TS we ave cmsnuhia ou use pes we get CuaeyC-SCSS Cone = (00.25 | so \uspect GOl-soz SS\ addihrout components, STAT 4714 FINAL EXAM December 16, 1998 Name__________________ 1. Suppose that 5 cards are selected at random from a standard deck of 52 cards. (a) Find the probability that 4 kings are selected. (b) Find the conditional probability that 2 hearts are selected, given that 4 kings are selected. 2. Consider a digital communication line, over which the bits 0 or 1are transmitted. Let tX represent the value of bit t, and assume that L,, 21 X are independent with pX XP t == )1( and pXP t −== 1)0( for L,2,1=t . Let the process tY be the total number of 1’s received by time t, so that L,2,1 , 1 =∑= = tXY t i it . (a) Is the state space for tY discrete or continuous? (b) Find the mean function, )(tYμ , for the process tY . (c) Find ),( 21 ttRY for the case of 4,2 21 == tt . (d) Find ),( 21 ttCY for the case of 4,2 21 == tt . 3. A new algorithm has been developed for solving a particular type of problem. A set of 8 problems of this type has been collected for testing the new algorithm, and it is believed that this set of 8 problems can be treated as a random sample of problems of this type. One variable of interest in testing the algorithm is the time required to obtain a solution. The number of minutes required on a particular system was recorded for the eight problems and the results were 7.6, 13.0, 9.4, 11.2, 6.1, 7.2, 8.3, 5.2. (a) Find a 95% confidence interval for μ, the mean time required for this type of problem. (b) Find a 95% confidence interval for σ, the standard deviation of the time required for this type of problem. (c) The developers would like to claim that the mean time is significantly below 10 minutes, so they decide to test H0: μ = 10 against H1: μ < 10. Find the p-value for this test. Based on this p-value, would you conclude that they have strong evidence that the mean is below 10? 4. Defects can occur at many places on a circuit board, and it is believed that the number of defects, X, on a board is a Poisson random variable with density L,2,1,0 , ! )( == − x x exf xλλ and moment generating function )1()( −= te X etm λ . (a) Use the moment generating function to derive the mean and variance of X. (b) If nX represent the numbers of defects for a random sample of n boards, drive an expression for the maximum likelihood estimator of λ. XX ,,, 21 L (c) If the actual value of λ is λ = 2.0 per board, find the probability that there will be at least one defect in each of the next 10 boards inspected. 5. The life length in hours, X, of a packaged magnetic disk exposed to corrosive gases has a Weibull distribution with density 0 ,)( 1 >= −− xexxf x βαβαβ , where α = 0.06 and β = 0.5. (a) Find P(X > 1000), the probability that a disk will last more than 1000 hours. (b) Find the expected life of a disk. (c) Suppose that three of these disks are used together in a system, and they operate in a parallel configuration. Assuming that the life lengths of the three disks are independent, find the reliability of the system at 1000 hours. 4. Defects can occur at many places on a circuit board, and it is believed that the number of defects on a board is a Poisson random variable. A manufacturing process for a particular circuit board is currently experiencing a mean rate of 2.0 defects per board. (a) If a board is inspected and found to contain more than 6 defects, then this is taken as an indication that the rate of defects may have increased for the manufacturing process. If the mean rate of defects is actually 2.0 per board, find the probability that a board will have more than 6 defects. (b) If the mean rate of defects is actually 2.0 per board, find the probability that, in the next 20 boards inspected, at least one board in will have more than 6 defects. (c) If the mean rate of defects is actually 2.0 per board, find the expected number of boards in the next 20 which will have more than 6 defects. 5. Let X represent the proportion of time in a week that an engineer spends on a particular project, and let Y represent the proportion of time that a second engineer spends on the same project. Suppose that X and Y are continuous random variables with joint density 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. f x y x y( , ) ,= + (a) Find P(X < 0.5), the probability that the first engineer spends less than half of the week on the project. (b) Given that the second engineer spends exactly one third of the week on the project, find the conditional probability that the first engineer spends less than half of the week on the project. (c) Given that the second engineer spends exactly one third of the week on the project, find the conditional expected proportion of the week that the first engineer spends on the project. 6. To estimate the mean resistance μ for a shipment of resistors, a sample of n = 25 of the resistors was selected and each resistor in the sample was tested. (a) If the population standard deviation is σ = 0.8 ohms, find the probability that the sample mean X will be within 0.2 ohms of μ.. (b) The mean resistance should be 10 ohms. The decision rule being used is to reject the null hypothesis that μ = 10 if X < 9.7 or if X > 10.3. If σ = 0.8 ohms, find the probability of a type I error. (c) If the actual value of μ is 10.5 ohms, find the probability of a type II error. Assume that σ = 0.8 ohms. 6. | STAT GT FINAL EXAM ANswER SHEET Man G , 998 tl i NN G@) The cbseurhone ane pave, Ho! My=Mr > Ho! My =o , Hi Ma #0 B-0 _ -6,125-0 woe, =e OE ee ee es ete CST | Cails3.43 vg CTNGIRRD bate Oe ee Ne ES SS ET (G8) _ GYR) a 1. CXP) | a @) “C®)- = Goo Gl (Rag) = 4a = 5 (me) Hoe PCprige avd mat Fish) CoY\CTYAB) (cH) || (0) Pomme |net 8) = SEG ty = 1 ew = 3H i el eka Nee) MC = s - ‘Hee G \( 2S) = O\/ 350 0 Sag sae cay @) Xe mh Kay oor contour so Yes Xe—Karr te cobisuons [| Ce) Me = EC) = E (Xe Xen) = EK) -E (Ker) = AW =O [oy Ry Ce BN) = EYE en) = E(Xe-Kenr Xue Ke)) i = ECke Kerr) — (Ser Ker) — BCE) + B(Ke Xe) =H Cop) yi =-¢ ° wo Cv) Yew, in 20 unl X76, PCY21) = I-PCvs0) = I- (7 )Go0s) CAS) = 0954 cc) ECV) =p = 20008) = 0.1 } | G) PCK 76) = 1-PLKEG) 21-498 =,005 Fron the Poisson table | la) Feo) 2 Sy (xt) by = wet oexet, PCxee) = qh Cet dn = 2375 {| PEELS = SFS9 Fon adegdx = 978 1 a. aaa | Fielad = 8, Crd = are, cf YEN. Fp lelad= “Gre 08 mEI rCeeb |e) = Se Faglelt) dx = Sle Ege dn = 95 ©) ECK[ Yeh) = Sl eFC de = 6! x “re dx = 6 lo Peeve X ptr) = eS e2é we) = Clas 6 2ELzS) = 7868 Ie d= POX S477 ov K > 103 | Mere) = PG <-161S ov Br l.BTS) = 0608 io p= PLATS K £ 10.3{ patos) = eC S108 ¢ 2¢ OSS) - est2é-|25) | a = 1056