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# SPTP HPC EARTH SCIENCE & PETE PETE 689

Texas A&M

GPA 3.72

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This 52 page Class Notes was uploaded by Lenore Medhurst on Wednesday October 21, 2015. The Class Notes belongs to PETE 689 at Texas A&M University taught by George Voneiff in Fall. Since its upload, it has received 24 views. For similar materials see /class/225884/pete-689-texas-a-m-university in Petroleum Engineering at Texas A&M University.

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Date Created: 10/21/15

TrxAs AM I39EtlllllEIIM ENGINEERING m SIIIIIIIIII Wins in IINIIIINIIIINEIIINIII Resume Reserves l tlll39 2 llasurintive Statistics llaserintive Statistics Part C Graphical Presentation of Data Fi i i h ENGINEERING l l39 i hj tiVBS 0 Construct histograms frequency curves and ogives cumulative and decumulative for data sets 0 Fit a normal distribution equation to an appropriate data set 0 Determine quartiles deciles and percentiles for grouped and ungrouped data sets frequency Ilistrihutiun IIiSt l lTl o Presents distribution of frequencies of values of variables Sometimes used for ungrouped data usually discrete variables More commonly used for grouped data either discrete or continuous variables 0 Absolute frequency distribution shows actual number of data elements in each class 0 Relative frequency distribution shows proportion of data items in each class Kinds IJf llistrihutiuns 0 Cumulative include number or proportion of data elements in and below each class 0 Decumulative include number or proportion of data elements in and above each class frequency lilll39VB 0 Frequency polygon in which data points are fitted to smooth curve 0 To t normal curve to histogram Number of observations 7 1 if drawn on Class interval used proportional frequency t0 draw histogram 2 XiX E T f x e s27r quotDive 0 Presents cumulative frequencies relative or absolute vs class boundaries Cumulative frequencies equal to or less than ogive has frequencies plotted at upper boundary of each class Decumulative frequencies equal to or greater than ogive has frequencies plotted at lower boundary of each class I39llJt Ilit Records Construct histogram and ogives both kinds Class Cum Mark Freq Freq Cum Rel Decum Class CM f CF Freq Rel Freq 5269 605 2 2 220010 2020100 7087 785 3 325 520025 1820090 88105 965 6 6511 1120055 1520075 106123 1145 6 61117 1720085 920045 124141 1325 3 31720 2020000 320015 20 I39l t Illt Records Construct histogram and ogives both kinds Histogram frequency curve 75 60 Freq Number 4395 of Bits 3n 157 00 52 69 87 105 123 141 Footage Drilled by Bits I39IIJt llit Records Construct histogram and ogives both kinds Ogives cumulative 100 80 Freq N um ber 60 of Bits Footage Drilled by Bits I39l t Illt Records Construct histogram and ogives both kinds Ogives decumulative 100 80 Freq Number 60 of Bits 20 0 Footage Drilled by Bits I39IIJt Ilit Records 0 Fit a normal curve to the histogram x p705gj e 36 ep ZOCI 542 40 35567 00285 01702 50 24805 00837 04992 60 15978 02024 12070 140 14972 02238 13347 150 23549 00949 05661 I39l t Ilit Records 0 Fit a normal curve to the histogram 7 6 Freq 5 Number 4 of Bits 3 2 1 0 40 60 80 100 140 160 Footage Drilled by Bits lluartiles o Quartiles symbol 0 divide data set into four equal parts Interquartile l lll tl lluartiles Analogous to a in equation for 39 computational median 539 Analogous to 7 Lower limits of classes in equation for containing quartiles Computational median th l S ll lVlSl S IJf llata Sets 0 Deciles symbol D divide data set into 10 equal parts 0 Percentiles symbol P divide data set into 100 equal parts Median second quartile 02 fifth decile 05 and 50th percentile P50 are identical I39usitiun Illlli t l f l ll til Fractile indicator Magnitude 1107 I p D I IS gt vvhic li gldtile is Of des39red deSired39 fractile 39 7 read from ungrouped data Maximum number of divisions Iletermine f t Ill lll ll 3t fractiles Use a r 39 Bit number Ft Drilled drilling records 1 2 69 Determine footage 3 n drilled for P50 0 Q and D5 20 139 FP 0201105 10 11 value 102 11th value 105 P50 1021052 1035 n t l39ll39llll ll f t Ill lll ll 3t fractiles Use a r 39 Bit number Ft Drilled drilling records 1 2 69 Determine footage 3 n drilled for P50 0 Q and D5 20 139 FD7201147 14Ih value 110 15 11 value 115 D7 110071151101135 n t l39ll39llll ll f t Ill lll ll 3t fractiles Use r 39 Bit number Ft Drilled drilling records 2 69 Determine footage 3 n drilled for P50 0 Q and D5 20 139 FQ32011575 15th value 115 16 11 value 116 Q3 11507511611511575 n t l39ll39llll ll f t Ill lll ll 3t fractiles Use a r 39 Bit number Ft Drilled drilling records 1 2 69 Determine footage 3 n drilled for P50 0 Q and D5 20 139 FD5201105 Since 05 P50 05 1035 f l nt f Peake ness 34 o Dimensionless value fourth central moment 4 4 s 94 fiXi X4 f l t IJf SKBWHBSS a third centralmoment m3 3 33 s3 lX if 1 n S3 When a3 0 curve is symmetric or bell shaped When 33 lt 0 curve is skewed to right When 33 gt 0 curve is skewed to left What WE VE l l li 0 Construct histograms frequency curves and ogives cumulative and decumulative for data sets 0 Fit a normal distribution equation to an appropriate data set 0 Determine quartiles deciles and percentiles for grouped and ungrouped data sets TrxAs MM llaserintive statistics End Part C i i39 iai i lih llaserintive statistics Part D Spreadsheet Applications Fimf iii i39 in ENGINEERING l l i hj tiVB 0 Use commercial spreadsheets to perform statistical analysis Ilsinu SDI BEIISHBBIS EXEBI 0 May need to install plugins 0 May enter data in random order No need to sort May need to enter data in single column or row 0 Excel uses kurtosis for peakedness fiml Means Median Mllll 0 Enter data in single column 0 Use Tools gt Data Analysis gt Descriptive Analysis Enter range of data Click Grouped by columns Click Summary Statistics Statistics fl lll EXBEI u unkz mm A i a i E9 9 72 123 39 B an EB 139 Ouline Collaboraunn lEIB 125 Formula waiting y E E Innis on the Wag i iEIZ 12 53 Macro 13 7E 1 m2 Add mg i an guszwmze i i is 11 m5 9mm WE Data mam 3 MS V 2D 1 lEI Statistics fl lll EXBEI Data Analy Anew w nnnva S ng e Factnv Anuva TwoFaclnv wmh Raphcauun Armva Wham Wxthaut Revhca an Cuns almn Cance LLHE Heb xpunen we MW mg F425 Ywmsamv e Fm Va antes Fwnev Ana vms Hxstagvam Statistics fl lll EXBEI 2 7 x n a g Kunkz DESGIDIIVE Statistics I Inm r n WWW El Cancel mm m J new Labels In rst raw gt New workbook lt l7 ummavy statwstics r Cngfidenre Leve for Mean 95 Kth Lygest 1 r Kth Smanest I Statistics fl lll EXBEI annkz A E 17 E9 A 72 i 123 Column I L 135 Mean mu 7 j 139 Standard Errnr 5 was i we MadIan 1035 ET 1 Made 102 7 m2 Standard Daviahan 2274 Tz 53 gamma Vanancz 517 2 173 13g Ku usss 43 35 E an SkewnesS Va 25 15 ME Renee 3395 1 m5 Mwmum 53 Maxwmum 139 7 Sum 2m if D Cuunt 20 Data Analysis 39 Ana vsls rang Swnqla nan Anwa Ywanatlm mh REdltatmn Armva TwuFaclnv mmuuwephmnn Vnn 2 F Yes warSamwe fur Vaneates Ilistnuram From EXBBI Input input Range I39 Lahe s mm a na Qutvut Range F New mkhaak gg l7 Cumu atwe Pevcentaq l7 Quay Output Range m m as2 sass M2 72 59 BM uem Cumua We 123 7 ES 2 ID EID 25 3 E B 39 More 3 quuency m a 0 ME ea 87 wuswzamana ms Bin What WE VE l l llescrintive Statistics End Part D TI39XM AS M BBSquot Pl llllilllililv Illlll39v Part A Classical and Modern Probability l l llill lljll tiVBS Calculate the probability of equally likely collectively exhaustive mutually exclusive events State the basis for empirically determined probabilities State and explain simple conditional joint and marginal probabilities l SSi l Alllll39 h 0 Approach based on a prior or abstract reasoning 0 Probability of occurrence of event expressed as P4 U avorable cases Pi A Ci l SSi l Alllll39 h o Probabiity of event not happening 132 where A is complement ofA n m PA Same concept stated as law of proportion or law of chance l SSi l Alllll39 h Whenever a particular action can have more than one outcome then if all possible outcomes have an equal chance of occurring the probability of any one of them occurring in a single test is the proportion that a particular outcome bears to all possible outcomes l lll t Probability IJf iVB llt lTlB Box contains 56 black balls 20 red 24 green Balls drawn one at 56 0 a time returned toBlack 056 or 56 A 100 box 20 Calculate chance of Red 020 or 20 drawing specific 100 color Green i 024 or 24 100 l SSl l Alllll39 ll ASSlllTlDtl S Events are equally likely collectively exhaustive and mutually exclusive Equally likely no outcome more or less likely than any other Collectively exhaustive sum of all favorable and unfavorable outcomes equals total number of outcomes Mutually exclusive different outcomes cannot occur simultaneously in a single event llll f Hime 3 l Pack of 52 cards Cards drawn replaced in pack Calculate chance of drawing Queen i i i 00769 or 769 Aqueen 52713 I A lb Club 13 1 025 or 250 0 cu 52 4 7 527137 3 39 N0 club NoClub Z075 or 75 l lll t llll f Hime 3 Ball Urn contains 12 white 10 red 5 green balls Calculate chance of drawing White 1227 04444 or 4444 REd 1027 03704 or 3704 Green 527 01852 or 1852 Not red 27 1027 06296 or 6296 ElTlllll l l Alllll39 ll 0 Approach based on experiments and observations Large number of trials required to establish chance of an event Probability of event represents proportion of times under identical conditions event can be expected to occur ElTlllll l l Alllll39 ll ASSlllTlDtl S Experiments or observations are random No bias in favor of any outcome so all elements have the same chance at selection Large number of observations ElTlllll l l Alllll39 ll Ilefinitiun If experiment is repeated a large number of times under essentially identical conditions then the limiting value of the ratio of the number of times the Event A happens to the total number of trials of the experiment as the number of trials increases indefinitely is called the probability of the occurrence of A PA 11m ngt 0 n Ilefinitiuns quotSBIl lll MIJIlBl Il Approaches Random trial or event An action event or operation that can produce any result or outcome Elementary event Each of the possible results in a single trial or experiment Sample space A set representing all possible outcomes from an experiment Event set Subset of the sample space Properties l AXlIJITIS A number between 0 and 1 called probability of that event is associated with each event set Sum of probabilities of all simple events or sample points constituting the sample space is equal to one Probability of compound event is sum of probabilities of simple events comprising the compound event MIJI B Ilefinitiuns 0 Simple probability Probability of occurrence that is independent of the occurrence of another event unconditional denoted by P64 or PB where A and B are events 0 Conditional probability Probability of occurrence that depends on another event denoted by PAB read as probability of A given that B has occurred MIJI B Ilefinitiuns 0 Joint probability Probability of more than one event occurring simultaneously or in succession denoted by P643 interpreted as the probability of both A and B o Marginal probability Sum ofjoint probabilities What WE VE Accomplished Calculate the probability of equally likely collectively exhaustive mutually exclusive events State the basis for empirically determined probabilities State and explain simple conditional joint and marginal probabilities BBSquot Pl llllilllililv Illlll39v End PartA TTXM AampM I i i i i h ENGINEERING BBSquot Pl llllilllililv Illlll39v Part B Operations With Probabilities l l llill lljll tiVIls 0 State and explain the following rules of probability complementation addition multiplication ueratiuns Ill Event Sets 0 Union of two events Event set that consists of all the outcomes sample points that belong to either A or Bor both denoted by AUB 0 Intersection of tWO events Event set that consists of all the outcomes sample points the two event sets A and B have in common denoted by A08 Complement of an event set Set of all sample points in sample space Snot contained in A denoted by A quot88 IN Venn Iliaurams o Illustrate characteristics of events 0 Mutually exclusive events Two or more events that cannot occur together Dry Hole 065 30 quot88 IN Venn Iliaur39ams o Partially overlapping events Part of one event and part of another event can occur together n tI t Events Occurrence of independent events in no way affects the occurrence of the others Coin toss always has same two outcomes Dependent events are affected by previous trials Removing a card from a deck changes number of cards remaining outcomes unless card is replaced before next draw 31 llules IJf Probability o Complementation rule Since an event either occurs or does not occur and since the sum of collectively exhaustive events must equal one then WP Marginal probability of no Event PA71 PA Additiun Ilule o If events are mutually exclusive the probability of outcomes can be calculated by adding the probabilities of the possible ways the outcome can occur PA orBPAPB 32 Additiun Ilule For N mutually exclusive events PA or B orC orN PAPBPCPN For events that are not mutually exclusive PA orB PAUB PABPA B Multiplication Rule 0 For two independent events PAB PA andB PA x PB o For n independent events PABCN PA and B and C and N PAgtltPBgtltPCgtlt gtltPN 33 Multiplication Ilule o For dependent events when the probability of A occurring given that B has occurred is different from the marginal probability of A BBSBI VIJiI Example 50 wells drilled in area with blanket sands Zone A has 8 productive wells Zone B has 11 productive wells 4 productive wells intersect both zones Determine productive wells in Zone A but not B Zone B but not A Either A or B Determine number of wells discovered m Determine number of dry holes 34 BBSBI V iI Example S 50 o quotEwells drilled in area with blanket sands nA8 nB 11 nAnB4 BBSBI VIJil Example 0 quotEwells drilled in area with blanket sands nA8 nB 11 nAnB4 0 Determine productive wells in Zone A but not B 39 ZoneB but notA A B B A B11 47 39 E iltherQorB nlt n m 39quotAUBquotAquotBWAHB811 415 35 BBSBI V iI Example 0 els drilled in area with blanket sands quot25 59214 8 nB 11 nAnB 4 0 Determine number of wells discovered nA nZ Bn1HB474 15 0 Determine number of dry holes nS nAUB 50 1535 l lll t Probability IJf t l i ti o 100 random samples of crude oil 45 contain sulfur 40 contain mercaptans 35 contain both 0 Calculate probability that sample contains Either sulfur or mercaptans Sulfur only V Neither sulfur nor mercaptans 36 l lll t Probability IJf t l i ti o 100 random samples of crude oil nC 100 nS 45 nM 40 nS M 35 0 Calculate probability that sample contains Either sulfur or mercaptans SuM nSi nM nS WM 100 45 40 35 100 05 or 50 l lll t Probability IJf t l i ti o 100 random samples of crude oil nC 100 nS 45 nM 40 nS M 35 0 Calculate probability that sample contains Sulfur only nS nS M S M p U 100 0lor10 100 37 l lll t Probability IJf t l i ti o 100 random samples of crude oil nC 100 nS 45 nM 40 nS M 35 0 Calculate probability that sample contains Neither sulfur nor mercaptans n C n SUM AS QM 100 100 50 100 05 or 50 Probability f Hime 3 ll llllt l o 17 counters in bag numbered 1 through 17 drawn individually and replaced 0 Calculate probability First number even second odd First number odd second even 0 Calculate effect of each if counter is not replaced into bag 38 Probability f Hime 3 ll llllt l o 17 counters in bag drawn individually and replaced 0 Calculate probability First number even second odd PA and BPAgtltPB gtlti 02491 or 2491 17 17 Probability f Hime 3 ll llllt l o 17 counters in bag drawn individually and replaced 0 Calculate probability First number odd second even PB and APBgtltPA 9 8 gtlt 02491 or 2491 17 17 39 Probability f Hime 3 ll llllt l o 17 counters in bag drawn individually and replaced 0 Calculate effect of each if counter is not replaced into bag PA and B PAx PBA 8 9 gtlt 02647 or 2647 17 16 llrillinu SUBBBSSiVB IliS VBl iBS 0 Ten prospective leases acquired All have equal chances of success Seismic shows 3 will be commercial One well to be drilled in each lease 0 Calculate probability of drilling first two wells as successive discoveries 4o llrillinu SUBBBSSiVB IliS VBl iBS 0 Ten prospective leases acquired 0 Calculate probability of drilling rst two wells as successive discoveries 3 2 gtlt 10 9 10 6 1 00667 or 667 90 15 PVKW2PVKxPWZIVV1 What WE39VE Accomplished 0 State and explain the following rules of probability complementation addition multiplication 41 BBSquot Pl llllilllililv Illlll39v End Part B BBSquot Pl llllilllililv Illlll39v Part C Tables Trees and es Theorem WAS ASMY i i i h ENGINEERING 42 l l39 i hj tiVBS Explain how a probability table is constructed and applied Explain how a probability tree is constructed and applied State and explain Bayes theorem or rule Probability Iahle 0 Alternative to Venn diagram for solving problems like previous Zone B Yes N0 otal Pivot Column 100 probability 43 Probability Iahle 0 Alternative to Venn diagram for solving problems like previous Zone B Yes No otal 4 39 Marginal Probabilities Pivot Column Probability Irees o Diagrams representing sequences of lines depicting probabilistic events Branch to all possible sequences that can occur in any situation 0 Pictorial presentation of conditional probabilities 0 Show all possible branches for each event 44 Iyuical Probability ll BB Producer Dry Hole Probability IPBBS Each branch is labeled with event taking place at branch and corresponding probability of occurrence given the prior sequence of events required to reach that point in the tree Probabilities on first branch of tree are simple or unconditional probabilities All probabilities after first branch are conditional probabilities because they assume branches to the left model events that have already taken place 45 Probability Irees 0 At sources of branches called nodes branches must be mutually exclusive and collectively exhaustive Probabilities at each node must sum to one When probabilities from origin of tree to any terminal point are multiplied joint probability of that particular sequence of events is obtained Iypical Probability TI BB Reserves Status 0 310 x 23 210 30 MMSTB Res 210 x 30 6 d 0 23 Pm Car 310 0 310 x 13 110 60 MMSTB Res 110 x 60 6 13 Dry Hoe 710 Simple Conditional Joint 46 Bayes IllBIJI BITI IJI lll 0 Used to revise probability of an earlier event given later event 0 Used when solving problems concerning value of additional information that will result in revised probabilities Bayes IllBIJI BITI IJI lll Product of Prior or source branch robabilities probabilities PB 5 leading to B k E through A i1 probabilities leading to B ZPlBlAvXP4W 47 Probability f III39EWili Pencils First box contains 3 green 2 red pencils Second box contains 1 green 3 red One die is rolled 1 or 2 a draw from first box 3 to 6 a draw from second box Pencil drawn is green Calculate probability of drawing from each box Probability f III39EWili Pencils 0 First box contains 3 green 2 red pencHs 0 Second box contains 1 green 3 red 48 Probability f Ili39 Wili Pencils First box contains 3 green 2 red pencils Second box contains 1 green 3 red 1 PAUBPAPB m E 3 MIN E Probability f Ili39 Wili Pencils First box contains 3 green 2 red pencils Second box contains 1 green 3 red P PGIBXP31 513 B IG 2 P 3 1 1 2 C43 3 EIEHZJQ E054550r5455 49 branches quotSl Probability ll39BB f l lll t Construct tree with branches representing all possible events write prior probabilities on Attach new branches to represent new information obtained or to be obtained Multiply prior probabilities by conditional probabilities Sum joint probabilities Divide each joint probability by sum ofjoint probabilities to obtain posterior probabilities quotSl Probability ll39BB f l lll t Prior Conditional Joint Posterior Event Prob Probability Probability Probability 1 2 3 4 2 x 3 5 424 A PA PgtltA PA X PgtltA PA X PgtltAZ4 B PB PgtltB PB X PgtltB PB X PgtltBZ4 N PN PgtltN W X PgtltN W X PgtltNE4 22 10 24 25 10 50 quotSl l hl f l lTl t Input all possible events in Column 1 Assess prior probability of each event and input in Column 2 total must equal 10 Input into Column 3 Calculate joint probabilities by multiplying Columns 1 and 3 sum in Column 4 Divide each joint probability by sum ofjoint probabilities to obtain posterior probabilities l l39lllll hj thBS Calculate the probability of equally likely collectively exhaustive mutually exclusive events State the basis for empirically determined probabilities State and explain simple conditional joint an marginal probabilities State and explain the following rules of probability complementation addition multiplication Explain how a probability table is constructed and applied Explain how a probability tree is constructed and applied State and explain Bayes theorem or rule 51 What WE VE Accomplished Explain how a probability table is constructed and applied Explain how a probability tree is constructed and applied State and explain Bayes theorem or rule BBSquot Pl llllilllililv Illlll39v End Part C TTXM AampM I i i ii i iih ENGINEERING 52

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