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Probability Theory Theory and Terminology Addition & Multiplication Rules The Binomial Distribution The Normal Distribution Testing for Normality 2 KEY CONCEPTS ***** Probability Theory Definition of probability Theoretical v relative frequency probability Statistical experiment Sample space A complement Probability of an event Relative frequency Mutually exclusive events Conditional probability Independent events Addition rule of probability Multiplication rule of probability Testing the independence of events Comparing observed and expected frequencies and probabilities Definition of a probability distribution Examples of probability distributions: Binomial distribution Normal distribution t distribution F distribution Chi-square distribution Poisson distribution The binomial distribution and its expansion The binomial distribution with equal and unequal probabilities Estimating binomial probabilities Pascal’s Triangle Blaise Pascal (1623-1662) Difference between the binomial and normal distribution Characteristics of the normal distribution Finding areas under the normal curve Standard score (Z) Determining if a variable is normally distributed: Pseudo standard deviation Histogram with normal curve over-lay Normal probability plot Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 3 Lecture Outline What is probability & the meaning of probability terminology Addition and multiplication rules of probability Using probability to test the independence of events Probability distributions The binomial distribution The normal distribution Testing variables for normality Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 4 Theory of Probability The probability of an event is expressed on a scale from 0.0 to 1.0 0.0 means it will never happen 1.0 means that it is certain to happen The probability of an event is the number of times that a specific event occurs relative to the sum of all possible events that can occur. Example The probability of rolling a 3 on a die is 1 out of 6, or 0.1667 Example The probability of two coins flipped once coming up heads is 1 out of 4, or 0.25 Theoretical probability Sometimes we know from the theory of the matter what the probability of an event is, e.g. rolling dice or flipping coins. Relative frequency probability In other cases we can only estimate the probability by observing how frequently a particular event occurs, e.g. jury acquittals, people quitting a job Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 5 Terminology Statistical experiment An empirical record of any phenomenon whose relative frequency of occurrence is uncertain. Example Data on sentencing outcomes: probability of acquittal, fine, deferred adjudication, probation, or prison Sample space All possible outcomes, Example Possible sentencing outcomes: acquittal, fine, deferred adjudication, probation, or prison An event A specific outcome or collection of outcomes; e.g. an acquittal A complement All possible events other than the one in question, e.g. if the event in question is acquittal, then the complements are fine, deferred adjudication, probation, and prison Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 6 Terminology (cont.) Probability of an event The proportion of times an event occurs divided by the frequency of all other events that can occur, e.g. 246 acquittals out of 14,573 cases acquitted, fined, deferred, probated, or sent to prison (246/14,573 = p = 0.0169) Relative frequency How often an event occurs relative to all other events that occurred in the experiment Mutually exclusive events Two or more events which can not happen together, .e.g. acquitted and sent to prison, the probability = 0.0 Conditional probability The probability of event A happening, given that event B has already occurred, e.g. probability of going to prison (A) given that the offender was put on probation (B). This is symbolized P(A B) Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 7 Terminology (cont.) Independent events Two events A & B are considered independent if the conditional probability P(A B) = P(A), e.g. probability of acquittal (A) given that it is raining outside (B) Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 8 Addition Rule of Probability Q What is the probability of A or B happening? If the two events are not mutually exclusive … P(A or B) = P(A) + P(B) – P(A and B) Example What is the probability of drawing either a Jack or a Heart from a deck of cards? P(J or H) = P(J) + P(H) – P(J and H) P(J or H) = P(4/52) + P(13/52) – P(1/52) P(J or H) = (0.0769 + 0.2453) – (0.01923) P(J or H) = 0.3077 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 9 Addition rule (con’d) If the two events are mutually exclusive … P(A or B) = P(A) + P(B) Example What is the probability of drawing a Jack or a King? P(J or H) = P(J) + P(H) P(J or H) = P(4/52) + P(4/52) P(J or H) = (0.0769 + 0.0769) = 0.1538 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 10 Multiplication Rule of Probability Q What is the probability of A and B happening together? The general rule P(B) P(AB) If the events are independent of each other, this simplifies to … P(A) P(B) Example What is the probability of drawing a Jack of Hearts? Since J and H are not mutually exclusive, therefore independent … P(J and H) = P(J) P(H) = (4/52) (13/52) = 0.01923 Notice that this is the same as (1/52) = 0.01923 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 11 Testing the Independence of Events These rules of probability allow us to test whether two events are in fact independent of each other Example Is success on probation (S) related to drug addiction (A)? Put the question the other way around … Is success independent of addiction? Cross-Tabulation of Probation Outcome & Drug Addiction (N=643) Not Outcome Addicted Addicted Total Success 115 277 392 Failure 194 57 251 Total 309 334 643 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 12 Testing independence (cont.) If success is independent of addiction, that is the variables are not related to each other… How many of the 643 probationers should have been successful in spite of their addiction? By the multiplication rule for independent events P(S) P(A) There are 392-successes (S) and 309-addicted (A) out of a total of 643 probationers … P(S) P(A) = (392/643) (309/643) = 0.2930 Interpretation Theoretically, 0.2930 proportion of the probationers (188 cases) should have been successful and addicted if these variables are independent In fact, 0.1788 (115/643) were successful and addicted (115 cases) Does the difference between these two proportions (0.2930 & 0.1788) suggest that success and addiction are not independent, i.e. they are related to each another? Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 13 Testing independence (cont.) Calculation of the probabilities & frequencies for the other cells under the assumption of independence. Successful / not addicted (392/643) (334/643) = 0.3167 (0.3167) (643) = 204 cases Failure / addicted (251/643) (309/643) = 0.1876 (0.1876) (643) = 121 cases Failure / not addicted (251/643) (334/643) = 0.2028 (0.2028) (643) = 130 cases These figures are the expected probabilities & frequencies, assuming independence … What we would expect to find in a sample of 643 probationers if success is unrelated to addiction (i.e. independent of). Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 14 Putting the Observed and Expected Frequencies Together Observed Results Not Outcome Addicted Addicted Total Success 115 277 392 (0.1789) (0.4308) (0.6096) Failure 194 57 251 (0.3017) (0.0886) (0.3904) Total 309 334 643 (0.4806) (0.5194) 1.0000) Expected Results Not Outcome Addicted Addicted Total Success 188 204 392 (0.2930) (0.3167) (0.6096) Failure 121 130 251 (0.1876) (0.2028) (0.3904) Total 309 334 643 (0.4806) (0.5194) 1.0000) Do the differences between the observed and expected results indicate that addiction is related to success? Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 15 An Algebraic Shortcut Recall how the expected cell frequencies were calculated, e.g. successful / not addicted Successful / not addicted (392 / 643) (334 / 643) = 0.3167 (0.3167) (643) = 204 cases This is algebraically the same as [(392 / 643) (334 / 643)] (643) = 204 cases By cancellation of the common term in the numerator and the denominator [(392 / 643) (334 / 643)] (643) = 204 cases The equation simplifies to [(392) (334 )/ 643] = 204 cases In short, to find the expected frequency in any cell … [(row total) (column total)] / (grand total) Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 16 Probability Distributions Probability distribution A theoretical model that indicates the probability of specific events happening for a phenomenon distributed in a particular manner. In statistics, numerous probability distributions are used to describe, explain, predict, and assist in decision making. Examples include Binomial distribution Normal distribution t distribution F distribution Chi-square distribution Poisson distribution Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 17 Binomial Distribution Consider flipping 3 coins once. What are the possible theoretical outcomes? 3 heads 3 combinations of 2 heads and 1 tail 3 combinations of 1 head and 2 tails 3 tails Let the probability of a head = p, a tail = q, n = the number of coins flipped, and the probability of a head or tail = 0.5 (p=q=0.5) This statistical experiment can be represented by the following binomial model (p + q) n (p + q)3 = (p + q) (p + q) (p + q) (p + q)3 = 1p3 + 3p2q + 3pq2 + 1q3 (p + q)3 = p3 + 3p2q + 3pq2 + q3 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 18 Interpretation of the Binomial Expansion (p + q)3 (p + q)3 = p3 + 3p2q + 3pq2 + q3 p3 = 1 way to get 3 heads 3p2q = 3 ways to get 2 heads and 1 tail 3pq2 = 3 ways to get 1 head and 2 tails q3 = 1 way to get 3 tails Combinations of outcomes = (1+3+3+1) = 8 Calculating probabilities The probability of 2 heads and 1 tail 3p2q = 3(0.5)2(0.5) = 0.375 This is the same as the coefficient 3 divided by the total number of combinations of outcomes; i.e. 8, (3 / 8) = 0.375 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 19 Interpretation (cont.) Calculating probabilities The probability of 3 heads p3 = (0.5)3 = 0.125 This is the same as the coefficient 1 divided by the total number of combinations of outcomes; i.e. 8, (1 / 8) = 0.125 Graph of the binomial distribution (p + q)3 Probability 0.375 0.250 0.125 0.0 p3 3p2q 3pq2 q3 Combinations of possible outcomes Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 20 Applications of the Binomial (p + q)3 This binomial probability distribution can be applied to any binary phenomenon in which there are three events (n = 3) Examples Pre-trial settlements (p) v trials (q) among 3 civil suits Combinations of sons (p) and daughters (q) in a family of three children Clearances (p) and failures to clear (q) among 3 criminal cases Escapes (p) and no escapes (q) among 3 trustees Probations (p) and incarcerations (q) among 3 criminal cases Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 21 Binomial with Unequal Probabilities In the binomial (p + q)3, the probability of p and q may be equal or unequal p = q = 0.5 or pq Example with unequal probabilities … Settlements (p = 0.7) and trials (q = 0.3) among 3 civil suits Calculation of the probabilities … p3 + 3p2q + 3pq2 + q3 2 settlements & 1 trial = 3p2q = 3(0.7)2(0.3) Probability = 0.441 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 22 Unequal probabilities (cont.) Summary of Probabilities Combination of Probability Events 3 p 0.343 3p2q 0.441 3pq2 0.189 q3 0.027 Total (c = 8) 1.000 Graph of the binomial (p + q)3 Probability (p=0.7, q=0.3) 00.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 p3 3p2q 3pq2 q3 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 23 Pascal’s Triangle Expanding the Binomial to (p + q)n The binomial distribution can be expanded to n = 4, 5, 6, … When n is large, multiplying out the binomial can be quite tedious. A quick way to determine the coefficients of a binomial of n is to use Pascal’s Triangle (Blaise Pascal 1623-1662) n C 1 1 1 2 2 1 2 1 4 3 1 3 3 1 8 4 1 4 6 4 1 16 5 1 5 10 10 5 1 32 6 1 6 15 20 15 6 1 64 7 1 7 21 35 35 21 7 1 128 Using Pascal’s Triangle: (p + q)6 1p6+6p5q1+15p4q2+20p3q3+15p2q4 +6p1q5 +1 q6 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 24 Chinese discovery of the binomial triangle from Chu- Shi-Chieh's Ssu Yuan Yü Chien (1303 AD) Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 25 A Shortcut for Determining Coefficients Given the binomial (p + q)n The 1st term is always 1pn. Where do we go from here? Example Expansion of the binomial (p + q)6 1st term 1p6 To get the coefficient for the second term Multiply the coefficient of the 1st term by the exponent of p and … Divide the product by the position of the term, which is 1, the 1st term E.g. [(1) (6)] / 1 = 6 The exponent of p in the 2nd term is 1 less than (i.e. 6 - 1) what it was in the previous term, i.e. 5, and the exponent of q increases by 1 (i.e. from 0 to 1). 2nd term: 6p5q1 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 26 A shortcut (cont.) Continue this process from the 3rd term to 7th 3rd term (6)(5)/2 = 15 15p4q2 4th term (15)(4)/3 = 20 20p3q3 5th term (20)(3)/4 = 15 15p2q4 6th term (15)(2)/5 = 6 6p1q5 7th term (6)(1)/6 = 1 1 q6 The result is the same as when expanded using Pascal’s Triangle 1p6+6p5q1+15p4q2+20p3q3+15p2q4 +6p1q5 +1 q6 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 27 An Equation for Calculating Probabilities of Binomial Events (p=q=0.5) P(f) = n! [ pf (1 - p) n-f ] f! (n - f)! P (f) Probability that f-number of the cases will fall in one of the categories n The number of events in the sample ! Factorial, e.g. 3! = 3x2x1 p 0.5 Example For the binomial (p + q)6, what is the probability of 4 cases falling in one category and 2 in the other, e.g. 4 boys among 6 children P(4) = (6)! [ (0.5) 4 (1 – 0.5) (6 – 4) ] (4)! (6 - 4)! P(4) = (6x5x4x3x2x1) (0.0625) (0.25) (4x3x2x1) (2x1) P(4) = 0.234 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 28 An equation (cont.) The same result is achieved by expanding the binomial (p + q)6 and calculating the probability of the term 15p4q2 1p6+6p5q1+15p4q2+20p3q3+15p2q4 +6p1q5 +1 q6 15 p4 q2 = 15 (0.5)4 (0.5)2 = 0.234 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 29 The Binomial v the Normal Distribution When p=q=0.5, the binomial distribution is a symmetric, discontinuous distribution. However, as n increases from a few events to a very large number, the binomial distribution approaches a normal distribution if p = q = 0.5. Shown below are three binomial distributions with an over-lay of a normal distribution. Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 30 Binomial & normal (cont.) Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 31 The Normal Distribution Johann Karl Fredrich Gauss developed the mathematics for the normal probability distribution. (1777-1855) [-x2/(2 S2)] Y= (N) (e) S 2 Frequency (Y) X The normal distribution is a continuous, symmetric distribution, also called the bell curve, the normal curve, and the Gaussian curve. Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 32 Areas Under the Normal Curve If the distribution of some phenomenon approaches a normal curve, the curve can be use to describe the probabilities associated with the phenomenon. Example A sample of 142 political corruption cases was analyzed to determine the time from case filing to final disposition. The cases are "near" normally distributed with a mean of 144.2 days and a standard deviation of S = 14 days. Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 33 The Standard Deviation & the Normal Curve Mean = 144.2 S = 14 Area = 0.3413 130.2 144.2 158.2 0.6826 (68.26%) One standard deviation (S) on either side of the mean is 0.3413 proportion of the area under the curve, or 34.13%. The mean plus and minus 1S is 0.6826 proportion, or 68.26% of the curve. Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 34 Calculating Normal Curve Probabilities If a variable approximates a normal curve, the curve may be used to estimate the probabilities associated with the variable. This can be done in two ways: Using the equation for the normal curve Using a normal curve table, which is the more convenient way To use a normal curve table, the event in question must first be converted to a standard score (Z) Example Consider the example of the time to process political corruption cases. (X = 144.2, S = 14, N = 142) What is the probability that a case will take more than 170 days to process? Convert 170 days to a Z score Z = (X – X) / S Z = (170 – 144.2) / 14 = +1.84 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 35 Probability of Cases Falling Above Z = +1.84 A standard score is the ratio of how far a score deviates from the mean, relative to 1 standard deviation. In effect, it standardized deviations from the mean Z = +1.84 is a point 1.84 times further from the mean than 1S Political Corruption Case Processing Time Area above +1.84 +1.84 The normal curve table indicates the area of the curve above Z of 1.84 is 3.29%. Therefore, the probability of a case taking longer than 170 days to process is 0.0329, or 3.29%. Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 36 Normal Probability Table ***** Data extracted from a normal probability table. Sometimes the entries are expressed as proportions, sometimes as percentages. In this case they are expressed as percentages Area between Area Z mean and Z beyond Z . . . . . . . . . 1.83 46.64 3.36 1.84 46.71 3.29 1.83 46.78 3.22 . . . . . . . . . Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 37 Probability of Cases Falling Below Z = +1.84 Given X = 144.2, S = 14, N = 142, Z = +1.84 Political Corruption Case Processing Time Area below Z = +1.84 +1.84 Since the area above +1.84 is 3.29% of the area under the curve, the area below +1.84 is (100% - 3.29%) = 96.71% The probability of a case taking less than 170 days is therefore 96.71% Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 38 Probability of Cases Falling Between the Mean and Z = +1.84 Given X = 144.2, S = 14, N = 142, Z = +1.84 Political Corruption Case Processing Time Area between X and Z = +1.84 +1.84 As indicated in the Table, the area under the curve between the mean and Z is 46.71%. Therefore, the probability of a case taking between 144 days and 170 days to process is 0.4671 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 39 Probability of Cases Falling Between Z Scores on Either Side of the Mean What is the probability of a case taking between 130 and 160 days to process, given X = 144.2, S = 14, & N = 142 (cf. attached table) Z1 = (130 – 144.2) / 14 = -1.01 Z2 = (160 – 144.2) / 14 = +1.13 Political Corruption Case Processing Time Area 1 = 34.38% Area 2 = 37.08% -1.01 +1.13 The area between the Z scores –1.01 and +1.13 is (34.38% + 37.08%) = 71.46%. The probability that a case will take between 130 and 160 days to process equals = 0.7146 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 40 Normal Probability Table Area between Area Z mean and Z beyond Z . . . . . . . . . 1.01 34.38 15.62 . . . . . . . . . 1.13 37.08 12.92 . . . Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 41 Probability of Cases Falling Between Z Scores on One Side of the Mean What is the probability of a case taking between 130 and 140 days to process, given X = 144.2, S = 14, & N = 142 (cf. attached table) Z1 = (130 – 144.2) / 14 = -1.01 Z2 = (140 – 144.2) / 14 = -0.30 Political Corruption Case Processing Time Area between –1.01 & -0.30 = 22.59% -1.01 -0.30 The area from the mean to –1.01 = 34.38% The area from the mean to –0.30 = 11.79% The area between –1.01 and –0.30 (34.38% – 11.79%) = 22.59% The probability of a case taking between 130 – 140 days = 0.2259 Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 42 Normal Probability Table Area between Area Z mean and Z beyond Z . . . . . . . . . 0.30 11.79 38.21 . . . . . . 1.01 34.38 15.62 . . . . . . Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 43 Cases falling between (cont.) Political Corruption Case Processing Time Area between –1.01 & -0.30 = 22.59% -1.01 -0.30 Area from mean to –0.30 = 11.79% 34.38% Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 44 Using the Normal Curve to Compute Percentile Ranks If a variable distribution approximates a normal curve, the curve can be used to determine the percentile rank of a particular value Example A sergeant received a 71 on the lieutenants examination in which the mean = 78 & the S = 7.3. What is her percentile rank? Convert 71 to a Z score Z = (71 – 78) / 7.3 = -0.96 Area below a Z of -0.96 = 16.85% -0.96 The area below a Z score of -0.959 is 16.85%. The sergeant scored at the 17th percentile. She scored above 17% of those taking the test. Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 45 How to Determine Whether a Variable is Normally Distributed Three techniques Compare the S with the IQR, but only if the variable is symmetrical (X Mdn). If S ≈ PSD, the distribution is normal (PSD = IQR/1.35). Plot a histogram with a normal curve over- lay based upon the mean and standard deviation of the variable Graph a cumulative normal probability plot, called a Normal P-P Plot in SPSS Two Examples Time from filing to disposition in cases of political corruption Sentences for offenders convicted of political corruption Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 46 The Standard Deviation & IQR ***** Sentences & Case Processing Time If a variable is normally distributed, the standard deviation (S) will be approximately equal to the pseudo standard deviation (PSD ≈ IQR / 1.35)) This technique should never be used if the variable is skewed, since the results will be very misleading. (Assume X Mdn) S PSD = (IQR / 1.35) Difference Variable IQR PSD S (S - PSD) Sentence 12.5 9.3 14.0 4.7 Process 6.0 4.4 5.0 0.6 Time It would appear that the variable "process time" is near normal since S PSD, but the distribution of sentences is not normal. Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 47 Testing for Normality with a Histogram Another technique for testing the normality of a variable involves … Constructing a histogram of the variable. Overlaying on the histogram a normal distribution that has the same mean and standard deviation is the variable. The two determinants of a normal distribution are its mean and standard deviation … [-x2/(2 S2)] Y= (N) (e) S 2 S = the standard deviation x = a deviation score (X - X) Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 48 Histogram & Normal Probability Plot Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 49 Normal P-P Plot A normal P-P Plot compares the cumulative rank ordered values of a variable with the cumulative expected normal values, given the sample size, mean and standard deviation of the variable. A straight line represents the normal distribution from the lower left corner to the upper right corner of the plot. The variable is represented by a series of “dots” from the lower left corner to the upper right corner of the plot. If the dots are synonymous with the line, the variable is normally distributed If the dots “bow-out” to the right, the distribution is skew right If the dots “bow-out” to the left, the distribution is skew left Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 50 Normal P-P Plot (cont.) Processing time is near normal, but sentences are skewed right. Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 51 The Wager on the Existence of God Blaise Pascal (1623-1662) In Pascal’s Book Pensées (published posthumously), a collection of meditations on human suffering and faith in God, he presents several “wagers” on the existence of God. The logic he used anticipated by almost 300 years the pioneering work of mathematicians John von Neumann and Oskar Morganstern on the theory of games and decision making under uncertainty. He rejected the Five Proofs (Quinque Die) of Aquinas and Aristotle, the ontological arguments of Anselm, and the cosmological arguments of Descartes. One of his “wagers” involves the lost opportunity costs involving decision making under uncertainty. The Decision Matrix The Reality The Decision God Exists God Does Not Exist Wager that God exists Gain All Status Quo Wager that God does not exist Misery Status Quo Pascal argued that wagering on the existence of God outweighs wagering against the existence of God, since the eternal payoff of being correct far out ways all the other alternatives. Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University 52 Case Studies Two case studies are associated with this module, which provide practice using the binomial expansion and testing whether sample distributions are normally distributed. These case studies, and the associated database, can be found on the statistical WebPage and are entitled: The Binomial Distribution The Normal Distribution Probability Theory: Charles M. Friel Ph.D., Criminal Justice Center, Sam Houston State University