Quantitative Methods for Lawyers - Class #9 - Professor Daniel Martin Katz - MSU College of Law

Binomial, Normal andPower Law Distributions Quantitative Methods for Lawyers Class #9 Daniel Martin Katz Michigan State University - College of Law

A Quick ReviewFrom Last Time

Random Sampling ErrorFor Simple Random Sample on alarge population, the Inverse of theSquare Root of the Sample SizeExample:a random sample of size 400 will give a margin oferror, at a 95% confidence level, of 0.98/20 or0.049 - just under 5%.A random sample of size 1600 will give a margin oferror of 0.98/40, or 0.0245 - just under 2.5%. Notice: Double the Precision Requires 4times the Sample Size!

Thinking of Data asa Distribution: HistogramHistogram of travel time, US 2000 census. Area underthe curve equals the total number of cases. Thisdiagram uses Q/width from the table.

Thinking of Data as a DistributionMoment 1 = MeanMoment 2 = VarianceMoment 3 = SkewnessMoment 4 = Kurtosis

Skewnessskewness is a measure of the asymmetry of a distribution

Calculating SkewnessTry This Problem:http://www.indiana.edu/~educy520/sec5982/week_12/skewness_demo.pdf

Kurtosiskurtosis is any measure of the"peakedness" of a distributionA high kurtosis distribution has asharper peak and longer, fatter tails,while a low kurtosis distribution hasa more rounded peak and shorter,thinner tails.

Calculating KurtosisThe moment coefficient of kurtosis of a dataset is computed almost the same way as thecoefficient of skewness:kurtosis = m4 / m22 and “excess” kurtosis: = Kurtosis−3 Note: the excess kurtosis is generally used because the excess kurtosis of a normal distribution is 0. where

Calculating KurtosisExample: n = 100 x̄bar = 67.45 variance m2 = 8.5275

Calculating KurtosisExample: n = 100 x̄bar = 67.45 variance m2 = 8.5275 kurtosis is = m4 / m2² = 199.3760/(8.5275)² = 2.7418 and the excess kurtosis is = 2.7418 − 3 = −0.2582

Binomial Distribution

Binomial Distribution“the binomial distribution is the discrete probabilitydistribution of the number of successes x (sometimeswritten as k # of successes) in a sequence of nindependent yes/no experiments, each of which yieldssuccess with probability pA binomial experiment (also known as a Bernoullitrial) is a statistical experiment that has thefollowing properties: The experiment consists of n repeated trials. Each trial can result in just two possible outcomes. The probability of success, denoted by P, is the same on every trial. The trials are independent”

Binomial DistributionConsider the following statistical experiment:You flip a coin 2 times and count the number oftimes the coin lands on heads.This is a binomial experiment because:The experiment consists of repeated trials.We flip a coin 2 times. Each trial can result injust two possible outcomes - heads or tails.probability of success is constant i.e. 0.5 onevery trial.independence

Binomial DistributionThe mean of the distribution(μ x) is equal to n*PThe variance (σ2x) is n*P*(1-P)The standard deviation (σx)is sqrt[ n * P * ( 1 - P ) ].

Binomial DistributionAgain, the binomial probability refers tothe probability that a binomialexperiment results in exactly xsuccesses.Here is the Formula for a binomialexperiment consisting of n trials andresults in x successes. If theprobability of success on an individualtrial is P, then the binomial probabilityis: b(x; n, P) = nCx * Px * (1 - P)n - x

Binomial Distribution b(x; n, P) = nCx * Px * (1 - P)n - xThis is Probably the Part of theFormula that is Least FamiliarThis is “N choose X”Remember Sometimes you will see aK in place of X (that is just notation)

Binomial Distribution b(k; n, P) = nCk * Pk * (1 - P)n - kLets Switch the X for a KThis is “N choose K” can also be written as Okay now we are going to have to do some factorials :)

Binomial Distribution ExamplesSuppose a die is tossed 5 times.What is the probability of gettingexactly 2 fours? b(x; n, P) = nCx * Px * (1 - P)n - x

Binomial Distribution ExamplesSuppose a die is tossed 5 times. What is theprobability of getting exactly 2 fours?Solution: This is a binomial experiment in which thenumber of trials is equal to 5,the number of successes is equal to 2,probability of success on a single trial is 1/6

Binomial Distribution ExamplesSuppose a die is tossed 5 times. What is theprobability of getting exactly 2 fours? b(2; 5, 0.167) = 5C2 * (0.167)2 * (0.833)3 5C2 * .0278 * .5780 5C2 * .0160 What is 5C2? http://www.mathsisfun.com/combinatorics/ Go Here combinations-permutations-calculator.html b(2; 5, 0.167) =10 * .0160 b(2; 5, 0.167) =.160

Binomial Distribution ExamplesImagine the probability that astudent is accepted to a prestigeouscollege is 0.3.If 5 students from the same schoolapply, what is the probability that atmost 2 are accepted?

Binomial Distribution ExamplesThe probability that a student is accepted to aprestigeous college is 0.3. If 5 students fromthe same school apply, what is the probabilitythat at most 2 are accepted?Solution: To solve this problem, we compute 3individual probabilities, using the binomialformula. The sum of all these probabilities isthe answer we seek.

Binomial Distribution Examples The probability that a student is accepted to a prestigeous college is 0.3. If 5 students from the same school apply, what is the probability that at most 2 are accepted? Solution: To solve this problem, we compute 3 individual probabilities, using the binomial formula. The sum of all these probabilities is the answer we seek.b(x < 2; 5, 0.3) = b(x = 0; 5, 0.3) + b(x = 1; 5, 0.3) + b(x = 2; 5, 0.3)b(x < 2; 5, 0.3) = 0.1681 + 0.3601 + 0.3087b(x < 2; 5, 0.3) = 0.8369

Binomial Distribution Examples Lets Just Calculate One of These: How do we get 0.3087 ?b(x < 2; 5, 0.3) = b(x = 0; 5, 0.3) + b(x = 1; 5, 0.3) + b(x = 2; 5, 0.3)b(x < 2; 5, 0.3) = 0.1681 + 0.3601 + 0.3087b(x < 2; 5, 0.3) = 0.8369

Binomial Distribution ExamplesCalculate One of These:How do we get 0.3087 for b(x = 2; 5, 0.3) ?Take a Minute and Work it OutWe Can Discuss This in Office Hours

Exact BinomialProbability Calculator http://faculty.vassar.edu/lowry/ch5apx.html

Normal Distribution“The normal (or Gaussian) distribution isa continuous probability distributionthat has a bell-shaped probabilitydensity function, known as the Gaussianfunction or informally the bell curve”where X is a normal random variable, μ is the mean, σ is thestandard deviation, π is approximately 3.14159, and e isapproximately 2.71828.

Normal DistributionThe graph of the normal distribution depends ontwo factors -the mean and the standard deviation.The mean of the distribution determines thelocation of the center of the graph, and thestandard deviation determines the height andwidth of the graph.When the standard deviation is large, the curveis short and wide; when the standard deviation issmall, the curve is tall and narrow. All normaldistributions look like a symmetric, bell-shapedcurve, as shown below.

Normal DistributionThe normal distribution is a continuousprobability distribution.The total area under the normal curve is equalto 1.The probability that a normal random variable Xequals any particular value is actually 0.

Normal DistributionThe probability that X is greater than a equals thearea under the normal curve bounded by a and plusinfinity (as indicated by the non-shaded area in thefigure below).The probability that X is less than a equals thearea under the normal curve bounded by a andminus infinity (as indicated by the shaded area inthe figure below).

Normal Distributionevery normal curve (regardless of its mean orstandard deviation) conforms to the following"rule".About 68% of the area under the curve falls within 1standard deviation of the mean.About 95% of the area under the curve falls within 2standard deviations of the mean.About 99.7% of the area under the curve falls within 3standard deviations of the mean.

Normal Distribution ExamplesAn average light bulb manufactured by the AcmeCorporation lasts 300 days with a standarddeviation of 50 days. Assuming that bulb life isnormally distributed, what is the probability thatan Acme light bulb will last at most 365 days? http://stattrek.com/Tables/Normal.aspx

Normal Distribution ExamplesSolution: mean score is 300 daysstandard deviation is 50 days,We want to find the cumulative probability that bulblife is less than or equal to 365 days. Thus, we knowthe following:The value of the normal random variable is 365 days.The mean is equal to 300 days. The standard deviationis equal to 50 days.We enter these values into the Normal DistributionCalculator and compute the cumulative probability.The answer is: P( X < 365) = 0.90. Hence, there is a90% chance that a light bulb will burn out within 365days.

Normal Distribution ExamplesSuppose scores on an IQ test are normallydistributed. If the test has a mean of 100 and astandard deviation of 10, what is the probabilitythat a person who takes the test will scorebetween 90 and 110?

Normal Distribution ExamplesYou need to Figure out the Probability Between theTwo Values of 90 and 110The Key to solving this problem is to realize thefollowing:P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 ) Work on This For a Few Minutes In Small Groups

Normal Distribution ExamplesWe use the Normal Distribution Calculator to computeboth probabilities on the right side of the aboveequation:• To compute P( X < 110 ), we enter the following inputs into the calculator: The value of the normal random variable is 110, the mean is 100, and the standard deviation is 10. We find that P( X < 110 ) is 0.84.• To compute P( X < 90 ), we enter the following inputs into the calculator: The value of the normal random variable is 90, the mean is 100, and the standard deviation is 10. We find that P( X < 90 ) is 0.16.

Normal Distribution ExamplesP( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )P( 90 < X < 110 ) = 0.84 - 0.16P( 90 < X < 110 ) = 0.68Thus, about 68% of the test scores willfall between 90 and 110.

Standard Normal DistributionThe standard normal distribution is a special case ofthe normal distribution. It is the distribution thatoccurs when a normal random variable has a mean ofzero and a standard deviation of one.The normal random variable of a standard normaldistribution is called a standard score or a z-score.Every normal random variable X can be transformedinto a z-score via the following equation: z = (X - μ) / σwhere X is a normal random variable, μ is the mean meanof X, and σ is the standard deviation of X.

Z Score and a Standardization ProblemMolly earned a score of 940 on a nationalachievement test. The mean test score was 850with a standard deviation of 100. Whatproportion of students had a higher score thanMolly?(Assume that test scores are normallydistributed.)

Z Score and a Standardization ProblemAgain, assume that test scores are normally distributed.Thus, we can use normal distribution as a model formeasurement.Given an assumption of normality, the solution involvesthree steps:• First, we transform Mollys test score into a z-score, using the z-score transformation equation. z = (X - μ) / σ = (940 - 850) / 100 = 0.90

Z Score and a Standardization ProblemSecond, using an online calculatorhttp://regentsprep.org/Regents/math/algtrig/ATS7/ZChart.htmto find the cumulative probability associated with thez-score. In this case, we find P(Z < 0.90) = 0.8159..90 ishere

Z Score and aStandardization ProblemFinally,P(Z > 0.90) = 1 - P(Z < 0.90) = 1 - 0.8159 = 0.1841.Thus, we estimate that 18.41 percentof the students tested had a higherscore than Molly.

Power Law Distribution (Scale Free) This is a Classic and Very Important Distribution A power law is a special kind of mathematical relationship between two quantities. When the frequency of an event varies as a power of some attribute of that event (e.g. its size), the frequency is said to follow a power law.

Power Law Distribution (Scale Free)Examples: Pareto distribution ( Wealth Distribution ) Zipfs law ( Natural Language Frequency ) Links on the Internet Citations Richardsons Law for the severity of violent conflicts (wars and terrorism) Population of cities Etc.

Power Law Distribution (Scale Free) What Makes it Scale Invariant?

Additional Conceptual Applications

Power LawDistributions Plotted on aLog-Log Scale

Log Log Plot Log Log PlotUS Supreme Court NY Ct of Appeals Smith (2007) Post & Eisen (2000)

Quantitative Methods for Lawyers - Class #9 - Professor Daniel Martin Katz - MSU College of Law

by Daniel Katz on Feb 09, 2012

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Quantitative Methods for Lawyers - Class #9 - Professor Daniel Martin Katz - MSU College of Law — Presentation Transcript