Continuous Probability DistributionsContinuous Uniform DistributionSlide 3Slide 4Slide 5Normal DistributionSlide 7Standard Normal DistributionSlide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Normal Distribution ProbabilitiesSlide 17Slide 18Slide 19Slide 20Normal Approx. to the BinomialSlide 22Slide 23Lognormal DistributionSlide 25Slide 26Exponential DistributionSlide 28Gamma DistributionSlide 30Slide 31Exponential and GammaSlide 33Slide 34Relationship to Poisson ProcessSlide 36Slide 37Chi-Squared (2) DistributionChi-Squared Distribution1Continuous Probability DistributionsContinuous Uniform DistributionNormal DistributionLognormal DistributionGamma and Exponential DistributionsChi-Squared Distribution2Continuous Uniform DistributionA continuous r.v. X with a “rectangular” p.d.f. over a specified range [A, B]•The p.d.f. and c.d.f depend on parameters A and BExpected ValueVariance 0)( o/w ;for 1),;(  xfBxAABBAxfXX2][BAXE12)()(2ABXVBxAABAxxXPBAxFX for ][),;(3Continuous Uniform DistributionA BAB1xf(x)4Continuous Uniform DistributionExample: Harry is never early and can be up to 20 minutes late. Assume that his lateness follows a uniform distribution.•A = 0, B = 20410200151)20,0;15(1]15[ XFXPminutes 102200][ XE31001240012)020()(2XV5Continuous Uniform DistributionExample: Observe events occurring in a fixed time frame, e.g. buses arriving at a bus stop. •Suppose we know that one such event has occurred in the time interval (a, b).•X = the actual time of occurrence of the event.•A very simple model assumes that X is equally likely to lie in any small subinterval within (a, b). X follows a Uniform(a, b) distribution.6Normal DistributionA continuous r.v. X with a “bell curve” p.d.f. •The most important distribution in statistics•The p.d.f. depends on mean  and variance  2Expected valueVarianceExcel function•p.d.f.: NORMDIST(x, , , 0)•c.d.f. : NORMDIST(x, , , 1)xexnxfor 21),;(221][XE2)(XV7Normal DistributionN(,  2)4 4x = 8Standard Normal DistributionLet X be distributed N(,  2)Then is distributed N(0, 1). •The p.d.f. for Z is•The c.d.f. for Z is tabulated in Table A.3zezzfor 21)(221XZ zzZPz for ][)(9Standard Normal DistributionHow to read Table A.3z .00 .01 .02 .030.0 0.5000 0.5040 0.5080 0.51200.1 0.5398 0.5438 0.5478 0.5517...1.2 0.8849 0.8869 0.8888 0.89071.3 0.9032 0.9049 0.9066 0.90820.9066)32.1(]32.1[ ZP0.09340.90661)32.1(1]32.1[ ZP5517.090490)13.0()31.1(]31.113.0[  .ZP10Standard Normal DistributionN(0, 1) (0) = 0.5 4 4x = 011Standard Normal Distribution P[Z  0] = P[Z  0] = 0.5 (0) = 0.50Z12Standard Normal DistributionP[1  Z  1] = P[Z  1]  P[Z  1] = (1)  (1) 1.0 1.01.01.013Standard Normal Distribution P[Z > z1] = 1  P[Z  z1] = 1  (z1)0z114Standard Normal Distribution P[Z  z1] = P[Z > z1] = 1  P[Z  z1] (z1) = 1  (z1)z1Z015Standard Normal Distribution P[z1  Z  z2] = P[Z  z2]  P[Z  z1] = P[Z  z2]  (1  P[Z  z1]) = (z2)  (1  (z1) )z1z2016Normal Distribution ProbabilitiesLet X be distributed N(,  2)1) Standardize X2) Convert range for X[ ]a X bP a X b Pa bP Zm m ms s sm ms s- - -� �� � = � �� �� �- -� �= � �� �� �XZms-� =17Normal Distribution ProbabilitiesLet X be distributed N(,  2)3) Rewrite using standard normal c.d.f.4) Look up c.d.f. probabilities for z1 and z2 in Table A.3, then calculate.a b b aP Zm m m ms s s s- - - -� � � � � �� � =F - F� � � �� �� � � � � �bzaz21;z18Normal Distribution ProbabilitiesExample: The length of a part follows a normal distribution with mean 100mm and standard deviation 2mm.•X = length of the part ~ N( = 100,  = 2)•What is the probability that the part length is above 103.3mm?[ ] [ ]( )103.3 100[ 103.3]21.65 1 1.651 1.65 1 0.9505 0.0495XP X PP Z P Zms- -� �> = >� �� �= > = - �= - F = - =19Normal Distribution ProbabilitiesExample (cont’d): length of a part.•X = length of the part ~ N( = 100,  = 2)•What proportion of the part output will have lengths between 98.5mm and 102mm?[ ]( ) ( ) ( ) ( )98.5 100 102 100[98.5 102]2 20.75 1.01.0 0.75 1.0 1 0.750.8413 (1 0.7734) 0.6147XP X PP Zms- - -� �� � = � �� �� �= - � �� �=F - F - =F - - F� �= - - =20Normal Distribution ProbabilitiesExample: Suppose Harry’s lateness is normally distributed with a mean of 2 minutes and a variance of 16.What is the probability that Harry will be between 1 and 5 minutes late?•X = Harry’s lateness ~ N( = 2,  2 = 16)    3721.0)5987.01(7734.025.075.075.025.0425421]51[ZPXPXP21Normal Approx. to the BinomialX follows a Bin(n, p) distributionIf n is large, np > 5 and n(1p) > 5, then is approximately N(0, 1))1()5.0()1()5.0()1()5.0()1()5.0(),;(pnpnpxpnpnpxpnpnpxZpnpnpxPpnxb)1( pnpnpXZ22Normal Approx. to the BinomialExample: n = 10, p = 0.523Normal Approx. to the BinomialExample: Flip a coin 40 times.•X = # heads ~ Bin(n = 40, p = 0.5)•np = (40)(0.5) = 20; np(1p ) = (20)(0.5) = 10     7074.0)9222.01(7852.042.1179.079.042.110205.2210205.15708275.0)5.0,40;15()5.0,40;22(]2216[ZPZPBBXP24Lognormal DistributionA continuous r.v. X such that r.v. Y = ln(X) is normally distributed•Y has mean  and variance  2•The p.d.f. depends on parameters  and  Expected valueVariance0)( o/w ;0for