Introduction to Probability - Chapter 2 docx

... to wait. Write a program to see if your conjecture is right. 2. 2. CONTINUOUS DENSITY FUNCTIONS 65 -1 -0 .5 0.5 1 1.5 2 0 .2 0.40.60.81 -1 -0 .5 0 0.5 1 1.5 2 0 .25 0.50.7511 .25 1.51.75 2 F ... - z1 - z1 - z1 - zEFigure 2. 19: Calculation of FZ. 20 40 60 80 100 120 0.0050.010.0150. 02 0. 025 0.03f (t) = (1/30) e - (1/30) tFigure 2. 20: Exponential density with λ =1/30. 52 ... occurs with probability p =π(1 /2) 2 π(1) 2 =14. 2. L>√3if|r| < 1 /2, which occurs with probability 1 /2 −(−1 /2) 1 −(−1)=1 2 .3. L>√3if2π/3 <α<4π/3, which occurs with probability 4π/3...

Introduction to Probability - Chapter 5 docx

... 28 92 7 3657 8 3 025 9 33 62 10 29 85 11 3138 12 304313 26 90 14 24 23 15 25 5616 24 56 17 24 79 18 22 7619 23 04 20 1971 21 25 43 22 26 78 23 27 29 24 24 14 25 26 16 26 24 26 27 23 81 28 20 59 29 20 39 30 22 9831 ... DENSITIESDemocrat RepublicanFemale 24 4 28 Male 8 14 22 32 18 50Table 5 .2: Observed data.Democrat RepublicanFemale s11s 12 t11Male s 21 s 22 t 12 t 21 t 22 nTable 5.3: General data table.nail ... 22 9831 20 81 32 1508 33 188734 1463 35 1594 36 135437 1049 38 1165 39 124 840 1493 41 1 322 42 1 423 43 120 7 44 125 9 45 122 4Table 5.7: Numbers chosen by contestants in the Powerball lottery.5 .2 Important...

Introduction to Probability - Chapter 8 docx

... 100. 02 0.040.060.080.10. 12 0.140 0 .2 0.4 0.6 0.8 100. 02 0.040.060.080.10. 12 0 0 .2 0.4 0.6 0.8 100.050.10.150 .2 0 .25 0 0 .2 0.4 0.6 0.8 100. 025 0.050.0750.10. 125 0.150.175n=10 n =20 n=40n=30n=60n=100Figure ... E(Xi)=10xdx=1 2 ,σ 2 = V (Xi)=10x 2 dx −µ 2 =13−14=1 12 .Hence,ESnn=1 2 ,VSnn=112n,and for any >0,PSnn−1 2 ≥ ≤112n 2 .This ... large to take n to get a desired accuracy. ✷ 8.1. DISCRETE RANDOM VARIABLES 3090 0 .2 0.4 0.6 0.8 100. 02 0.040.060.080.10 0 .2 0.4 0.6 0.8 100. 02 0.040.060.080 0 .2 0.4 0.6 0.8 100. 02 0.040.060.080.10. 12 0.140...

Introduction to Probability - Chapter 1 pps

... DISCRETE PROBABILITIES 3 .20 3309 .7 620 57 .151 121 . 623 868.9 320 52 .415178 .716719 .9674 12 .069664 .6709 82 .3 523 20 .049 723 .75 021 6 .784810 .089734 .966730.946708 .380365 . 027 381 .900794Table 1.1: ... PROBABILITIES 5510 15 20 25 30 35 40 -1 0 -8 -6 -4 -2 2 46810Figure 1.1: Peter’s winnings in 40 plays of heads or tails.One can understand this calculation as follows: The probability that no ... more appropriate to assign a distributionfunction which assigns probability .513 to the outcome boy and probability .487 to the outcome girl than to assign probability 1 /2 to each outcome. This...

Introduction to Probability - Chapter 3 ppt

... ways to assign the ﬁrst element, for 94 CHAPTER 3. COMBINATORICSn = 0 110 1 10 45 120 21 0 25 2 21 0 120 45 10 1 9 1 9 36 84 126 126 84 36 9 1 8 1 8 28 56 70 56 28 8 17 1 7 21 35 35 21 7 ... {a1,a 2 ,a3,a4} can be written in theformσ = 123 4 21 43,indicating that a1went to a 2 , a 2 to a1, a3 to a4, and a4 to a3.Ifwealwayschoosethetoprowtobe 123 4then, to prescribe ... Binomial distributions. 78 CHAPTER 3. COMBINATORICSNumber of people Probability that all birthdays are diﬀerent 20 .5885616 21 .5563117 22 . 524 3047 23 .4 927 028 24 .4616557 25 .4313003Table 3.1:...

Introduction to Probability - Chapter 4 doc

... 156 CHAPTER 4. CONDITIONAL PROBABILITY Y -1 0 12 X -1 0 1/36 1/6 1/ 12 01/18 0 1/18 010 1/36 1/6 1/ 12 21/ 12 0 1/ 12 1/6Table 4.6: Joint distribution.36 A die is thrown twice. Let X1and X 2 denote ... SmithUnconditional probability bbbbgggb1/41/81/81/81/81/41/41/41/41/411 /2 1 /2 1 /2 1 /2 1bggbggbbbbb1/41/81/81/41/41/411 /2 1 /2 bggbConditional probability 1 /2 1/41/4Figure 4.13: Tree for Example 4 .26 . 148 CHAPTER 4. CONDITIONAL PROBABILITY .001cannot.01.95.05+ - .0010.05.949+ - .051.949+ - .98110cannot.001.050.949cannot.019Original ... andY 2 = φ 2 (X 2 ).(a) Show thatP (Y1= r, Y 2 = s)=φ1(a)=rφ 2 (b)=sP (X1= a, X 2 = b) .(b) Using (a), show that P (Y1= r, Y 2 = s)=P (Y1= r)P (Y 2 = s) so thatY1and Y 2 are...

Introduction to Probability - Chapter 6 doc

... Y ) 2 ) −(a + b) 2 = E(X 2 )+2E(XY )+E(Y 2 ) −a 2 − 2ab − b 2 .Since X and Y are independent, E(XY )=E(X)E(Y )=ab. Thus,V (X + Y )=E(X 2 ) −a 2 + E(Y 2 ) −b 2 = V (X)+V (Y ) .✷ 24 6 CHAPTER ... value 0:pX= 2 −10 12 3/11 2/ 11 1/11 2/ 11 3/11;pY= 2 −10 12 1/11 2/ 11 5/11 2/ 11 1/11. 25 6 CHAPTER 6. EXPECTED VALUE AND VARIANCE37 The reader is referred to Example 6.13 for ... any random variable with E(X)=µ, thenV (X)=E(X 2 ) −µ 2 .Proof. We haveV (X)=E((X − µ) 2 )=E(X 2 − 2 X + µ 2 )= E(X 2 ) 2 E(X)+µ 2 = E(X 2 ) −µ 2 .✷Using Theorem 6.6, we can compute the variance...

Introduction to Probability - Chapter 7 pptx

... fZiffX(x)=1√ 2 σ1e−(x−µ1) 2 /2 2 1fY(x)=1√ 2 σ 2 e−(x−µ 2 ) 2 /2 2 2.*7 Suppose that R 2 = X 2 + Y 2 . Find fR 2 and fRiffX(x)=1√ 2 σ1e−(x−µ1) 2 /2 2 1fY(x)=1√ 2 σ 2 e−(x−µ 2 ) 2 /2 2 2.8 ... with λ =1 /2, β =1 /2 (see Example 7.4). Now letR 2 = X 2 + Y 2 . ThenfR 2 (r)=+∞−∞fX 2 (r − s)fY 2 (s) ds=14π+∞−∞e−(r−s) /2 r − s 2 −1 /2 e−ss 2 −1 /2 ds ,=1 2 e−r 2 /2 , if ... havefX(x)=fY(y)=1√ 2 e−x 2 /2 ,and sofZ(z)=fX∗ fY(z)=1 2 +∞−∞e−(z−y) 2 /2 e−y 2 /2 dy=1 2 e−z 2 /4+∞−∞e−(y−z /2) 2 dy=1 2 e−z 2 /4√π1√π∞−∞e−(y−z /2) 2 dy.The...

Introduction to Probability - Chapter 9 pps

... E(X)=7 /2 and σ 2 = V (X)=35/ 12. Thus, E(S 420 )= 420 ·7 /2 = 1470, σ 2 (S 420 ) = 420 ·35/ 12 = 122 5, and σ(S 420 ) = 35. Therefore,P (1400 ≤ S 420 ≤ 1550) ≈ P1399.5 −147035≤ S∗ 24 0≤1550.5 ... .0000 1.0 .3413 2. 0 .47 72 3.0 .4987 .1 .0398 1.1 .3643 2. 1 .4 821 3.1 .4990 .2 .0793 1 .2 .3849 2. 2 .4861 3 .2 .4993 .3 .1179 1.3 .40 32 2.3 .4893 3.3 .4995.4 .1554 1.4 .41 92 2.4 .4918 3.4 .4997.5 ... .4997.5 .1915 1.5 .43 32 2.5 .4938 3.5 .4998.6 .22 57 1.6 .44 52 2.6 .4953 3.6 .4998.7 .25 80 1.7 .4554 2. 7 .4965 3.7 .4999.8 .28 81 1.8 .4641 2. 8 .4974 3.8 .4999.9 .3159 1.9 .4713 2. 9 .4981 3.9 .5000...

Introduction to Probability - Chapter 10 pptx

... 10 .2. BRANCHING PROCESSES 387GeometricpjData Model0 .20 92 .18161 .25 84 .3666 2 .23 60 .20 283 .1593 .1 122 4 .0 828 .0 621 5 .0357 .03446 .0133 .01907 .00 42 .01058 .0011 .00589 .00 02 .00 32 10 ... FUNCTIONSZ1Z 2 Z3Z4Z5Z6Z7Z8Z9Z10Z11Z 12 Proﬁt 126 778119766 520 0100000000000 -5 0100000000000 -5 0111000000000 -5 000000000000 0-1 0011111 124 9797300 23 3 420 000000 0100000000000 -5 0 21 0000000000 ... lims→∞u(t/√s)s−1= lims→∞u(t/√s)t2s−1 /2 = lims→∞ut√st 2 2= σ 2 t 2 2=t 2 2.Hence, g∗n(t) → et 2 /2 as n →∞. Now to complete the proof of the Central LimitTheorem,...

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