Question

Can anyone please help me with this problem asap?

g) For each of parts ( 21 ), Wily all the tou 8.27 Refer to Example 8.5 on page 409. Use the CDF obtained in that example to mine Pla <Z <b), P(a < Z <b), P(a < Z <b), and P(a < Z <b) for the spe values of a and b. a) a = 0.2, b = 0.8 b) a = 0,b=0.8 c) a = 0.2, b = 1.5 d) a = -1, b = 1.5 e) a = -2, b = -1 f) a = 1, b = 2 g) For each of parts (a)-(f), why are the four probabilities identical?

Answer 1

Page 1 Given that, the C-Dif of Random variable z is PIZ-2) (E) 50, 6230 - | 2², if o Sall To, if 221 here, the edif of the Randam of a vaniable is continuous so, the pandom vaniable Z is continuus. Lue know, Fz (2-) =P (242) , whene, Z- = the points bekering bebent. ?* probability using ein [ (K) = the wake of r at the point whene a =02,5=0.8 is less thank L 2 • P(aL2 = b) = P(0.22220-8) = P(240.8)- P(250-2) | FAC?)= P(257) 7 FSince, ed lisdehiedy FFCO.8) - F(02) L& F&Ca)=2" = 0.8 -0.2? idoczu = 0.6 p(azza) = P(0.22 Z £0.8) = P(Z 2018) - PC220-2) = F(0-8-) - F(0.23 = 0.8 -0.22 . E 18")= P(2220) 1 le Face 2² if oszal Pa_24b = P(0-249 20.8) = F 0.8-- FO2) • Platz2b) z pro.22 2 20.8) = 0.81-0.2=06 SP(2 20:8)-P(Z <0.2) E. F(0-2-) = the wake at 2 F (08) - F (0.2-) I ELE) at the point where 0-8 -0.2 0.6 z less than 2 7 o az z =0.8, calculate probability using edif. •PCOLZ 308)= P(240-8)–p(?<0- F(-9) F (OJ .PCOZZ3093 0809-0.69 asos simeesnifat est •PCOLE208) =P (2.2008) - P(320) - FC®.8-) - 70 I Pa (27=2", id = 0.82-0: 04241 = 6-69 P (04240-8) P220-8) - PCt20) = F(0.8-) - FO -0.64-0 =0.69

Page 2 'PCOsz 50-8) - PC2 30-3) - P(240-) since fiere; f(0-8)-F10, L & fluoizce icoz4 20-64 & a20.2, b=1.5; putting the vale in als we get. tasyonP (-2<Z 515)=P(25105)- P(750.27 = F (1.5) - F(-2) 31– 0.2 20.96 P0-24Z41:5)= F (15-) - F(0.2) 1 =18 0.22 20.96 OP 0:2 422):52 F (1.5) - F(0-2) -1- 0.22 0.56 PO-2-ZE15) P(Z515)- P(120-2) - Fclis) - FC0.2-) 2 0 .-|-0-22 30.96 1a asof & 6=15; putting the vale of a do weget. • P(-1<2 21:5)= P(Z 515)-8 (2=-1)/Sines FE (3)=1, ideyil - FC.5) - FOD ( CD=01d ZLO = op (12741,5) = P(241.5) - P(Z <yje = F(1.5) - F(-) - P(-45.721.5) - F (1.5-) - FE- 31-03) • PEILZE 1.5) = P(Z £15)- P(Z2-1-) = R(1-5)-F(-) le aa-2, 6=-1, putting the role inaf 6 weget o P(-2222-1) = P(z 6-1)-P (22-2) = f(-1)-F (-2) ( fz (7)=0,if 220

thank ) To P(-2222-1) = P(ZL-1)- P (32-2) / Sine; F2 (2) 70, 8220 = (-ity - F2(-3) (& Ę (ka) - the 30-0 vake of Efts of the point whipe zis less supilamp (-2522-1) = 0 ANA & PC-242 -1) = P(24-1)-P(22-2) i Sime Fz (2) = F(-1) - F(*2-) ( fake '0' whene - zis negaline hen, -1409 €2-}<0 . ro) we notice here for each parts of the fours probabilities and some. Suppose: of a founction is continuous ies: fos is continuous within the interval a & b. i pents? - there exists infinite numbent between the intental [as]. So, the chane e of getting a function within this interval close to zeno, cie, the probability of finding a v a particular point within this intental is zero. mathematically; - Let x be a continuous random variable within the interval Laib] and its poded is given by fix (ru) s as if it is azxc6 (o, otherwise Here ,z be the continuous handom variable and its p.did is given by 27, 05241 &z (2) 2 = ) o other wise

olan target is vectity the placzes), Pla&Elb), Placzuby & PCOEZEb) foon probability and here . We now, werkennt, [a, b]= [aus) U {f}={a}uCo.6Ju{6) =(0,5)U{5? per la su, plat 745) = P(2.4745)1 : [a,s) & 41 these two - Sieradz + S person sit one this cod ) - Senda t° 17'02)), 52083.02 +4$(5)= f'(57 25% flandz t0i =3fc27.dz$ 27.07 Her pfat o : 2419 = Her, Plasz Lb)=P(azz Ls)= (6=22) es the function is continus so, the probabitie at any constant point is O. Tpx=);4043drs 3 pesso these implics, the four probabilities as eit Same.