Question

(1 point) Find the value of the standard normal random variable z, called Zo such that: (a) P(Z <zo) = 0.8319 20 (b) PC-Zo <z<zo) = 0.5508 20 = (c) P(-20 <2<zo) = 0.748 zo = (d) P(z > Zo) = 0.2823 20 = (e) P(-20 <z<0) = 0.0283 Zo = (1) P(-1.5 <2<zo) = 0.7108 zo Note: You can earn partial credit on this problem.

Answer 1

a) P(z<z) = 0.8319 Zo ENORMSINV(0.8319)=0.9617=0.96 (Using excel) b) P(-Zo Sz Szo) = 0.5508 Since P(-Zo <z z)=2*P(Z <z)-1 2*P(Z <z)-1 = 0.5508 P(z<zo) = (1+0.5508)/2 P(z<z) = 0.7754 Zo=NORMSINV(0.7754)=0.756749=0.76 C) P(-Zo <z <zo) = 0.748 Since P(-20 <z <zo)=2*P(z<z)-1 2*P(z<z)-1 = 0.748 P(z<zo) = (1+0.748)/2 P(z<zo) = 0.874 Zo =NORMSINV(0.874)=1.1455 = 1.15

d) P(Z > zo) = 0.2823 Since P(Z <zo) = 1-(Z > zo) P(z<z) = 1-0.2823 P(z<zo) = 0.7177 z = 0.576022=0.58 e) P(-Zo <z<0=0.0283 P(Z<0)-P(Z<-zo) = 0.0283 (Since P(z<0) = 0.50) 0.50 - P(Z<-Z) = 0.0283 P(Z<-Z) = 0.50 -0.0283 P(Z<-z) = 0.4717 -Z0 =NORMSINVO.4717)= -0.071 Zo = 0.07 f) P(-1.5<2<z) = 0.7108 P(z<z)-P(Z<-1.5)=0.7108 (Since P(Z<-1.5)= 0.0668) P(z<z) -0.0668=0.7108 P(z<zo) = 0.7108+0.0668 P(Z<zo) = 0.7776 zo =NORMSINV(0.7776)= 0.764113=0.76