Question

6. [-/4 Points) DETAILS JMODD8 4.4.011. MYN Find the following probabilities. (Round your answers to four decimal places.) (a) p(o <z < 1.63) (b) p(1.40 <z < 1.86) (c) P(-0.36 < z < 1.58) (d) plz < -1.92) (e) P(-1.33 < z < -0.86) (0) P(Z < 1.29)

Answer 1

questions Answers answer all this. using excel for F(a) use command Fra) = = NORMS DISCO) where fras is cdf of standard Normal disth at pt a ploczc1.63) F 4.63) - NORMSDIST (163) & Fco)=0.5 plocz< 1.63)= 0.9484 -0.5=0.4484 2) P(1.40 <z (1.86) F (1.86) - F (1.40) = Norms DIST(1086) - (= NORMDIST (1.40) 0.9686 -0.9192 O 0494 p(1.4 <2L 1.86) = 0.0494. 3 pl-0.36 C1'58) Fli.rs) F(-0.36) oog429 -0.3594 O-5835 PC-0.36 <2 (1.58) 0.5835 - plz<-1.92) F(-1.92) 0.0274 plz L-1.92) 0.0274 pl-1.33 c2c-0.86) FC=0.86) Fl-1.33) o 1949 0.0918 Plac1-29) F (1.29) 0.9015 p(22-29) = 0.90ls

o posza 1.42) 8 (1.42) Flo) -0-9222 OS 0.4222 plocal 1.42) = 0.4222 p(1.05<a c 1.66) F11.66) F(1:05) =o.g5ls 0.8531 = 0.0984 PU:05 <2<1.60) - 0.0984 p(-0.85 <2 <175) = E(1175) - FC -0.85) -0.9599 -0.1976 -0.7623 P(-0.85 <2<1-75) = 0.7623 Plz<-2.08) - F (-2-08) -0.0188 P2<-2008) 0.0188 P1-2.34 < 22-1:18) Fl-1:18) - El-2:34). 0·1190 -0.0096. o 1094 Pc-2.34 <2 2-1 18 ) = 0.1094 P( ZL1.54) 9392 Plz <1-54) = 0.9382 9 2 find a Prolecc)=0.1336

F(c)- flo) = 0.1336. - Flo) - 0.5 = = 0.1336 =) fro= 0.1336 to 5 - F(c) = 0.6336 - c = f (0.6336) use command = NORMS INV (0 6336) 0.3414 b Plc< zco)= 0.4846 - FCO)- FIC) = 0.4846 - Flo= 0.5 -0.484 6 Flo) 0.0154 c = = NORMS INV (0.0154) c = - 2.1596 pl-cc2cc)=0.4648 F(c) F(-C) = 0.4648 we know z z is symmetric fco- - F(-6) from above ean n (1-fl-()) - F(-6)=0.4648 1- El-o-fr-) = 0.4698 1-2 f(-) = 0.4 648 2 El-c) = -0.4648 Ec-c- o 535 2 - FC-0 = 0.2676

:.-c = f'co 2676) -0.6200 & 0.62 ds P( 27 ) = 0.6071 1 - P(Z <C) = 0.6071 Przec) = I-0.607 L F(c) = 0.3929 :c= ico.3929) use Commons 1. = -0.2718 -> P1z7c = 0.0525 I-Plzcc) = 0.0525 przc(): -0.0525 Pize() = o 9475 c = Fco.9475) c = 1.6210 Plz (c) = 0.1068 = 0.1048 c = c = -1.25 47