Question

Calculate the required probabilities for the normal distributions with the parameters specified in parts a through e. a. u = 5,0 = 4; calculate P(0<x< 6). P(0 < x < 6) = (Round to four decimal places as needed.) b.u = 5,0 = 5; calculate P(0<x< 6). PO < x < 6) = . (Round to four decimal places as needed.) C. M = 4,0 = 4; calculate P(0 < x < 6). P(O< x < 6) = (Round to four decimal places as needed.) d. = 6,0 = 2; calculate P(x > 1). Plx > 1) = (Round to four decimal places as needed.) e.p=0,0 = 2; calculate P(x > 1). P(x > 1) = (Round to four decimal places as needed.)

Answer 1

Solutions a H=5; O=4 P(0<2<5) - U14 .4 P(-2 5274 z2825) cum P(-1.25 ZZ P(Z <0.25) P(Z <-1-25) from mulative nomal distribution table P(Z <0.25) 0.5987 P(22- Plo2X46) 035987 -0-1056 0.4931 P(3) P (04X<6) 0:493] =0.1056 1:25) 1025 Z=0025 b 6-5 5 5 H - 5,0 oby PC02826) = P (<24 Z20.2 P(220.2). - P(22-1) golom CND table CND - Comulative Normal Distribution P (z 20.2) p(22-1) 0.1587 0-5793

AP 97%7]dE = 0.5713 - 1587 04206 0.21 PLZ6) ==Z - 0. 4206 C 114) نے را 7 5 (i-zz) - (0414) - ? ) P - <z 2055 (ZZ05) Im CND table P(ZZ055) = 0.65, P(zL-1) = 6-15 37 PO ZIA 0 6115 0:1587 0 5328 P2 Pol) 0 5328 T 0.5 아2 여 H = 2 a 1-6 P (11) ( > (z> -2-5) PZZ-25 A-P (ZZ-25) 0.0062 from CND CND table table

Plz7-25) 1- P(22-25) 1 0.0062 0.9938 APCZ) | P(x=1) 0.9938 Z=-25 Z=0 co H=0,0=2 > P(1-1) = P(2 P(270.5) -l-plzzon 1-0.6915 (:P(2205) 0-6915) 0.3085 PG=1) -0.3085 pre 790.5