Question

(1 point) Find the following probabilities for the standard normal random variable z. (a) P(-0.81 <<0.42) (b) P(-1.14 <z < 0.5) (c) P(Z < 0.69) a (d) P(Z > -0.6)

Answer 1

a) Cure Honiin Irma required probabilit * -0.81 00.42 Z-scove on

6 P(-1014 € 220.5) P(Z <0.5) - P(ZL-1014) (: plac zeb) = P(Z <b) - P(Z <a)) P(220.5) From the normal distribution table vawes to the left of z p(240.5) 0.6915 P(Z <10 14) = 1- P(Z 21014) (: P(Zcat + p (ze-a)=1) P(Z < 1.14) = 0.8729 (from table values) > p (Z < -1014)= 1-0.8729 P(Zs1014) = 0.1271 => PC-lo!4 ?Z? 0:5)= 0.6915 - 0:12H -0.56 44 in P(-lolt ez 0.5) = 0.5644

b Curve required probability x -10l4 0 0.5 Z-score

© P(Z <0.69) from normal distribution table values for z = 0.69 probability = p (240.69) = 0.7549 .. p (Z 40.69) - 0.7549 d @ P (z 70.6) (: P(Z La) + p (279) =1) >> P (27-0.6) = 1-p (22-0.6) (= P(al-a)+p(2.20) = 1) = P(ZL-0:6)= 1 - P(Z L0.6) => p (27-0.6) = 1-(1-p (2 2006) 1 = P(Z <0.6) from the table values for Z=0.6 probability - Plzcoof) = 0.7257 PLZ 70.6) = 0.7257

C curve required probability 0 0.69 z-score > (a cure required probability X -Out o Z-slove