Question

eBook Video Given that z is a standard normal random variable, find z for each situation (to 2 decimals). a. The area to the left of z is 0.209. (Enter negative value as negative number.) -0.81 b. The area between – z and z is 0.905. 1.1553 c. The area between – z and z is 0.2128 . d. The area to the left of z is 0.9951 . -2.58 e. The area to the right of z is 0.6915. (Enter negative value as negative number.) 0.50

Answer 1

(a) The area to the left of z is 0.209. That is, P(Z <=)=0.209 .-(1) From (1). (Using Excel function = = [= NORMSINV (0.209] =-0.81 (b) The area between -- and z is 0.905. That is, P(-25Z :)=0.905 PZS:)-P(ZS-=)=0.905 P(Z <:)-[1-P(Z <:)]=0.905 2P(Z <=)-1=0.905 2P(Z <=)=1-0.905 2P(Z <=)=0.095 PIZ33) -0.095 P(Z <=)=0.0475 (2)

From (2) ==1.67 Using Excel function, |(= NORMSINV(0.0475))=1.67 The area between -- and z is 0.2128 That is, P(-:SZS:)=0.2128 PZS:)-P(Z <--) = 0.2128 P(Z <:)-[1-P(Z <:)]=0.2128 2P(Z <:)-1=0.2128 2P(Z <=)=1-0.2128 2P(Z <=)=0.7872 P(Z <:) - 0.7872 P(Z <=)=0.3936 (3)

From (3) ==0.27 Using Excel function, (=NORMSINV (0.3936)) = 0.27 (d) The area to the left of z is 0.9951. That is, P(Z<=)=0.9951 From (4). (Using Excel function == [= NORMSINV (0.9951)] = 2.58

The area to the right of z is 0.6915. That is P(Z >:)=0.6915 1-P(Z <=)=0.6915 P(Z <:)=0.3085 From (5). ==T=NORMSINV (0.3085) =-0.50 (Using Excel function