Respuesta :
Answer:
(a) The area that lies between z=-1.36 and z=1.36 is 0.82616
(b) The area that lies between z=-1.98 and z=0 is 0.47615
(c) The area that lies between z=1.14 and z=-0.33 is 0.50215
Step-by-step explanation:
Determine the area under the standard normal curve that lies between:
(a) z=-1.36 and z=1.36
This is:
P(-1.36<=z<=1.36)=P(z<=1.36)-P(z<=-1.36)
P(z<=1.36)=0.91308
P(z<=-1.36)=P(z>=1.36)=1-P(z<=1.36)=1-0.91308→P(z<=-1.36)=0.08692
P(-1.36<=z<=1.36)=P(z<=1.36)-P(z<=-1.36)=0.91308-0.08692→
P(-1.36<=z<=1.36)=0.82616
(b) z=-1.98 and z=0
This is:
P(-1.98<=z<=0)=P(z<=0)-P(z<=-1.98)
P(z<=0)=0.5
P(z<=-1.98)=P(z>=1.98)=1-P(z<=1.98)
P(z<=1.98)=0.97615
P(z<=-1.98)=1-P(z<=1.98)=1-0.97615→P(z<=-1.98)=0.02385
P(-1.98<=z<=0)=P(z<=0)-P(z<=-1.98)=0.5-0.02385→
P(-1.98<=z<=0)=0.47615
(c) z=1.14 and z=-0.33
This is:
P(-0.33<=z<=1.14)=P(z<=1.14)-P(z<=-0.33)
P(z<=1.14)=0.87285
P(z<=-0.33)=P(z>=0.33)=1-P(z<=0.33)
P(z<=0.33)=0.62930
P(z<=-0.33)=1-P(z<=0.33)=1-0.62930→P(z<=-0.33)=0.3707
P(-0.33<=z<=1.14)=P(z<=1.14)-P(z<=-0.33)=0.87285-0.3707→
P(-0.33<=z<=1.14)=0.50215
