Determine the area under the standard normal curve that lies between (a) z=-1.36 and z=1.36 (b) z=-1.98 and z=0, and (c) = 1.14 and z=-0.33. The area that lies between z=-1.36 and z=1.36 is

(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) = 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) = 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) = 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