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)=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

5x+2x=49

7x=49

x=7. 35 apples