Find the value of the probability of the standard normal var

Find the value of the probability of the standard normal variable Z corresponding to this area for problems 1-3.

1. (Z < 0.58)
a. 0.8962
b. 0.7190
c. 0.1317
d. 0.7489
e. 0.5412

2. P ( Z> ?1.07)
a. 0.0582
b. 0.7612
c. 0.1978
d. 0.8577
e. 0.3578

3. P(Z?0.14 < < 1.63)
a. 0.5041
b. 0.6531
c. 0.8714
d. 0.6406
e. 0.8012

Let Z be the standard normal variable. Find the values of z if z satisfies the following problems 4-6

4.  P( Z< z ) = 0.9251
a. -0.57
b. 0.98
c. 0.37
d. 1.44
e. -1.16

5.P(Z > ?z) = 0.3372
a. -0.42
b. 0.43
c. -0.21
d. 0.78
e. -056.

6. (?z < Z <z ) = 0.8882
a. 1.81
b. 1.59
c. 1.04
d. 1.44
e. 1.68

An explanation of how you got the answer would be much appreicated!

Solution

From:

     STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score.

1. (Z < 0.58)
   answer is: b) 0.7190

2. P ( Z> ?1.07)

   p(Z > 1.12) = 1 - p(Z < 1.07)

                        = 1-0.8577=0.1423 this is not accepting so
   ansrer is: d) 0.8577

3. P(Z?0.14 < < 1.63)
a. 0.5041
b. 0.6531
c. 0.8714
d. 0.6406
e. 0.8012  

find p(Z < 0.14) - p(Z < 1.63).

4.  P( Z< z ) = 0.9251
answer is d) 1.44 from   cumulative Probabilities of the Standard Normal Distribution N(0, 1)

5.P(Z > ?z) = 0.3372
answer is:a) -0.42 from    Cumulative Probabilities of the Standard Normal Distribution N(0, 1)
  6. (?z < Z <z ) = 0.8882
   answer is:e. 1.68

      z=1.8882/2=0.4442