What percentage of the area under a normal curve lies within the following boundaries? (use two decimal places)

a. between the mean and a z of 1.27
b. between the mean and a z of -0.91
c. between the z scores of -1.84 and 1.39
d. between the z scores of -258 and 2.58
e. above a z of 1.96
f. below a z score of -2.58
g. above a z score of -1.72
h. below a z of 1.96

All you have to do is to find the following from the z-score table:
On the 1st row of the table, you can read the value of z & you complete the value by reading the corresponding horizontal line. However watch out which table you are using & what is requested from you. For example in the following question:

a. Between the mean and a z = 1.27: , if you use a normal table it will give you the value of ==> 0.8990==>89.8%, which is the area from the EXTREME LEFT TO z=1.27., whereas yo have to calculate the area from the mean to
 to z=1.27 0r 0.8990 -.5=0.398 or 39.85 (.5, because you start counting from the mean):
so:
a. Between the mean and a z = 1.27 ==> 39.8%

b. Between the mean and a z = -0.91 ==> 0.31859==>18.31%
c. Between the z scores of -1.84 and 1.39 . In this case you will find the value of each z & you will add up both: for z= -1.84 ==> 0.88482 or 88.48%
d. Between the z scores of -2.58 and 2.58: same as above,==>99.1%
e. Above a z of 1.96 , means from z to the end. The area from the mean to z=1.96 ==> 0.475, what wee need is from this point to the end
that is 0.5-0.475=2.5%
f. Below a z score of -2.58 same as above but -2.58 to the extreme left
for z=-2.58 ==>0.49506 , & below this area==> 0.5-0.49506==> 0.49%
g. Above a z score of -1.72 ==> 95.72%
h. Below a z of 1.96 ==>97.5%

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