Given that z is a standard normal random variable, find z for each situation.
a. the area to the left of z is .9750
b. the area between 0 and z is .4750
c. the area to the left of z is .7291
d. the area to the right of z is .1314
e. the area to the left of z is .6700
f. the area to the right of z is .3300 (hint: normal probability dist)

You have to use your standard normal table (or calculator) for these. The area under the curve should be 1, a z-value corresponds to the probability and is area to the left of that z.

a. A simple reverse lookup of 0.9750. z=1.96

b. Total area up to z would be 0.5+0.4750, so reverse look up 0.9750 and find z=1.96

c. A simple reverse lookup. z=0.61

d. Area to the left of z would b 1-0.1314, so lookup 0.8686 and find z=1.12

e. A simple reverse lookup. z=0.44

f. Just invert the area: 1-0.33 = 0.67 and reverse lookup: z=0.44