6. Assume that z scores are normally distributed with a mean of 0 and a standard deviation of 1. Solve for each variable  

a. P(0 < z < a) = 0.3907

b. P(-b < z < b) = 0.8664

c. P(z > c) = 0.0643

d. P(z > d) = 0.9922

e. P( z < e) = 0.4500