Question 913353
Population:μ = 100 and σ = 30 
sample size 36
a) SE = 30/sqrt(36) = 5
b) P(z > (92-100)/5) = normalcdf( -1.6, 100)
c) P(z > (108-100)/5) = normalcdf( 1.6, 100)
d) P(z < -4/5) - P( z < 4/5) = normalcdf(-.8,.8)
e) 5invNorm(.99) + 100 = X (probability is .01 of obtaining one as high or higher than X)