Evaluate the following for a standard normal probability distribution, drawing an appropriate picture ... Evaluate the following for a standard normal probability distribution, drawing an appropriate picture.
A.) Find the standard normal z-value so that the area that lies to the right of z is 0.2119.
For part A, I did P(Z > 0.8) = 1 - P(Z < 0.8) = 0.2119 and shaded everything to the right of 0.8 on my drawing.
B.) Find the standard normal z-value so that the area between –z and z is 0.9030.
My professor ended lecture early and did not cover this material but I had this question for homework. For part A I am not sure if I am on the right track as far as my thought process and answer go. On part B I am completely lost. Any help would be much appreciated.
 A: As @Callculus42 has commented, you already have the answer to Part A. In R statistical software, where qnorm is the standard normal quantile function (invese CDF), we see that $P(Z > 0.8) = 0.2110.$ I hope you know how to get this result from a printed table of the
standard normal CDF.
qnorm(1-.2119)
[1] 0.799846

This is the area under the standard normal density curve to the right of the solid vertical black line.

R code for figure:
hdr = "Standard Normal Density"
curve(dnorm(x), -3, 3, lwd=2, col="blue", 
      ylab="Density", xlab="z", main=hdr)
 abline(h=0, col="green2")
 abline(v=0, col="green2")
 abline(v=.8)

For Part B, the lower boundary $-1.659575$ should cut probability
$(1 - .9030)/2 = 0.0485$ from the left tail of the standard
normal distribution and (by symmetry) the upper boundary $1.659575$ should cut the
same probability from the right tail.
qnorm(0.0485)
[1] -1.659575

Thus $P(-1.659575 \le Z \le 1.659575) = 0.9030.$
diff(pnorm(c(-1.659575,1.659575)))
[1] 0.903

In the figure below, $0.9030$ is the area under the
standard normal curve between the dotted red lines at $\pm 1.659575.$

hdr = "Standard Normal Density"
curve(dnorm(x), -3, 3, lwd=2, col="blue", 
      ylab="Density", xlab="z", main=hdr)
 abline(h=0, col="green2")
 abline(v=0, col="green2")
 abline(v = c(-1.659575,1.659575), col="red", lwd=2, lty="dotted")

Note: Both of these computations are of types you will need to
do repeatedly as you learn to use the standard normal distribution
in probability and statistics. It is a good idea to begin by
making sketches.
