Relevance of z-score in the age of computers In this the modern age of computers, standard normal probabilities don't need to be looked up in a table. What is more, any normal probability can be easily calculated by a computer. So are normal probability tables still relevant? If not, are z-scores still relevant? To what extent?
 A: Insightful question: As one can see by the comments, this is a matter of opinion. As a textbook author in
'this modern age', I have given this issue some thought. So here is my opinion.
It is true that many routine problems involving normal distributions are easily solved using software. To an extent that statistics and probability texts of the future will start to have different kinds of drill problems. 
With access to suitable software, printed normal tables
would no longer be needed. But some problems may be
easier to solve using standard scores--as in (4) below.
Also, some non-computational conceptual discussions will
essentially use standard scores:  "In a normal distribution about 68% of the probbility lies within one standard deviation of the mean.
Four problems using normal distributions:
(1) If scores on a national test are normally distributed with mean 200 and standard deviation 30, what fraction of those taking the test will score below 173? In R, where pnorm is a normal CDF:
pnorm(173, 200, 30)
[1] 0.1840601

(2) If State U undertakes to admit applicants scoring in the top 15% on this exam, what score should be published in the Catalog as the minimal acceptable score? In R, where qnorm is an inverse CDF or quantile function:
ceiling(qnorm(.85, 200, 30))
[1] 232                      # Publish 232
1 - pnorm(231.5, 200, 30)
[1] 0.1468591                # Check: < 15%

(3) Boxes of breakfast cereal are filled automatically by machine, with an average of 15.1 oz and a standard deviation of 0.2 oz.
What percentage of the boxes will contain less than 15 oz.? From R, we get $0.3085,$ almost 31%.
pnorm(15, 15.1, .2)
[1] 0.3085375

(4) Suppose government regulations require no more than 15% of boxes to contain less than the "15. oz." printed on the box. To what average number $\mu$ of ounces should the filling machine be set? Round up to the nearest 0.01 of an ounce.
This question
may be most easily solved using standard scores:
$$ 0.15 = P(X \le 15)
= P\left(Z =\frac{X - \mu}{\sigma} \le \frac{15 - \mu}{.2}\right).$$ 
We still don't need printed normal tables. In R:
qnorm(.15, 0, 1)
[1] -1.036433

Then set $(15-\mu)/.2 = -1.036433$ to get $\mu = 15.21.$
And use R to check:
pnorm(15, 15.21, .2)
[1] 0.1468591

Using R directly would require a 'grid search' using vectors in R. (Either that or tedious trial and error.)
mu = seq(15, 17, by = .01) # guessing 17 is too high
pr = pnorm(15, mu, .2)
min(mu[pr <= .15])
[1] 15.21

