How do I solve this probability problem? Here is the full question:
The lengths of all machine parts made by a company have a distribution that is skewed to the right with a mean of 68 mm and a standard deviation of 4 mm. Find the probability that the mean length of a random sample of 100 parts produced by this company would be:
(a) Less than 66.8 mm.
(b) Between 66.5mm and 67.7mm.
(c) Within 0.6mm of the populations mean.
(d) Lower than the population mean by 0.5mm or more.
I answered parts (a) and (b), and got them to be 0.13% and 22.65% respectively. I just don't get what parts (c) and (d) mean. In part (c), I don't get what 'within' implies, and in part (d) I am not sure if I am supposed to decrease by 0.5mm then solve it as in part (a) or not. 
 A: In spite of the skewness of the population, we hope that the Central Limit Theorem guarantees that sample mean of $n = 100$ observations is approximately
distributed as $\mathsf{Norm}(\mu = 68, \sigma = 4/\sqrt{100} = 0.4).$
If you have access to statistical software or a statistical calculator you
can find the desired probabilities without "standardizing," which is
necessary if you use printed normal tables. For example, in R statistical
software the normal CDF is pnorm with appropriate parameters to show
$\mu$ and $\sigma.$
For example, to check your answer to part (a): $P(\bar X < 66.8) = .00135.$
 pnorm(66.8, 68, .4)
 ## 0.001349898

For part (c), you want $P(68-.6 < \bar X < 68 + .6) = P(67.4 < \bar X < 68.6) = 0.8664.$
 pnorm(68.6, 68, .4) - pnorm(67.4, 68, .4)
 ## 0.8663856

In the sketch below, you want the area under the normal curve between
the vertical broken red lines. By hand, you can't make such an accurate
sketch, but you should always try to make a rough sketch to keep you
from making needless errors.

This is, indeed, the same as the probability that a standard normal random
variable is within 1.5 standard deviations of its mean 0. Using printed
tables, you should be able to get very close to this answer. In R, it looks like this (no extra parameters are needed for the standard normal distribution):
pnorm(1.5)-pnorm(-1.5)
## 0.8663856

For part (d), you want $P(\bar X < 68-.05 = 67.5):$
pnorm(67.5, 68, .4)
## 0.1056498

Of course, you should use whatever method your course requires,
but I hope you will come close to the numbers I have given.
Also, you should understand that the mean of a sample from a
skewed population distribution is only approximately normal.
Unless the population is extremely skewed, I would trust the first two decimal places to be correct.
A: Part C, I assume, means that you take the 0.6mm it meantions and divide that by the s.d. of 4, to get 0.15 standard deviations from the mean, then use phi on 0.15, to get a probability over 50%
BUT, that probability assumes you want everything under 0.15 s.d. above the mean, but you only want the 0.15 s.d. either side of the mean, so you take 1 - 2*phi(0.15), this removes everything bar the 0.15 s.d. in the middle.
Then part D is similar, 0.5/4 to get 1/8 standard deviations, then 1- phi(1/8), as you are looking exclusively for data below the mean, by 1/8 of a  standard deviation.
