Confidence Interval Question - Confused on Approaching the Problem My question is almost exactly the same as another one here on math stack exchange, but it isn't as explanatory as I'd like it to be with some parts. I was unsure of whether or not it'd be "okay" to repost the question, especially since the other one was asked about three years ago.
Question is:
An electric scale gives a reading equal to the true weight plus a random error that is normally distributed with mean 0 mg and standard deviation = 0.1 mg. Suppose that the results of five successive weighings (in mg) of the same object are as follows:
    3.142, 3.163, 3.155, 3.150, 3.141.
a) Compute a 95 percent confidence interval estimate of the true weight.

b) Compute a 99 percent confidence interval estimate of the true weight.

Based on the answer that was given, we first find the sample mean, which is just:
$(3.142+3.163+3.155+3.150+3.141)/5.$
Then, we find the standard deviation, which is where I get some confusion. We're supposed to use the "other" standard deviation, but I don't know what the differences would be. The apparent "new" standard deviation would be:
std deviation = $(1/\sqrt5) * .1$
Then, we multiply that by plus or minus 1.96. However, I am completely confused as to how 1.96 is obtained. There is a table we can consult, but it seems that there are different tables depending on whether or not the confidence interval is two-sided - in which case, how would we know if the CI is two-sided?
I hope I'm being clear enough here, a slow explanation would be very helpful.
 A: I believe you are supposed to take $\sigma = 0.1$ and compute
a z-interval $\bar X \pm 1.96\sigma/\sqrt{n}.$
This formula for a 95% confidence interval arises from the normality of the data, whence
$$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim Norm(0,1).$$
Then from tables of the standard normal distribution, you
have $$P(-1.96 \le Z \le 1.96) = 0.95.$$
The number 1.96 cuts 2.5% of the area from the upper tail
of the standard normal density curve, and -1.96 cuts
2.5% from its lower tail, leaving 95% of the area between
$\pm$ 1.96.
Notes: 
(1) I think your data are intended to be results
of five successive weighings of the same object on a scale
that does not give exactly the same result every time.
(2) A two sided confidence interval (CI) gives an upper and a lower
bound. It can be written as $\bar X \pm 1.96\sigma/\sqrt{n}$
or as 
$$(\bar X - 1.96\sigma/\sqrt{n},\;\bar X + 1.96\sigma/\sqrt{n}).$$
(3) For a 99% CI use 2.576 instead of 1.96. Hope you can find these numbers in your normal table.
(4) If you were not given the value of the population standard deviation $\sigma$ you would need to
compute the sample deviation $S =  0.0092$ and find a CI using
Student's t distribution.
