Normal Distribution / Central Limit Theorem I have a problem with some questions that say "calculate an interval within which 90% of the __ will lie". 
I understand the how to find 90% confidence INTERVAL, but somehow when I subbed 1.645 into the equation, it does not give the limit of $μ $. When I did it backwards, the values of z was $±$5.46, which was out of the charts I read off z-scores. 
Can anyone explain why? Am I missing a theory? Thanks
Here's the question for reference, just the last bit, 95% I can do

 A: The source of confusion seems to be over the application of
a "confidence interval," so I'll try to start from basic principles.
In the first part of the problem (which explicitly asked for a $95$%
confidence interval),
you know some facts about the population of pink bars:
you know their masses are normally distributed, and you even know the
standard deviation of that distribution,
but you do not know the mean of that distribution.
Hence the problem is to make an intelligent guess as to what that mean might be.
Since you are unlikely to guess the mean exactly
(especially as you have only twelve measurements to go on),
your guess is a range of values, and that's your confidence interval.
A key fact about a confidence interval is that there is actually only
one value (the unknown actual mean of the population) it is designed to contain.
The fact that other values might fall within that range is not really important.
For the second part, you have a known mean and standard deviation;
in fact, you know essentially everything there is to know about the
weights of the general population of white bars.
In particular, you know (in principle, at least)
the cumulative distribution function $F(x)$ of the weights of white bars.
One particularly nice way to "bracket" the weights of $90$% of the bars
is first to choose a weight $x_H$ such that $F(x_H) = 0.95$.
This means that $95$% of the bars weigh less than $x_H$.
That's more than we need, so we next choose 
a weight $x_L$ such that $F(x_L)=0.05$;
then $5$% of the original population of bars weigh less than $x_L$
if we discard those bars from the set that weigh less than $x_H$,
we're left with $90$% of the original population.
In other words, the interval we're looking for is $(x_L, x_H)$
such that $F(x_L)=0.05$ and $F(x_H)=0.95$.
To figure out what $x_L$ and $x_H$ might be, we observe that
if $X$ is a random variable distributed like the weights of white bars,
then $X = 6.46\,Y + 176.2,$ where $Y$ is a standard normal random variable.
We know (or can determine from a table)
that $95$% of the standard normal distribution is less than $1.645$
and $5$% is less than $-1.645$, so we can plug in $-1.645$ and $1.645$
for $Y$ in the previous equation and read off the values of $x_L$ and $x_H$.
By the way, this particular interval is symmetric around the mean,
which gives it some nice properties. It is also possible to describe intervals
that are not symmetric around the mean, but unless we have a particular reason
to do that, I would use the symmetric interval.
