Central Limit theorem solving for n with absolute value First off, I'd like to apologize for the lack of latex in this question. I'm not understanding a crucial step in a homework problem I have. It states: 


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*Suppose that a measurement has mean $\mu$ and variance $\sigma^2 = 25$. Let $X$ be the average of $n$ such independent measurements. How large should $n$ be so that $P(|X − \mu| < 1) = .95$?


I've worked on it extensively and got stuck. I found a couple of solutions online which determined that $n=96$. I obviously understand that using the CLT is critical, therefore you get, from the above inequality:
$.95 = P(|X − \mu|/(\sigma/\sqrt{n}) < 1/(\sigma/\sqrt{n}))$, simplifying to:
$$.95 = P(|Z| < 1/(5/\sqrt{n}))$$
Somehow, the solution jumps from that last bolded step to:
$$.95 = P(|Z| < 1.96)$$
I have a feeling that this somehow involves a Z-table but for the life of me, I can't understand how these two steps are connected. Past this step, I completely understand the steps to solve: you just set $1.96 = 1/(5/\sqrt{n}))$ and get $n=96.04$, essentially $96$. 
What am I missing here? And is there a way to do it without a Z-table?
Thanks so much for the help! Cheers!
 A: You have a reasonable start on this, so let me try to nudge you over the finish line. First, a restatement of the problem to make sure the lack of TeX notation did not lead to confusion. You are going to sample at random enough observations $n$ from a normal distribution with unknown population mean $\mu$ and known population variance $\sigma^2 = 25$ (population standard deviation $\sigma = 5$), so that
$P\{|\bar X -\mu| < 1\} = .95,$ where $\bar X$ is the sample mean of your $n$ observations.
We  express your key event $A$ in a form that contains a standard normal random
variable $Z$ as follows
$$A = \{|\bar X - \mu| < 1\} = \{-1 < \bar X - \mu < 1\} = 
\{\sqrt{n}/\sigma < \frac{\bar X - \mu}{\sigma/\sqrt{n}} = Z < \sqrt{n}/\sigma\}.$$
Now, in order for $P(A) = .95$, you need  $\sqrt{n}/\sigma = \sqrt{n}/5 = 1.96.$
I see you have found that $n = 96.04.$ (Some people might want this "rounded up" to $n = 97$ just to make sure the original inequality holds (with a bit of room to spare.)
Addendum: Problem seems to have been edited. If you will look for 1.96 in the margin of a normal table, you will find the corresponding value in the body of the table is .9750 (or maybe .4750 depending on the style of the table).
This indicates there is 1-.9750 = .0250 = 2.5% (or .5 - .4750 = .0250) in the right
tail of the standard normal distribution beyond 1.96. By symmetry, there is
another 2.5% area below - 1.96. That leaves 95% of the area between -1.96 and 1.96, or $P\{|Z| < 1.96\} = .95$. The Empirical Rule says there is about 95% of the area under a "mound shaped" (nearly normal) curve between $\pm$ 2 (close to 1.96). In R software 'pnorm(-1.96)' returns 0.02499790 (essentially 2.5%). There is no
way other than 'numerical integration' (intensive computation) to find areas under the normal curve. (Standard methods of calculus don't work.)
