Find size $n$ of a normal sample for a given error in estimating $\mu$ Distribution of the asset level of T city's workers household is known to follow the normal distribution from past investigation. The standard deviation of this distribution is 3.6 million yen. When estimating the average of this population by sample mean, we want to set the probability that the estimate error does not become larger than 100,000 yen to 0.8. How much sample is necessary for that
i want to know 
in inequalities
$P( |\overline{X}-  \mu| \le 0.1)=0.8$ 
why $|\overline{X}-  \mu|  $has to be in absolute? 
thanks!
 A: If $X_i$ are independently and identically distributed as $\mathsf{Norm}(\mu, 
\,\sigma),$ then $\bar X \sim \mathsf{Norm}(\mu,\, \sigma/\sqrt{n}).$
Thus $Z =\frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim \mathsf{Norm}(0,1).$
Then from printed tables of the standard normal distribution (or software),
we have
$$0.80 = P\left(-1.282 < \frac{\bar X - \mu}{\sigma/\sqrt{n}}< 1.282\right)\\
= P\left(-1.282\frac{\sigma}{\sqrt{n}} < \bar X - \mu <1.282\frac{\sigma}{\sqrt{n}}\right)\\
= P\left(| \bar X - \mu | < \frac{1.282\sigma}{\sqrt{n}}\right).$$
You know $\sigma,$ so set $\frac{1.282\sigma}{\sqrt{n}} = 0.1$ and solve for $n.$
Notes: (1) An 80% confidence interval (CI) for $\mu$ is of the form $\bar X \pm 1.282 
\frac{\sigma}{\sqrt{n}},$ where $\pm 1.282 
\frac{\sigma}{\sqrt{n}}$ is sometimes called the 'margin of error' of the CI, so your question is equivalent to finding $n$ so that an 80% CI has margin of error $\pm 0.1.$ [Perhaps see also.] (2) In R statistical software the computation to find a cut-off value $c$ such that $P(Z < c) = 0.9$ and $P(-c < Z < c) = 0.8,$ for standard normal $Z,$ is as follows:
qnorm(.9)
## 1.281552

