# How do you find the z-score for which 98% of the distribution's area lies between -z and z?

$z = 2.33$
You need to look this up from a z-score table (e.g. http://www.had2know.com/academics/normal-distribution-table-z-scores.html) or use a numerical implementation of the inverse normal distribution cumulative density function (e.g. normsinv in Excel). Since you desire the 98% percent interval you desire 1% on each side of $\pm z$, look up 99% (0.99) for $z$ to obtain this.
The closest value for 0.99 on the table gives $z = 2.32$ on the table (2.33 in Excel), this is your $z$ score.