How do you find the z-score for which 95% of the distribution’s area lies between -z and z?

Using a z-score table: The area of the tails must be .05 (which is 1 – .95), and each tail must be .05/2, or .025. Find .025 in the interior part of a z-score table. See that it corresponds to a z-score (the numbers in the margins) of 1.96. That is the answer. Using a TI-83 or TI-84 graphing calculator: 2nd VARS > invNorm(.025,0,1) gives you -1.96, which is -z. So, z = 1.96.