Given a 95% confidence interval why are we using 1.96 and not 1.64? I'm studying for my test and am reviewing the solution to some example problems. The problem is:

You are told that a new standardized test is given to 100 randomly selected third grade students in New Jersey. The sample average score is $\overline Y$ on the test is 58 and the sample standard deviation is $s_y$ = 8.

The first part of the question asks you to construct a 95% confidence interval for the mean score of all New Jersey third graders.
First I found $$\text{SE}_y = \frac{s_y}{\sqrt{n}} = \frac{8}{\sqrt{10}} = 0.8.$$
Then I did
$$58 \pm (0.8)(1.64).$$
However, instead of using $1.64$ they used $1.96$. Can someone explain why?

Two reasons:
1) Students are in New Jersey.
2) the $z$-value associated with 95% is z=1.96; this can be seen as part of the 1-2-3, 65-95-99 rule that tells you that values within $ p.m 1-, 2- or 3$- deviations from the mean comprise 65- 95- or 99% of all values in the distribution  . Or look it up in a table.
