Compare performance using standard scores

The model’s second parameters are variances. The two standard scores are

$z_1=\frac{60-55}{\sqrt{15}}\approx 1.291,\qquad z_2=\frac{65-63}{\sqrt{12}}\approx 0.577.$

These reproduce both source calculations; the square roots are visible in the printed equations even though plain text extraction may drop them.

Both z-scores are positive, so both results are above their respective mean scores. The first score is about 1.291 standard deviations above its mean; the second is about 0.577 above its mean. The higher z-score in test 1 indicates better standing relative to peers under the stated normal models, despite the lower raw percentage.

The source assumes the two tests are independent and normally distributed. Independence is not needed to calculate and compare these two individual z-scores; it would matter for a later calculation involving their joint distribution. The numerical models supplied here answer the earlier opening question without inventing missing parameters.