Answer:
0.621%
Step-by-step explanation:
First of all let's find the z-score which is given by the formula;
z = (x' - μ)/σ
We are given;
Mean; μ = 80 g
Standard deviation; σ = 2 g
Test score; x' = 85 g
Thus;
z = (85 - 80)/2
z = 5/2
z = 2.5
From the z-distribution table attached, at z = 2.5, the probability is:
P(z > 2.5) = 1 - 0.99379 = 0.00621 or 0.621%