Answer:
2.275% 
Explanation:
We start by calculating the z-score 
Mathematically;
z-score = (x-mean)SD
here, x = 70.5
mean = 65.5
SD = 2.5
Substituting these values;
z = (70.5-65.5)/2.5
= 5/2.5 = 2
So we want to calculate the probability that;
P( z> 2)
We check the standard normal distribution table for this 
That will be 
0.02275 
In percentage, this is 2.275 %