Question 1109233
Assuming we know the population sd, use z - test, where z=(x-mean)/sd
z> (3.5-3.11)/0.3 or > 0.39/0.3 or > +1.3
That probability is 0.0968 so the top 9.68% or 9.7%
The 95% percentile is z=1.645
1.645=(x-3.11)/.3
0.4935=x-3.11
x=3.60 GPA.