Given:
n = 30 students
sample mean (X) = 95
standard deviation (s) = 6.6
We use the t distribution.
99% interval means 0.99 = 1 - α ; α = 0.01
tα/2 = t0.005 = invT(0.995,29) = 2.75638
X - t a/2 * (s/√n) < μ < X + t a/2 * (s/√n)
95 - 2.75638*(6.6/√30) < μ < 95 + 2.75638*(6.6/√30)
95 - 3 < μ < 95 + 3
92 < μ < 98
Hope this helps :)