Thirty randomly selected students took the calculus final. if the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval of the mean score of all students. assume that the population has a normal distribution

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

X> -2 is the simplified version.