Hi
mean of 35 and standard deviation of 7 minutes
Sample size 9, s = 7/sqrt(9) = 7/3
a) P(x > 40), z = 5/(7/3) = 15/7 = 2.1429, P(z > 2.1429) = .0161 0r 1.61%
b) P(32 ≤ x ≤ 37) = normalcdf(32,37,35, 7/3)
c) invNorm(.10) = -1.282
-1.282 = (X - 35)/(7/3)
(2.333)(-1.282) + 35 = X