Question 871685
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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 &#8804; x &#8804; 37) = normalcdf(32,37,35, 7/3)
c) invNorm(.10) = -1.282
-1.282 = (X - 35)/(7/3)
(2.333)(-1.282) + 35 = X