Question 922148
  mean of 74 inches, and a standard deviation of 12 inches, n = 36
{{{z =blue (x - mu)/blue(sigma/sqrt(n))}}}
x= 76.8, z = 2.8//(12/sqrt(36)) = 1.4
P(xbar> 76.5) = P(z >1.4) = normalcdf(1.4, 100) = .0808  0r 8.08%
z-test was used as Population SD is known and sample size > 30
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A NEED to know is what sample size is thought to be sufficient
for using z-test WHEN Population SD is known...check with Your sources.
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Population SD unknown...t-test is generally called for.
{{{t =blue (x - mean)/blue(s/sqrt(n))}}} and tcdf function used.