Since we want to be 95% confident that our interval contains the true value of mu, we set alpha = 0.05. Then the 100(1-alpha)% confidence interval for mu is found by the formula:
bar x +- t_(alpha//2, n–1)sigma/sqrtn
=137 " ft"+-t_(0.025,31)(7.0" ft")/(sqrt 32)
The t-value comes from Student's t-distribution that has n-1 degrees of freedom; the area under the t-curve to the right of that t-value is 0.025. The value for t_(alpha//2, n–1) is usually found by lookup in a t-table, but for values of n>30, we know t_(alpha//2, n–1) ~~ z_(alpha//2), so we can look up this z-value instead. (As n grows larger, t approaches the standard normal distribution Z.)
So our 95% confidence interval for mu becomes
137 " ft"+-z_(0.025)(7.0" ft")/(sqrt 32)
~~137 " ft"+-(1.96)(1.237" ft")
~~137 " ft"+-2.425 " ft".
