Answer
31.2 < $\mu$ < 31.6
Work Step by Step
c =0.95, $\bar{x}$ = 31.39, $\sigma$ = 0.8, n = 82
Formula for CI:
$\bar{x}$ $\pm$ ( z$\times$$\frac{\sigma}{\sqrt n}$)
1) The z score that corresponds to a 95% confidence level is 1.96
2) Find the margin of error
1.96 x $\frac{0.8}{\sqrt 82}$ $\approx$ 0.173
3) Find the bounds of the interval
31.39 - 0.173 = 31.217
31.39 + 0.173 = 31.563
Therefore the Confidence interval is
31.217 < $\mu$ < 31.563
$\approx$ 31.2 < $\mu$ < 31.6
