Answer
20 < $\mu$ < 21.2
Work Step by Step
c =0.80, $\bar{x}$ = 20.6, $\sigma$ = 4.7, n = 100
Formula for CI:
$\bar{x}$ $\pm$ ( z$\times$$\frac{\sigma}{\sqrt n}$)
1) The z score that corresponds to a 80% confidence level is 1.28
2) Find the margin of error
1.28 x $\frac{4.7}{\sqrt 100}$ $\approx$ 0.602
3) Find the bounds of the interval
20.6 - 0.602 = 19.998
20.6 + 0.602 = 21.202
Therefore the Confidence interval is
19.998 < $\mu$ < 21.202
$\approx$ 20 < $\mu$ < 21.2
