Answer
9.7 < $\mu$ < 11.3
Work Step by Step
c =0.99, $\bar{x}$ = 10.5, $\sigma$ = 2.14, n = 45
Formula for CI:
$\bar{x}$ $\pm$ ( z$\times$$\frac{\sigma}{\sqrt n}$)
1) The z score that corresponds to a 99% confidence level is 2.58
2) Find the margin of error
2.58 x $\frac{2.14}{\sqrt 45}$ $\approx$ 0.823
3) Find the bounds of the interval
10.5 - 0.823 = 9.677
10.5 + 0.823 = 11.323
Therefore the Confidence interval is
9.677 < $\mu$ < 11.323
$\approx$ 9.7 < $\mu$ < 11.3
