Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) below. TInterval \[ \begin{array}{l} (13.046,22.15) \\ X=17.598 \\ S x=16.01712719 \\ n=50 \end{array} \] a. Express the confidence interval in the format that uses the "less than" symbol. Given that the original listed data use one decimal place, round the confidence interval limits accordingly. 13.05' Mbps $<\mu<22.15$ Mbps (Round to two decimal places as needed.) b. Identify the best point estimate of $\mu$ and the margin of error. The point estimate of $\mu$ is $\square$ Mbps. (Round to two decimal places as needed.)

Step 1 ：Given the confidence interval is (13.046,22.15), we round the limits to one decimal place as per the original data. This gives us the interval (13.0, 22.1). So, the confidence interval in the format that uses the 'less than' symbol is \(13.0 \text{ Mbps} < \mu < 22.1 \text{ Mbps}\).

Step 2 ：The best point estimate of the mean (\(\mu\)) is simply the sample mean (\(\bar{X}\)) which is given as 17.598. Rounding this to one decimal place gives us 17.6 Mbps.

Step 3 ：The margin of error can be calculated as half the width of the confidence interval. The width of the interval is the difference between the upper limit and the lower limit, which is 22.15 - 13.046 = 9.104. Half of this is 4.552. Rounding this to two decimal places gives us 4.55.

Step 4 ：Final Answer: The confidence interval in the format that uses the 'less than' symbol is \(13.0 \text{ Mbps} < \mu < 22.1 \text{ Mbps}\). The best point estimate of \(\mu\) is \(\boxed{17.6 \text{ Mbps}}\) and the margin of error is \(\boxed{4.55}\).