Problem
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 73.5 Mbps. The complete list of 50 data-speeds has a mean of $\bar{x}=17.37 \mathrm{Mbps}$ and a standard deviation of $\mathrm{s}=17.33 \mathrm{Mbps}$. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. a. The difference is $\square$ Mbps. (Type an integer or a decimal. Do not round.)
Solution
Step 1 ：Given that the highest data speed is 73.5 Mbps and the mean of all 50 data speeds is 17.37 Mbps.
Step 2 ：Calculate the difference between the highest data speed and the mean of all 50 data speeds by subtracting the mean from the highest data speed: \(73.5 - 17.37 = 56.13\) Mbps.
Step 3 ：Final Answer: The difference between the carrier's highest data speed and the mean of all 50 data speeds is \(\boxed{56.13}\) Mbps.
