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 $\bar{x} = 17.37 Mbps$ and a standard deviation of $s = 17.33 Mbps$ .
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 $◻$ Mbps.
(Type an integer or a decimal. Do not round.)

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Final Answer: The difference between the carrier's highest data speed and the mean of all 50 data speeds is $56.13$ Mbps.

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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 $56.13$ Mbps.

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