Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 74.4 Mbps. The complete list of 50 data speeds has a mean of x=18.26 Mbps and a standard deviation of s=22.09 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. d. If we consider data speeds that convert to z scores between −2 and 2 to be neither significantly low nor significantly​ high, is the​ carrier's highest data speed​ significant?

a.

Given that:

The highest speed =74.4 Mbps and mean of all data speeds x=18.26 Mbps

The difference = 74.4 - 18.26 = 56.14 Mbps

Answer: The difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 56.14 Mbps

b.

We divide the difference obtained in (a) above by the standard deviation to obtain the number of standard deviations

Given that the difference = 56.14 and standard deviation σ= 22.09

Then we have:

$=\frac{56.14}{22.09}$ = 2.5414

Therefore, the difference found in part​ (a) is approximately 2.5414 standard deviations above the mean

c.

z = $\frac{X-x}{s}$

Where:

X =  the​ carrier's highest data speed

x = sample mean

s = sample standard deviation

z = z score

Now,

Given that X = 74.4, x=18.26, and s = 22.09

Z = $\frac{74.4 - 18.26}{22.09}$ = 2.54

Answer: z = 2.54, which is equivalent to the number of standard deviations the value of carrier's highest data speed is from the mean

d.

Since z=2.54 is not between -2 and 2 standard deviations, then we can consider the​ carrier's highest data speed​ significantly high.