Question

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 72.6 Mbps. The complete list of 50 data speeds has a mean of x=18.55 Mbps and a standard deviation of s=19.69 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?

Answer 1

Answer:

a. 54.05 Mpbs.

b. 2.745... standard deviations.

c. The z-score is 2.745....

d. The carrier's highest data speed is significantly high.

Step-by-step explanation:

a. The difference between the highest measured data speed and the mean is 72.6 - 18.55 =  54.05 Mbps.

b. The amount of standard deviations of 54.05 Mbps is equal to this value divided by the standard deviations, so we yield $\frac{54.05}{19.69} = 2.745...$ standard deviations.

c. The z-score is equal to the difference between the mean and a data point in standard deviations, so the z-score is 2.745....

d. 2.745... is not between -2 and 2, so the carrier's highest data speed is not insignificant - so it's significantly high.