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=17.98 Mbps and a standard deviation of s =35.53 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

a) The difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is of: 54.62 Mbps.

b) This difference represents 1.54 standard deviations.

c) The z-score is of z = 1.54.

d) The carrier's highest data speed​ is not significant, as it is less than 2 standard deviations from the mean.

How to obtain the measures?

The parameters to this problem are given as follows:

• Highest speed: 72.6 Mbps.

• Mean: 17.98 Mbps.

• Standard deviation: 35.53 Mbps.

Hence the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is obtained as follows:

72.6 - 17.98 = 54.62 Mbps.

Then we calculate the z-score, which is the division of this difference by the standard deviation, giving how many standard deviations the data is from the mean.

z = 54.62/35.53

z = 1.54

Hence the measure is 1.54 standard deviations from the mean, which is not unusual, as it is less than 2 standard deviations from the mean.

More can be learned about z-scores at https://brainly.com/question/25800303

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