Question

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 74.9 Mbps. The complete list of 50 data speeds has a mean of x = 17.97 Mbps and a standard deviation of s = 30.01 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 minus2 and 2 to be neither significantly low nor significantly​ high, is the​ carrier's highest data speed​ significant?

Answer 1

Given The highest speed measured was 74.9 Mbps.

Given The complete list of 50 data speeds has a mean of x = 17.97 Mbps

Given standard deviation S = 30.01 Mbps

Number of airports n = 50

Question (a)

Difference between​ carrier's highest data speed and the mean of all 50 data​ speed = 74.9 - 17.97

= 56.93 mbps

Question (b)

Standard deviation S = 30.01 Mbps

Difference between​ carrier's highest data speed and the mean of all 50 data​ speed = 56.93

So it is =  (56.93/30.01) standard devations

= 1.897034

So Difference between​ carrier's highest data speed and the mean of all 50 data​ speed is 1.897034 times standard deviation

Question (c)

Carrier's high speed measured was 74.9 Mbps which is X

Z-score = (X - Mean) / Standard Deviation

= (74.9 - 17.97) / 30.01

= 56.93 / 30.01

= 1.897034

So z-score for carrier's highest data speed is 1.897034

Question (d)

Given that we consider data speeds that convert to z scores between -2 and 2 to be neither significantly low nor significantly​ high

Here our z-score of 1.897034 is between -2 and 2. Hence the carrier's highest data speed is neither significantly low nor significantly high