Answer:
Step-by-step explanation:
Researchers measured the data speeds for a particular smartphone carrier at 50 airports.
The highest speed measured was 76.6 Mbps.
n= 50
X[bar]= 17.95
S= 23.39
a. What is the difference between​ the carrier's highest data speed and the mean of all 50 data​ speeds?
If the highest speed is 76.6 and the sample mean is 17.95, the difference is 76.6-17.95= 58.65 Mbps
b. How many standard deviations is that​ [the difference found in part​ (a)]?
To know how many standard deviations is the max value apart from the sample mean, you have to divide the difference between those two values by the standard deviation
Dif/S= 58.65/23.39= 2.507 ≅ 2.51 Standard deviations
c. Convert the​ carrier's highest data speed to a z score.
The value is X= 76.6
Using the formula Z= (X - μ)/ δ= (76.6 - 17.95)/ 23.39= 2.51
d. If we consider data speeds that convert to z scores between minus−2 and 2 to be neither significantly low nor significantly​ high, is the​ carrier's highest data speed​ significant?
The Z value corresponding to the highest data speed is 2.51, considerin that is greater than 2 you can assume that it is significant.
I hope it helps!
