Answer:
The highest speed measured was 78.2 Mbps.
n = 50
[tex]\bar {x}=18.22[/tex]
[tex]s= 23.87[/tex]
a)What is the difference between carrier's highest data speed and the mean of all 50 data speeds?
= 78.2 - 18.22
=59.98
b)How many standard deviations is that [the difference found in part (a)]
= [tex]\frac{difference }{s} = \frac{59.98}{23.87}=2.5127[/tex]
c) Convert the carrier's highest data speed to a z score.
[tex]z=\frac{78.2 - 18.22}{23.87}[/tex]
[tex]z=2.512[/tex]
d) If we consider data speeds that convert to z scores between minus2 and 2 to be neither significantly low nor significantly high
Yes the carrier's highest data speed is significant because it is greater than 2.
