Question

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 78.2 Mbps. The complete list of 50 data speeds has a mean of x overbarequals18.22 Mbps and a standard deviation of sequals23.87 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​ carr

Answer 1

Answer:

The highest speed measured was 78.2 Mbps.

n = 50

$\bar {x}=18.22$

$s= 23.87$

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)]

= $\frac{difference }{s} = \frac{59.98}{23.87}=2.5127$

c) Convert the​ carrier's highest data speed to a z score.

$z=\frac{78.2 - 18.22}{23.87}$

$z=2.512$

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.