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
Accepted Solution
A:
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=59.98b)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 highYes the carrier's highest data speed is significant because it is greater than 2.