Question

researchers measured the data speeds for a particular smartphone carrier at 50 airports. the highest speed measured was 71.5 mbps. the complete list of 50 data speeds has a mean of $\bar{x}=18.44$ mbps and a standard deviation of $s = 20.17$ mbps.
a. what is the difference between carriers 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 carriers highest data speed to a z - score.
d. if we consider data speeds that convert to z scores between - 2 and 2 to be neither significantly low nor significantly high, is the carriers highest data speed significant?
a. the difference is $square$ mbps.
(type an integer or a decimal. do not round )

Explanation:

Step1: Calculate difference in part a

Subtract mean from highest speed.
$71.5 - 18.44$

Step2: Calculate number of standard - deviations in part b

Divide the difference by standard deviation. Let the difference be $d = 71.5 - 18.44$, then $\frac{d}{s}=\frac{71.5 - 18.44}{20.17}$

Step3: Calculate z - score in part c

The z - score formula is $z=\frac{x-\mu}{\sigma}$. Here, $x = 71.5$, $\mu=18.44$, $\sigma = 20.17$, so $z=\frac{71.5 - 18.44}{20.17}$

Step4: Determine significance in part d

Check if the z - score is within the range of - 2 and 2.

Answer:

a. $53.06$
b. $\frac{53.06}{20.17}\approx2.63$
c. $\frac{71.5 - 18.44}{20.17}\approx2.63$
d. Yes, since the z - score ($\approx2.63$) is greater than 2, the carrier's highest data speed is significant.