Question

researchers measured the data speeds for a particular smartphone carrier at 50 airports. the highest speed measured was 75.6 mbps. the complete list of 50 data speeds has a mean of $\bar{x}=16.24$ mbps and a standard deviation of $s = 32.78$ 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 for part a

Subtract mean from highest speed.
$75.6 - 16.24$

Step2: Calculate number of standard - deviations for part b

Divide the difference by standard deviation.
$\frac{75.6 - 16.24}{32.78}$

Step3: Calculate z - score for part c

The z - score formula is $z=\frac{x-\bar{x}}{s}$, where $x$ is the value, $\bar{x}$ is the mean and $s$ is the standard deviation. Here $x = 75.6$, $\bar{x}=16.24$ and $s = 32.78$, so $z=\frac{75.6 - 16.24}{32.78}$

Step4: Determine significance for part d

Compare the z - score with the range $- 2$ to $2$.

Answer:

a. $59.36$
b. $\frac{59.36}{32.78}\approx1.81$
c. $\frac{75.6 - 16.24}{32.78}\approx1.81$
d. No, since the z - score ($z\approx1.81$) is between $-2$ and $2$, the carrier's highest data speed is not significant.