Question

researchers measured the data speeds for a particular smartphone carrier at 50 airports. the highest speed measured was 76.9 mbps. the complete list of 50 data speeds has a mean of $\bar{x}=15.66$ mbps and a standard deviation of $s = 21.84$ 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 61.24 mbps. (type an integer or a decimal. do not round.)
b. the difference is 2.8 standard deviations. (round to two decimal places as needed.)
c. the z score is $z=square$. (round to two decimal places as needed.)

Explanation:

Step1: Recall z - score formula

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the data - point, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step2: Identify values

We know that $x = 76.9$ Mbps (highest data speed), $\mu=15.66$ Mbps (mean of data speeds), and $\sigma = 21.84$ Mbps (standard deviation of data speeds).

Step3: Calculate z - score

$z=\frac{76.9 - 15.66}{21.84}=\frac{61.24}{21.84}\approx2.80$

Step4: Determine significance

Since the z - score $z = 2.80>2$, the carrier's highest data speed is significantly high.

Answer:

a. 61.24
b. 2.80
c. 2.80
d. Yes, the carrier's highest data speed is significant.