Question

researchers measured the data speeds for a particular smartphone carrier at 50 airports. the highest speed measured was 78.8 mbps. the complete list of 50 data speeds has a mean of x = 18.71 mbps and a standard deviation of s = 19.99 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 60.09 mbps.
(type an integer or a decimal. do not round.)
b. the difference is standard deviations.
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the difference in part (a)

The formula for the difference between two values is \( \text{Difference}=\text{Highest value}-\text{Mean}\).
Given the highest value \(x = 78.8\) Mbps and the mean \(\mu=18.71\) Mbps.
\(78.8 - 18.71=60.09\) Mbps.

Step2: Calculate the z - score in part (b)

The z - score formula is \(z=\frac{x-\mu}{s}\).
Here \(x = 78.8\), \(\mu = 18.71\), and \(s = 19.99\).
\(z=\frac{78.8 - 18.71}{19.99}=\frac{60.09}{19.99}\approx3.01\)

Step3: Analyze the significance in part (d)

A z - score is considered significant if \(|z|\geq2\).
Since \(z = 3.01>2\), the carrier's highest data speed is significant.

Answer:

a. \(60.09\) Mbps
b. \(3.01\)
d. Yes, the carrier's highest data speed is significant.