Question

In Exercises 5–8, express all z scores with two decimal places.

ATL Data Speeds For the Verizon airport data speeds (Mbps) listed in Data Set 32 “Airport Data Speeds” in Appendix B, the highest speed of 77.8 Mbps was measured at Atlanta’s (ATL) international airport. The complete list of 50 Verizon data speeds has a mean of x = 17.60 Mbps and a standard deviation of s = 16.02 Mbps

a. What is the difference between Verizon’s data speed at Atlanta’s international airport and the mean of all of Verizon’s data speeds?

b. How many standard deviations is that [the difference found in part (a)]?

c. Convert Verizon’s data speed at Atlanta’s international airport 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 Verizon’s speed at Atlanta significant?

Short answer

a. The difference between Atlanta airport’s speed and the mean speed is equal to 60.20 Mbps.

b. The difference between Atlanta airport’s speed and the mean speed is 3.76 standard deviations

c. The z-score value for the given speed equal to 77.8 Mbps is obtained as 3.76.

d. The speed at Atlanta airport equal to 77.8 Mbps is significantly high.

Step-by-step solution

Given information

Airport data speeds are measured in Mbps for a set of 50 airports.

The mean speed is equal to 17.60 Mbps.

The standard deviation of speeds is equal to 16.02 Mbps.

Formula of  z-score

Az-score for a certain value of the data is calculated as follows:

$z=\frac{x-\overline{x}}{s}$

Here,

x is the value from the data.

$\overline{x}$is the sample mean.

s is the sample standard deviation.

Calculation

a.

The difference between the speed at Atlanta airport (x = 77.8 Mbps) and the mean speed is obtained as follows:

$$\begin{gathered} x-\overline{x}=77.8-17.60 \\ =60.20 \end{gathered}$$

Thus, the difference between the speed at Atlanta airport and the mean speed is equal to60.20 Mbps.

b.

The difference between Atlanta airport’s speed and the mean speed is obtained using the z-score as shown below:

$$\begin{gathered} z=\frac{x-\overline{x}}{s} \\ =\frac{77.80-17.60}{16.02} \\ =3.76 \end{gathered}$$

Therefore, the difference is3.76 standard deviations.

c.

The z-score can be calculated as:

$$\begin{gathered} z=\frac{x-\overline{x}}{s} \\ =\frac{77.80-17.60}{16.02} \\ =3.76 \end{gathered}$$

Therefore, the calculated value of the z-score is equal to 3.76.

d.

The calculated z-score is 3.76.

Since the z-score value is greater than 2, it can be concluded that the speed at Atlanta airport is significantly high.