Question

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

PHL Data Speeds Repeat the preceding exercise using the Verizon data speed of 0.8 Mbps at Philadelphia International Airport (PHL).

Short answer

a. The difference between Philadelphia international airport’s speed and the mean data speed is equal to -16.80 Mbps.

b. The difference between Philadelphia international airport’s speed and the mean data speed is 1.05 standard deviations.

c. The z-score value for the given speed equal to 0.8 Mbps is obtained as -1.05.

d. The speed at Atlanta airport equal to 0.8 Mbps is not significant.

Step-by-step solution

Given information

Data is provided on the Verizon data speed for a group of 50 airports.

The mean data speed is equal to 17.60 Mbps.

The standard deviation of data speeds is equal to 16.02 Mbps.

Formula of z-score

Az-score for an observation shows the number of standard deviations by which the observation is above or below the mean. Its formula is as follows:

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

Here, x is the observation.

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

s is the standard deviation of the sample.

Calculation

a.

The difference between the speed at Philadelphia airport (x = 08 Mbps) and the mean speed is obtained as follows:

$$\begin{gathered} x-\overline{x}=0.8-17.60 \\ =-16.80 \end{gathered}$$

Thus, the difference between the speed at Philadelphia airport and the mean speed is equal to -16.80 Mbps.

b.

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

$$\begin{gathered} z=\frac{x-\overline{x}}{s} \\ =\frac{0.8-17.60}{16.02} \\ =-1.05 \end{gathered}$$

Therefore, the difference is1.05 standard deviations.

c.

The z-score is calculated as follows:

$$\begin{gathered} z=\frac{x-\overline{x}}{s} \\ =\frac{0.8-17.60}{16.02} \\ =-1.05 \end{gathered}$$

Here, the calculated value of the z-score is equal to -1.05.

d.

Since the z-score value is between -2 and 2, it can be concluded that the speed at Philadelphia airport is not significant.