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).

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.

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.

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.