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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.

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

$z=\frac{x-\stackrel{炉}{x}}{s}$

Here,

x is the value from the data.

$\stackrel{炉}{x}$is the sample mean.

s is the sample standard deviation.

a.

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

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

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

b.

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

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

Therefore, the difference is3.76 standard deviations.

c.

The z-score can be calculated as:

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

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.