91影视

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:

$z=\frac{x-\stackrel{炉}{x}}{s}$

Here, x is the observation.

$\stackrel{炉}{x}$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:

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

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鈥檚 speed and the mean speed is obtained using the z-score as shown:

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

Therefore, the difference is1.05 standard deviations.

c.

The z-score is calculated as follows:

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

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.