Question

researchers measured the data speeds for a particular smartphone carrier at 50 airports. the highest speed measured was 76.8 mbps. the complete list of 50 data speeds has a mean of \\( \overline { x } = 15.47 \\) mbps and a standard deviation of \\( s = 40.78 \\) mbps.
a. what is the difference between carriers highest data speed and the mean of all 50 data speeds?
b. how many standard deviations is that the difference found in part (a)?
c. convert the carriers highest data speed 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 the carriers highest data speed significant?
a. the difference is 61.33 mbps.
(type an integer or a decimal. do not round.)
b. the difference is \\( \square \\) standard deviations.
(round to two decimal places as needed )

Explanation:

Step1: Recall the formula for number of standard deviations

The number of standard deviations is calculated by dividing the difference (from part a) by the standard deviation. The formula is \(n=\frac{\text{Difference}}{s}\)

Step2: Substitute the values

We know the difference is \(61.33\) Mbps (from part a) and \(s = 40.78\) Mbps. So \(n=\frac{61.33}{40.78}\)

Step3: Calculate the value

\(n=\frac{61.33}{40.78}\approx1.50\)

Answer:

\(1.50\)