Questions: Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 71.2 Mbps. The complete list of 50 data speeds has a mean of 17.41 Mbps and a standard deviation of 20.12 Mbps. a. What is the difference between carrier's 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 carrier's 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 carrier's highest data speed significant? a. The difference is Mbps. (Type an integer or a decimal. Do not round.)

Solution Steps

To solve the given questions, we will follow these steps:

a. To find the difference between the carrier's highest data speed and the mean, subtract the mean from the highest data speed.

b. To determine how many standard deviations the difference is, divide the difference by the standard deviation.

c. To convert the highest data speed to a z score, subtract the mean from the highest data speed and then divide by the standard deviation.

To find the difference between the carrier's highest data speed and the mean of all 50 data speeds, we use the formula:

\[ \text{Difference} = \text{Highest Speed} - \text{Mean Speed} = 71.2 - 17.41 = 53.79 \, \text{Mbps} \]

To determine how many standard deviations the difference is, we divide the difference by the standard deviation:

\[ \text{Standard Deviations} = \frac{\text{Difference}}{\text{Standard Deviation}} = \frac{53.79}{20.12} \approx 2.673 \]

To convert the highest data speed to a z score, we use the formula:

\[ z = \frac{\text{Highest Speed} - \text{Mean Speed}}{\text{Standard Deviation}} = \frac{71.2 - 17.41}{20.12} \approx 2.673 \]

Final Answer