Question

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.7 Mbps. The complete list of 50 data speeds has a mean of x overbarequals17.02 Mbps and a standard deviation of sequals38.03 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.

Answer 1

(a) The difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 58.68 Mbps.

(b) The number of standard deviations the highest data speed is from the mean is 3.45.

(c) The z-score for the carrier's highest data speed is 3.45.

Explanation:

The random variable X is defined as the data speeds for a particular smartphone carrier.

The highest speed measured was
`X_(max.)=75.7\ \text{Mbps}`.

The mean of X is,
`\bar X=17.02\ \text{Mbps}` and the standard deviation is,
`s=38.03\ \text{Mbps}`.

(a)

Compute the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds as follow:

`d=X_(max.)-\bar X`

`=75.7-17.02\\\\=58.68`

Thus, the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 58.68 Mbps.

(b)

Compute the number of standard deviations the highest data speed is from the mean as follows:

`\text{Number of standard deviations}=(d)/(s)`

`=(58.68)/(17.02)\\\\=3.44771\\\\\approx 3.45`

Thus, the number of standard deviations the highest data speed is from the mean is 3.45.

(c)

In statistics, a standardized score is the number of standard deviations an observation or data point is from the mean.

Thus, z-scores are a type of standardized scores.

So, the z-score for the carrier's highest data speed is 3.45.