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.

Respuesta :

Answer:

(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.

Step-by-step explanation:

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

The highest speed measured was [tex]X_{max.}=75.7\ \text{Mbps}[/tex].

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

(a)

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

[tex]d=X_{max.}-\bar X[/tex]

  [tex]=75.7-17.02\\\\=58.68[/tex]

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:

[tex]\text{Number of standard deviations}=\frac{d}{s}[/tex]

                                                 [tex]=\frac{58.68}{17.02}\\\\=3.44771\\\\\approx 3.45[/tex]

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.