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.

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warylucknow warylucknow · Answer 1 · 2020-07-02T06:37:40+02:00

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.