contestada

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 71.9 Mbps. The complete list of 50 data speeds has a mean of x=15.95 Mbps and a standard deviation of s=34.75 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?

Respuesta :

Answer:

a.55.95Mbps

b.1.61 standard deviations

c.1.61

d. Not significant

Step-by-step explanation:

a. What is the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds?

Carrier's highest data speed = 71.9 Mbps

Mean of all 50 data speeds = 15.95 Mbps

Difference = 71.9Mbps - 15.95Mbps

= 55.95Mbps

b. How many standard deviations is that​ [the difference found in part​ (a)]?

55.95Mbps/34.7Mbps

= 1.61 standard deviations

c. Convert the​ carrier's highest data speed to a z score.

z score formula = z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

z = (x-μ)/σ

x = 71.9Mbps

μ = 15.95 Mbps

σ = s=34.75 Mbps.

z= 71.9 - 15.95/34.75

z = 1.6100719424

Approximately, z scores ≈ 1.61

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?

From the calculation in part c.

The z score = 1.61

It is less than 2

Therefore, the carrier's highest data speed is not significant because is neither significantly low nor significantly high

Otras preguntas