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
