Answer:
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I believe the answer is C
That makes no sense what are we supposed to figure out
Answer:
15.866%
Step-by-step explanation:
We solve using z score formula
z = (x-μ)/σ, where
x is the raw score = 3,282 gallons of kerosene
μ is the population mean = 3,087 gallons of kerosene
σ is the population standard deviation = 195 gallons
For x > 3282 gallons
z = 3282 - 3087/195
z = 1
Probability value from Z-Table:
P(x<3282) = 0.84134
P(x>3282) = 1 - P(x<3282) = 0.15866
Converting to percentage
0.15866 × 100 = 15.866%
The percent of flights on these particular routes that will burn more than 3,282 gallons of kerosene is 15.866%
