Suppose flights on standard routes between two given cities use on average 3,087 gallons of kerosene with standard deviation of 195 gallons, and the distribution of fuel use is bell-shaped. What percent of flights on these particular routes will burn more than 3,282 gallons of kerosene?

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%

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