Question

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?

Answer 1

15.866%

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%