Question

Demand per hour for gasoline at a local station is normally distributed with a mean of 875 gallons and std deviation of 55 gallons. What is the probability that demand is greater than 1800 gallons over a 2-hour period

Answer 1

The probability that demand is greater than 1800 gallons over a 2 hour period is : 0.5

Given data :

Mean value of gasoline per hour = 875 gallons

Standard deviation = 55 gallons

Determine the probability of demand being greater than 1800 gallons over 2 hours

Demand for gas in 1 hour = X₁

Demand for gas in 2 hours = X₁ + X₂

Therefore ; ( X₁ + X₂) ~ N ( u₁+u₂, sd₁² + sd₂² )

In order to  calculate probabilities for normals apply the equation below

Z = ( X- u ) / sd

where : u = 1800, sd = √ ( 55² + 55² ) = 77.78

using the z-table

P( Y > 1800) = P( Z > ( 1800 - 1800 ) / 77.78)

                    = P( Z>0 ) = 0.5

Hence we can conclude that The probability that demand is greater than 1800 gallons over a 2 hour period is : 0.5.

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