The probability that demand is greater than 1800 gallons over a 2 hour period is : 0.5
<u>Given data :</u>
Mean value of gasoline per hour = 875 gallons
Standard deviation = 55 gallons
<h3>Determine the probability of demand being greater than 1800 gallons over 2 hours </h3>
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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