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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<span>Assuming that we consider the standard workday of eight hours, each part would take 18.75 minutes per part completion. In order to complete 1,250 parts, it would take 8.33 days to complete. In order to complete the 1,250 parts, the manufacturer would have to create at least eight work stations to meet the daily demand.</span>
Answer:
the answer is natural;human
"30" is the answer you're looking for.
