At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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.
Learn more about probability : https://brainly.com/question/24756209
#SPJ1
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.