SOLUTION: A certain size box of a common cereal is supposed to contain 18 ounces of cereal, but there is some variation in the weight of the contents causes by the packaging machinery. Assum

Question 972176: A certain size box of a common cereal is supposed to contain 18 ounces of cereal, but there is some variation in the weight of the contents causes by the packaging machinery. Assume the weight of the cereal in a box is normally distributed with mean 18 ounces and standard deviation .2 ounces. What is the probability that the contents of a box of this cereal will weigh less than 17.67 oz? Between 17.45 and 17.95 oz?
So far all I know is that this type of problems deals with bell curves but I do not have one to show so I do not know how to put this into a problem so an explanation would be really helpful!
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
This is a normal distribution and you use z-values, z-tables and the corresponding probabilities(P) to solve this problem.
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a) P( X < 17.67 ) is calculated by finding its z-value
z-value = (X - mean) / (std dev) = (17.67 - 18.00) / (0.2) = −1.65
we check the z-tables for the probability corresponding to −1.65
P( X < 17.67 ) = 0.0495
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b) P( 17.45 < X < 17.95 ) = P( X < 17.95 ) - P( X < 17.45 )
P( X < 17.95 ) is calculated by finding its z-value
z-value = (17.95 - 18.00) / (0.2) = −0.25
P( X < 17.95 ) = 0.4013
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P( X < 17.45 ) is calculated by finding its z-value
z-value = (17.45 - 18.00) / (0.2) = −2.75
P( X < 17.45 ) = 0.0030
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P( 17.45 < X < 17.95 ) = P( X < 17.95 ) - P( X < 17.45 )
P( 17.45 < X < 17.95 ) = 0.4013 - 0.0030
P( 17.45 < X < 17.95 ) = 0.3983

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