Question

The weights of a certain brand of candies are normally distributed with a a mean weight of 0.8554 g and a standard deviation of.0511 g for the. A sample of these candies came from a package containing 459 candies and the package label stated that the net weight is 392.1 g If every package has 459 candies the mean weight of the candies must exceed 392.1/459= 0.8543 for the net contents to weigh at least 392.1 g

a. if 1 candy is randomly selected, find the probabily that it weighs more than .08543 g

b if 459 candies are randomly selected find the probablity that their mean weight is least .08543 g

Answer 1

a)

X ~ N ( µ = 0.8554 , σ = 0.0511 )
P ( X > 0.8543 ) = 1 - P ( X < 0.8543 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 0.8543 - 0.8554 ) / 0.0511
Z = -0.02
P ( ( X - µ ) / σ ) > ( 0.8543 - 0.8554 ) / 0.0511 )
P ( Z > -0.02 )
P ( X > 0.8543 ) = 1 - P ( Z < -0.02 )
P ( X > 0.8543 ) = 1 - 0.4920
P ( X > 0.8543 ) = 0.5080

b)

X ~ N ( µ = 0.8554 , σ = 0.0511 )
P ( X > 0.8543 ) = 1 - P ( X < 0.8543 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 0.8543 - 0.8554 ) / ( 0.0511 / √ ( 459 ) )
Z = -0.46
P ( ( X - µ ) / ( σ / √ (n)) > ( 0.8543 - 0.8554 ) / ( 0.0511 / √(459) )
P ( Z > -0.46 )
P ( X̅ > 0.8543 ) = 1 - P ( Z < -0.46 )
P ( X̅ > 0.8543 ) = 1 - 0.3228
P ( X̅ > 0.8543 ) = 0.6772