Beer bottles are filled so that they contain an average of 340 ml of beer in eac
ID: 3176535 • Letter: B
Question
Beer bottles are filled so that they contain an average of 340 ml of beer in each bottle. Suppose that the amount of beer in a bottle is normally distributed with a standard deviation of 7 ml. Use Table 1 (http://lectures.mhhe.com/connect/0078020557/Table/table1.jpg)
What is the probability that a randomly selected bottle will have less than 333 ml of beer? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
What is the probability that a randomly selected 6-pack of beer will have a mean amount less than 333 ml? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
What is the probability that a randomly selected 12-pack of beer will have a mean amount less than 333 ml? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
What is the probability that a randomly selected bottle will have less than 333 ml of beer? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
Explanation / Answer
Mean ( u ) =340
Standard Deviation ( sd )=7
Normal Distribution = Z= X- u / sd ~ N(0,1)
a.
P(X < 333) = (333-340)/7
= -7/7= -1
= P ( Z <-1) From Standard Normal Table
= 0.1587
b.
Mean ( u ) =340
Standard Deviation ( sd )= 7/ Sqrt(n) = 2.8577
Number ( n ) = 6
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)
P(X < 333) = (333-340)/7/ Sqrt ( 6 )
= -7/2.8577= -2.4495
= P ( Z <-2.4495) From Standard NOrmal Table
= 0.0072
c.
Mean ( u ) =340
Standard Deviation ( sd )= 7/ Sqrt(n) = 2.0207
Number ( n ) = 12
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)
P(X < 333) = (333-340)/7/ Sqrt ( 12 )
= -7/2.0207= -3.4641
= P ( Z <-3.4641) From Standard NOrmal Table
= 0.0003
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