Beer bottles are filled so that they contain an average of 450 ml of beer in eac

Question

Beer bottles are filled so that they contain an average of 450 ml of beer in each bottle. Suppose that the amount of beer in a bottle is normally distributed with a standard deviation of 8 ml.

a. What is the probability that a randomly selected bottle will have less than 444 ml of beer? (Round final answer to 4 decimal places.)

b.What is the probability that a randomly selected 6-pack of beer will have a mean amount less than 444 ml? (Round final answer to 4 decimal places.)

c. What is the probability that a randomly selected 12-pack of beer will have a mean amount less than 444 ml? (Round final answer to 4 decimal places.)

Beer bottles are filled so that they contain an average of 450 ml of beer in each bottle. Suppose that the amount of beer in a bottle is normally distributed with a standard deviation of 8 ml.

a. What is the probability that a randomly selected bottle will have less than 444 ml of beer? (Round final answer to 4 decimal places.)

b.What is the probability that a randomly selected 6-pack of beer will have a mean amount less than 444 ml? (Round final answer to 4 decimal places.)

c. What is the probability that a randomly selected 12-pack of beer will have a mean amount less than 444 ml? (Round final answer to 4 decimal places.)

Explanation / Answer

Mean is 450 and s is 8

a) z here is given as (x-mean)/s

P(x<444)=P(z<(444-450)/8)=P(z<-0.75) or 1-P(z<0.75), from normal distribution table we get 1-0.7734 =0.2266

b) here, standard error is s/sqrt(N) =8/sqrt(6)=3.27

P(xbar<444)=P(z<(444-450)/3.27)=P(z<-1.83) or 1-P(z<1.83), from normal distribution table we get 1-0.9664=0.0336

c) here standard error is 8/sqrt(12)=2.31

P(xbar<444)=P(z<(444-450)/2.31)=P(z<-2.6) or 1-P(z<2.6), from normal distribution table we get 1-0.9953 =0.0047

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