SOLUTION: Statistic students weighed a sample of 9 randomly chosen cans of coke, and found the sample weight to be 12,06 oz. The weight of the population of cans of coke are normally distrib

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Question 1113378: Statistic students weighed a sample of 9 randomly chosen cans of coke, and found the sample weight to be 12,06 oz. The weight of the population of cans of coke are normally distributed with a mean of 12.00 oz and a population standard deviation of 0.11 oz.
What is the standard error?
What is the probability that a sample of 9 cans will have a mean amount of at least 12.06?
What is the probability that a sample of 9 cans will have a sample mean weight between 11.99 oz and 12.04 oz?
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
standard error(SE) = population standard deviation / square root (sample size)
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SE = 0.11/square root(9) = 0.0367
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since we know the standard deviation of the population, we use the normal distribution's z-score calculation
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probability(P) that a sample of 9 cans will have a mean amount of at least 12.06 = 1 -P ( X < 12.06 )
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z-score = (12.06 - 12.00) / 0.0367 = 1.6349
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P associated with 1.6349 (using table of z-values or a normal distribution calculator) = 0.949
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P that a sample of 9 cans will have a mean amount of at least 12.06 = 1 -P (X < 12.06) = 1 - 0.949 = 0.051
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P that a sample of 9 cans will have a sample mean weight between 11.99 oz and 12.04 oz = P (X < 12.04) - P (X < 11.99)
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z-score(12.04) = (12.04 - 12.00) / 0.0367 = 1.0899
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P (X < 12.04) = 0.862
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z-score(11.99) = (11.99 - 12.00) / 0.0367 = -0.2725
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P (X < 11.99) = 0.393
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P that a sample of 9 cans will have a sample mean weight between 11.99 oz and 12.04 oz = 0.862 - 0.393 = 0.469
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