Question 1049944: The mean of a random sample of n=100 is going to be used to estimate mean daily production of a very large herd of dairy cows.Given that the standard deviation of the population sampled is 3.6 quarts,what can we assert about probabilities that the error of this estimate will be,
(a)more than 0.72 quarts
(b)less than 0.45 quarts?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! std dev of sample = std dev of the population/ square root of sample size
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std dev of sample = 3.6 / square root(100)
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std dev of sample = 3.6 / 10 = 0.36
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we use the 68-95-99.7 rule for 1, 2 , 3 standard deviations
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(a) asks us what is the probability that the error of the estimate for the mean of the sample is greater than 2 standard deviations
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this will be (99.7 - 95) / 2 = 2.35%
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note why did I have to divide by 2?
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(b) asks us what is the probability that the error of the estimate for the mean of the sample is less than 0.45
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0.45 = 0.36 (1 standard deviation) + 0.09 (1/4 of 1 standard deviation)
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this will be (99.7/2) + (68/2) + (95/(4*2)) = 95.725%
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