Question 1147746
First, you want to find the standard deviation of the sample. You do this by taking the standard deviation of the population and dividing it by the square root of the number of items in the sample:
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{{{84/sqrt(10)}}} = {{{84/3.1623}}} = 26.5629
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Zscore = {{{(x - mean)/SD sample}}}
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Zscore = {{{(543 - 506)/26.5629}}} = {{{37/26.5629}}} = 1.39
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Go to a z-table and look up +1.39.  At a z-score of +1.39, the area to the left of the curve is 0.9177.  However, we want to find the probability that  the mean weight is GREATER than 543 lbs., so we want to find the area to the RIGHT of the curve. To do this, we subtract 0.9177 from 1.  This comes out to 0.0823.
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So, there is a 0.0823 probability the mean weight is over 543 lbs.  (Or, to answer the original question, there is a 0.0823 probability the cows' combined weight will exceed 5430 lbs.)