Question 1147813
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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{{{15/sqrt(7)}}} = 15/2.6458 = 5.6694
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Zscore = {{{(x - mean)/SD sample}}}
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Zscore = {{{(83 - 76)/5.6694}}} = {{{7/5.6694}}} = 1.23
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Go to a z-table and look up +1.23.  At a z-score of +1.23, the area to the left of the curve is 0.8907.  However, we want to find where the mean is LONGER than 83 minutes, so we want to find the area to the RIGHT of the curve. To do this, we subtract 0.8907 from 1.  This comes out to 0.1093.
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So, there is a 0.1093 probability the mean of the sample is longer than 83 minutes.