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Bunuel
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It is given that the standard deviation of the nine scores is 19.51.
To find how many scores are greater than one standard deviation ABOVE the mean, we have to calculate the mean of the values.

Mean of the given scores = \(\frac{Sum of the scores }{ 9}\)

Sum of the scores = \(\frac{675 }{ 9}\) = 75

The scores that we are looking for should be greater than one standard deviation above the mean; therefore, the value that they should be above = 75 + 19.51 = 94.51

The values above 94.51 are 97, 98 and 99. Therefore, there are 3 scores greater than one standard deviation above the mean.

The correct answer option is C.
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