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Question 378:



No answer provided yet.Lets assume the data is normally distributed. We can use the properties of the normal curve and calculate a normalized or z-score. Subtract the data-point in question (86) from the mean and divide that result by the standar deviation: (86-71)/9= a z-score of 1.66. Now we can look this value up using the z-score to percentile calculator . We want the area above the z-score of 1.66 so select 1-sided area. You should get 4.8% of class members would be expected to score above this amount.

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