Asked by sue
                Find the Z-score that separate the middle 63% of the distribution from the area in the tails of the standard normal distribution
            
            
        Answers
                    Answered by
            Steve
            
    you want z such that P(Z>z) = .63/2
z = .896
So, the scores are ±0.896
    
z = .896
So, the scores are ±0.896
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