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OpenStudy (anonymous):
Use a table to find a z-score that fits the given conditions. Interpolate if necessary. 48% of the area under the standard normal curve is above the score.
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OpenStudy (anonymous):
Z = 0.05
OpenStudy (anonymous):
http://2.bp.blogspot.com/-j2t1gky9beI/TfXu4QXNw3I/AAAAAAAAEe4/jX6bmeXuvvQ/s1600/percentilesTable%255B13%255D.png looking at this table, z = .05 means 52% less than z, so that's 48% greater than z
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