Attachment D .gov
From table or calculator the 90th percentile for standard normal distribution is at z=1.2816. So answer is 90 + 1.2816*38 mg/dL =138.7 mg/dL Suppose that the “abnormal range” is defined to be glucose levels which are 1.5 standard deviations above the mean or 1.5 standard deviations below the mean. ................
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