Find percentage of a sample given the mean and standard deviation

samedi 31 août 2013

1. The problem statement, all variables and given/known data

People drink on average, 78L of beer per year with a standard deviation of 25L. at least what percentage of people would drink between 20 and 136L per year?



3. The attempt at a solution



I found the z-scores of both and they came up to be +/- 2.32. Compared these values to the z-table and got P(Z<2.32)=0.0102 and P(Z<2.32)=0.9898. Then i found the percentage as followed... 0.9898-0.0102 which is around 98%. Which is wrong. Is my assumption that the graph is symmetrical correct?






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