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Statistics

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Assume the body temperature of healthy adults are normally distributed with a mean of 98.2 degrees F and a standard deviation of .62 degrees F. If you have a temperature of 99.0 what is your percentile score?

  • Statistics -

    99 - 98.2 = .8

    .8 divided by .6 = 1.29...z-score

    z-score of 1.29...on table
    =.9015 = 90.15 %

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