Body temperatures of healthy adults are normally distributed with a mean of 96.20 degrees F and a standard deviation of 0.82 degrees F, what is the percentile score if your temperature is 97 degrees F

Standard z-score = (97 -96.2)/0.82= 0.9756

We now want area to the left

P( z=<0.9756) = 0.8354 = 83.54 percentile score

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