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Question 870:



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We need to find the z-scores (normal scores) for the percentiles at 95 and 5, then find the raw values that associate with these z-scores given the mean and standard deviation.

To find z-scores from percentiles you can use the Excel function =NORMSINV(.95) or lookup the value in a normal table for 1-sided area. You get the z-scores of 1.645 and -1.645.  The z-score formula is

z= (datapoint- mean)/ sd

To solve for the datapoint we re-arrange the formula  to get datapoint = z*sd+mean.

Plugging in the values for the 95th percentile we get 1.644*.62+98.2 = 99.22 and for the 5th percentile we get -1.644*.62+98.2 = 97.18.  So the body temperatures are 99.2 and 97.18.

For the second part we use the z-score formula to first find the z-score then the percentile associated with it. Plugging in the values we get (100.6-98.2)/.62 = 3.87. Looking up this value in a normal table or using the Excel function =NORMSDIST(3.87) we get .999946 or the 99.99th percentile.  So less than .01% of adults would be considered to have a fever.

This cutoff might be appropriate if the hospital really doesn't want to admit healthy adults. However, it is so high then they increase the chances that they miss some adults that really have a fever. It would likely make more sense to lower the cutoff to just the 99th percentile or a temperature of 99.6.

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