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wrote...
Posts: 233
2 weeks ago
Human body temperatures taken through the ear are typically 0.5°F higher than body temperatures taken orally. Making this adjustment and using the 1992 Journal of the American Medical Association article that reports average oral body temperature as 98.2°F, we will assume that a Normal model with an average of 98.7°F and a standard deviation of 0.7°F is appropriate for body temperatures taken through the ear.

a. An ear temperature of 97°F may indicate hypothermia (low body temperature). What percent of people have ear temperatures that may indicate hypothermia?
b. Find the interquartile range for ear temperatures.
c. A new thermometer for the ear reports that it is more accurate than the ear thermometers currently on the market.  If the average ear temperature reading remains the same and the company reports an IQR of 0.5°F, find the standard deviation for this new ear thermometer.
Textbook 

Stats: Modeling the World


Edition: 4th
Authors:
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wrote...
2 weeks ago
a. z = = -2.43, so P(z < -2.43) = 0.0075
About 0.75% of people have ear temperatures that may indicate hypothermia.
b. The z-scores associated with the IQR are z = -0.67 and z = 0.67. So, we need to solve for y in each of the following equations: -0.67 = and 0.67 = .  We get y = 98.7 − 0.67(0.7) = 98.2 and y = 98.7 + 0.67(0.7) = 99.2.  The interquartile range is IQR = 99.2°F - 98.2°F = 1.0°F.
c. The new IQR is 0.5°F, while the old IQR was 1.0°F.  So, we want
IQR = [98.7 + 0.67σ] − [98.7 - 0.67σ] = 0.5, or 1.34σ = 0.5.  Thus, σ = = 0.37. Our new standard deviation is 0.37°F.
wrote...
2 weeks ago
Good timing, thanks!
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