Denote by
X= the body temperatures random variable and by
Z = the standard normal distribution with mean 0 and standard deviation 1.
Also denote by
mu(X) = 98.20°F -- the mean of X and by sigma(X)=0.62°F – the standard deviation of X.
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Assume the body temperatures of healthy adults are normally distributed with a mean of
98.20 °F and a standard deviation of 0.62 °F (based on data from the University of Maryland
researchers). If you have a body temperature of 99.00 °F, what is your percentile score?
Convert 99.00 °F to a standard score (or a z-score). I
Answer: the z-score is (99-mu(X))/sigma(X) = (99-98.20)/0.62 = 1.29.
Looking in the standard normal distribution table we see that about 90.15% of the observations
fall below this value. This means that the percentile corresponding to a temperature of 99.00
°F is the 90th percentile.
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Is this body temperature unusual? Why or why not?
Answer: Somewhat unusual. According to the data only about 10% of the adults have a body
temperature of 99.00 °F or higher. In other words only about one in ten people registers a
body temperature this high.
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Fifty adults are randomly selected. What is the likelihood that the mean of their body
temperatures is 97.98 °F or lower?
Answer: Denote by Y = the sample mean random variable. It is known that the mean of Y is the
same as the mean of X that is, mu(X)=mu(Y)=98.20°F.
On the other hand the standard deviation of Y,
sigma(Y)=sigma(X)/squareroot(50)=0.67/7.071=0.0877.
Notice that the standard deviation of Y is much smaller than the standard deviation of X. This is
due to the large number of people in the sample (50 subjects)
The problem asks to compute
P(Y<97.98°F) = P((Y-mu(Y))/sigma(Y)<(97.98-98.2)/0.0877) =P(Z<-2.5)= 0.0062 – less that 1%!!
It is very unlikely this will happen.
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A person’s body temperature is found to be 101.00 °F. Is the result unusual? Why or why not?
What should you conclude?
Answer: P(X>101.00)=P((X-98.2)/0.62>(101.0-98.2)/.62)=P(Z>4.51) = 0.000003
Such a temperature is highly unusual. Only about three people out of 1 million have as high of a
body temperature. Two conclusions are possible: a. the person requires immediate medical
assistance or b. the assumption according to which the body temperatures are being normally
distributed with a mean of 98.2 and standard deviation of 0.62 is erroneous.
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What body temperature is the 95th percentile? What body temperature is the 5th percentile?
Answer:
The z-score corresponding to the 95th percentile is 1.645
This implies that (X95%-98.2)/.62=1.645 from which X95%= 99.22°F – this is the 95th percentile
(95% of the body temperatures are below this number)
The z-score corresponding to the 5th percentile is -1.645
This implies that (X5%-98.2)/.62=-1.645 from which X5%= 97.18°F – this is the 5th percentile
(5% of the body temperatures are below this number)
Bellevue Hospital in New York City uses 100.6 °F as the lowest temperature considered to
indicate a fever. What percentage of normal and healthy adults would be considered to have
a fever? Does this percentage suggest that a cutoff of 100.6 °F is appropriate?
Answer: P(X>100.6°F)=P(Z>(100.6-98.2)/.62)= P(Z>3.87) = 0.000054 = approx 1/20,000
According the Bellevue Hospital only about 1 person in every 20,000 normal and healthy adults
has fever. The number is probably much higher. The cutoff of 100.6 °F is definitely
inappropriate.
We may suggest a cutoff of 100.0 °F as much more appropriate. In this case
(X>100.0°F)=P(Z>(100.0-98.2)/.62)= P(Z>) = 0.001866 = approx 1/500
It seems more reasonable to assume that at least 1 person in every 500 normal and healthy
adults has fever.