Download 1. Which of the following statements are correct? a. A normal

1. Which of the following statements are correct?
a. A normal distribution is any distribution that is not unusual.
b. The graph of a normal distribution is bell-shaped.
c. If a population has a normal distribution, the mean and the median are not equal.
d. The graph of a normal distribution is symmetric.
Both b and d are correct.
Using the 68-95-99.7 rule: Assume that a set of test scores is normally distributed with a
mean of 100 and a standard deviation of 20. Use the 68-95-99.7 rule to find the following
quantities: Suggest you make a drawing and label first…
a. Percentage of scores less than 100
50%
b. Relative frequency of scores less than 120
0.84
c. Percentage of scores less than 140
97.5%
d. Percentage of scores less than 80
16%
e. Relative frequency of scores less than 60
0.025
f. Percentage of scores greater than 120
16%
2. 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).
a. If you have a body temperature of 99.00 °F, what is your percentile score?
90th percentile (p = 0.9015)
b. Convert 99.00 °F to a standard score (or a z-score).
z = (99.0 – 98.2)/0.62 = 1.29
c. Is a body temperature of 99.00 °F unusual? Why or why not?
Because it ranks as the 90th percentile, only 10% of the population would be
expected to have a body temperature this high or higher. That makes this
body temperature fairly unusual.
d. Fifty adults are randomly selected. What is the likelihood that the mean of their body
temperatures is 97.98 °F or lower?
z = (97.98 – 98.2) / (0.62/√50) = -2.51
p(z<-2.51) = 0.006 = 0.6%
e. A person’s body temperature is found to be 101.00 °F. Is the result unusual? Why or
why not? What should you conclude?
z = (101.0 – 98.2) / 0.62 = 4.516
This temperature would be very unusual, as a very, very small percentage
of the population would be expected to have a temperature this high.
f. What body temperature is the 95th percentile?
z = 1.645
T = 98.2 + 1.645(0.62) = 99.2º
g. What body temperature is the 5th percentile?
z = -1.645
T = 98.2 - 1.645(0.62) = 97.2º
h. 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?
(100.6 – 98.2)/0.62 = 3.87
z = 3.87  p = 0.0001 = 0.1%
Given that such a small percentage of the population would be
considered to have a fever at that cut-off, it seems like an inappropriately
stringent standard.