week 2 quiz
University of Phoenix Material
Week Four Quiz
1. Please indicate whether each of the statements below is true or false.
a. A normal distribution is any distribution that is not unusual. False
b. The graph of a normal distribution is bell-shaped. True
c. If a population has a normal distribution, the mean and the median are not equal. False
d. The graph of a normal distribution is symmetric. True
Using the 68-95-99.7 rule:
2. 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
34%+13.5%+2.35%+0.15%= 50%
b. Relative frequency of scores less than 120
.34+0.34+0.135+0.0235+0.0015 = .84
c. Percentage of scores less than 140
50%+47.5%=97.5%
d. Percentage of scores less than 80
50%-34%= 16%
e. Relative frequency of scores less than 60
50%-47.5%= 2.5%
f. Percentage of scores greater than 120
50%-34%= 16%
3. 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? z(99) = (99-98.2)/0.62 = 1.2903
P(z < 1.2903) = normalcdf(-100,1.2903) = 0.9015
90.15 percentile
b. Convert 99.00 °F to a standard score (or a z-score). (99-98.2)/.62= 1.29
c. Is a body temperature of 99.00 °F unusual? Why or why not?
Not unusual, it is 1.29 standard deviation above the mean.
d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower? Y ~ N(98.2 , 0.62 / root(50) ) P(Y < 97.98) = P(Z < (97.98 - 98.2) / 0.0876...) = P(Z < -2.509...) = 1 - phi(2.509...) = 1 - 0.99395 = 0.0060 = 6% correct
0.0065
e. A person’s body temperature is found to be
Week Four Quiz
1. Please indicate whether each of the statements below is true or false.
a. A normal distribution is any distribution that is not unusual. False
b. The graph of a normal distribution is bell-shaped. True
c. If a population has a normal distribution, the mean and the median are not equal. False
d. The graph of a normal distribution is symmetric. True
Using the 68-95-99.7 rule:
2. 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
34%+13.5%+2.35%+0.15%= 50%
b. Relative frequency of scores less than 120
.34+0.34+0.135+0.0235+0.0015 = .84
c. Percentage of scores less than 140
50%+47.5%=97.5%
d. Percentage of scores less than 80
50%-34%= 16%
e. Relative frequency of scores less than 60
50%-47.5%= 2.5%
f. Percentage of scores greater than 120
50%-34%= 16%
3. 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? z(99) = (99-98.2)/0.62 = 1.2903
P(z < 1.2903) = normalcdf(-100,1.2903) = 0.9015
90.15 percentile
b. Convert 99.00 °F to a standard score (or a z-score). (99-98.2)/.62= 1.29
c. Is a body temperature of 99.00 °F unusual? Why or why not?
Not unusual, it is 1.29 standard deviation above the mean.
d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower? Y ~ N(98.2 , 0.62 / root(50) ) P(Y < 97.98) = P(Z < (97.98 - 98.2) / 0.0876...) = P(Z < -2.509...) = 1 - phi(2.509...) = 1 - 0.99395 = 0.0060 = 6% correct
0.0065
e. A person’s body temperature is found to be