Friday

April 18, 2014

April 18, 2014

Posted by **Andrew** on Friday, March 11, 2011 at 4:04pm.

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.

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

b. Relative frequency of scores less than 120

c. Percentage of scores less than 140

d. Percentage of scores less than 80

e. Relative frequency of scores less than 60

f. Percentage of scores greater than 120

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?

b. Convert 99.00 °F to a standard score (or a z-score).

c. Is a body temperature of 99.00 °F unusual? Why or why not?

d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower?

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?

f. What body temperature is the 95th percentile?

g. What body temperature is the 5th percentile?

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?

- Statistics -
**PsyDAG**, Saturday, March 12, 2011 at 3:16am1. B and D

http://www.oswego.edu/~srp/stats/6895997.htm

2. a-c, e-h. Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.

2d. Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.

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