# Week 4 Quiz

**Topics:**Normal distribution, Standard deviation, Percentile rank

**Pages:**4 (677 words)

**Published:**October 2, 2012

Week Four Quiz

1. WHICH OF THE FOLLOWING STATEMENTS ARE CORRECT?

a. A normal distribution is any distribution that is not unusual. (Correct)

b. The graph of a normal distribution is bell-shaped. (Correct)

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. (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 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 2.5%

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?

(99-98.2)/0.62=1.29

In table 5-1, 1.29 corresponds to 90th percentile.

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

1.29. (99 - 98.2) / 0.62 = 1.29

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

Not unusual. Follow the 68-95-99.7 rule, 99.00 °F is within two standard deviations of the mean (96.96 - 99.44°F)

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

(97.98 - 98.2) / 0.62 = - 0.35

In table 5-1, - 0.35 corresponds to 36%.

50 x 36% = 18

So a 36% or 18 adults are likelihood that the mean of...

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