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Week 4 Quiz

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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 …show more content…

Is the result unusual? Why or why not? What should you conclude?

Unusual. Because (101.00 – 98.2) / 0.62 = 4,5

In table 5-1, the z-score 4.5>3.5 so it is out from the normal distribution, so this body temperature is unusual.

f. What body temperature is the 95th percentile?

In table 5-1, 95th percentile is corresponds to a z-score 1.6 so

(x - 98.2) / 0.62 = 1.6

X = 99.19 °F or 99.2 °F

g. What body temperature is the 5th percentile?

In table 5-1, 5th percentile is corresponds to a z-score -1.6 so

(x - 98.2) / 0.62 = -1.6

x = 97.21 °F or 97.2 °F

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

In table 5-1, a z-score of 3.5 is corresponds to 99.98% as normal and healthy adults,

100 - 99.98 = 0.02 so 0.02% adults would be considered as having an abnormal body temperature, because we assume the body temperatures of healthy adults are normally distributed, then 0.02% / 2 = 0.01%.

So 0.01% of normal and healthy adults would be considered to have a fever

This percentage suggests that a cutoff of 100.6 °F is not appropriate. Because from table 5-1, the 99.98% is corresponds

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