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P3.3
SOLUTION
A.
The mean, or sample average is 3.90 years of work experience. By inspection of a rankorder from highest to lowest values, the Amiddle@ or median value is four years= work
experience. The mode is three years experience, enjoyed by four workers. In this
instance, each measure of central tendency offers a highly similar view of the typical
observation.
B.
The range for these ten workers is from three to five years experience. The sample variance is
0.767 (years squared) with a standard deviation of 0.876 years.
In class problem assigned with Expected Value:
1. EV= .15(1000) + .20(800) +.40(500) + .25(0) = 150+160+200+0=510
2. Deviations: (1000-510)=490
(800-510)=290
(500-510)=-10
(0-510)=-510
Square the deviations and weight by probability:
.15(240,100) + .20(84,100) + .40(100) + .25(260,100)
36,015 + 16, 820 + 240+ 65,025=118,100
Take square root: 343.66 is standard deviation
3. With same expected value as first lottery you can simply compare standard deviations. The
first lottery is 343 compared to the new lottery of 30. The first lottery has more risk.
Remember, if the EV was not the same you’d use the coefficient of variation to determine
relative risk.