Find CV

Best way to do it is using Z = (0-10)/sigma

You want to find the Z from the table whose value corresponds to a probability of 16%...if you look at the table, you will see that this is Z = -1. Look at the negative Z table, and just search for 0.16. (or look for 0.84 in teh normal table, and take that Z), then use Z, and plug above:

-1 = (0-10)/sigma, so sigma = -10/-1 = 10

CV = 10/10 = 1.0

Use this method with any other numbers, check the second question I posted above.
 
.16 = 1 - 1/k^2
K = st. deviation = 1.091
cv = 1.091/10 = .109.
 
This says that at least 16% of the observations fall within about 1 std from the mean. Whereas teh problem says that 16% of the observations fall below 0. Completely different.
 
0.5-0.16 = 0.34
0.34*2 = 0.68

0.68=1-1/k^2

k=1.78

k=10/stdev-->stdev = 10/1.78 = 5.65

CV=stdev/mean=5.65/10=0.57

dont know if its right??
 
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