|Jan2-11, 09:58 AM||#1|
Suppose that X is a normal Variate with mean 5.
Suppose that X is a normal Variate with mean 5. If P(X>9)=0.2 approximately, what is Var X?
Here [tex]\mu[/tex]=5 and P(X>9) = 0.2
0.2 = 9 - 5 / V(x)
V(x) = 4 / 0.2 = 20 [Is this correct]
|Jan2-11, 04:00 PM||#2|
but, as written, it is a meaningless statement. If you were trying to compare .2 to a Z-score, that won't work: they aren't the same thing.
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