Suppose X and Y are independent. X and Y are both normally distributed with mean 100 and 96, respectively. The standard deviation of X is 6 and the standard deviation of Y is 8. What is the probability that X is greater than Y?

The standard deviation of X+Y = sqrt(8^2 + 6^2) = sqrt(100) = 10

Now we need the Z score for difference of X and Y..

Z = (X - Y)/(st dev) = (100 - 96)/10

Z = .4

Using a Z table you get probability = .6554

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