Question 268868
a certain operation on a time- shared computer takes anywhere from 1/100 to 4/100 of a second, and the operation time is no more likely to be in any sub interval in this range than it is to be in any other sub interval of the same length. what is the probability that the operation will take over 2/100 of a second?
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Assuming the times are normally distributed:
mean = [(4-1)/100]/2 = (1.5)/100 = 0.015
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Assume the data range is 6sigma wide
Then 6sigma = 3/100
sigma = 0.5/100 = 0.005
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z(2/100) = (0.020-0.015)/0.005 = 0.005/0.005 = 1
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P(x > 0.02) = P(z > 1) = 0.1587..
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Cheers,
Stan H.