The central limit theorem can be used to analyze round-off error. Suppose that the round-off error is represented as a uniform random variable on [−1 2 , 1 2 ]. If 100 numbers are added, approximate the probability that the round-off error exceeds (a) 1, (b) 2, and (c) 5.

The mean of the sum of 100 variables is 100*0 since all have mean 0.  The variance is the sum of the variances, which is 100 * 1/12 = 8.333.  The standard deviation is the square root, 2.887.  

a. The probability that the sum is greater than 1 is Prob[x > 1] = Prob[(x - 0)/2.887 > (1 – 0)/2.887] = .3645     If you interpret this to mean that the absolute value of the sum is > 1, then the probability is doubled.  

b.  Prob[x > 2] = Prob(z > 2/2.887) =  .245  (or .490).  

c.  Prob[x > 5] = Prob(z > 5/2.887)  = .0416 (or .0892).

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3/6   and 9/18

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answer: 5

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