Fifty numbers are rounded off to the nearest integer and then summed. If the individual round-off errors are uniformly distributed over (−.5, .5), approximate the probability that the resultant sum differs from the exact sum by more than 3.

Short Answer

Expert verified

PXi150>3=0.1416

Step by step solution

01

Step 1. Given information

Let Xirepresents the round-off error of ithnumber. It is given that the random variables Xi, i = 1,...,50 are uniformly distributed over (-.5,.5).

02

Step 2. Calculating mean and variance 

Mean = E(Xi)=0.5+(-0.5)2=0

Variance = V(Xi)=(-0.5-0.5)212=112

03

Step 3. Using Central limit theorem

PXi150>3=2×PXi150>3PXi150>3=2×PXi-50(0)15050×112>3-50(0)50×112PXi150>3=2×P{Z>1.47}PXi150>3=2×0.0708=0.1416

04

Final answer

The probability that the resultant sum differs from the exact sum by more than 3 is 0.1416.

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