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Old 07-31-2007   #1
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Default Can someone help me solve this statistics problem?

80 numbers are rounded off to the nearest integer and then summed. If the individual round-off error are uniformly distributed over (-0.5, 0.5) what is the probability that the resultant sum differs from the exact sum by more than 3?Please help me solve this problem and tell me the steps as well. Thanks....it will be helpful of you!!!
Old 07-31-2007   #2
Patrick S
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Old 07-31-2007   #3
dead inside
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Say X1 is one observation of the 60.E(X1) is 0 and V(X1) is 1/12 (common properties of a uniform dist)Then if S = Sum of all Xs 1 -> 60E(S) = E(X1+X2+...+X60) = 60E(X1) = 0V(S) = V(X1 + X2 +...+ X60) = 60V(X1) = 5Since N = 60 you are able to use the normal distribution with mean = 0 and Variance = 5One note is that you need to multiply your resulting probability by 2 as it is a 2-tailed test. You need to find (Given that T = true sum value) P( S > T + 3) and P( S < T - 3)
Old 07-31-2007   #4
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''math guy'' is correct

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