70 numbers are rounded off to the nearest integer and then summed. If the individual round-off error are uniformly distributed over (-.5,.5) what is the probabilty that the resultant sum differs from the exact sum by more than 3?
2006-10-22 17:51:38 · 2 answers · asked by NONE N 1 in Science & Mathematics ➔ Mathematics
Hmmm, I think the probability is zero because the round-off errors are UNIFORMLY distributed over (-.5, .5) That means that 35 numbers will be rounded down and 35 numbers will be rounded up. In otherwords, if one number is rounded down by 0.4, there will be a corresponding number that is rounded up by 0.4. So the net change in the total sum will be zero.
2006-10-22 18:16:41 · answer #1 · answered by z_o_r_r_o 6 · 0⤊ 0⤋
One approach is to use the standard deviation of a uniformly distributed error, which is 1/sqrt(12).
Then 3/stdev = 3*sqrt(12) = 6*sqrt(3) ~ 10, which makes the probability of such an error smaller than 10^-15, which is the error for 8 standard deviations.
2006-10-23 01:50:12 · answer #2 · answered by arbiter007 6 · 0⤊ 0⤋