Question

Fifty numbers are rounded off to the nearest integer and then summed. If the individual round-off errors are uniformly distributed over (−0.5, 0.5), approximate the probability that the resultant sum differs from the exact sum by more than 3.

Answer 1

Here,

Individual round-off errors are uniformly distributed over (-0.5,0.5)

Expected value = (0.5+(-0.5))/2=0

Variance = (0.5-(-0.5))² / 12 = 1/12

Standard deviation = sqrt(Variance ) 0.289

Standard deviation of error, S = 0.289*sqrt(50) = 2.0435

Now, The probability that the resultant sum differs from the exact sum by more than 3 is

= 1- P(-3< S < 3)

= 1-P( (-3-0)/2.041<Z<(3-0)/2.041 ) ( By converting to standard normal variable)

= 1- P(-1.4699 < Z < 1.4699)

= 1- P(Z<1.4699) + P (Z<-1.4699)

= 1- 0.9292 + 0.0708

=0.1416