Question

(c) We taka a sum of 10000 numbers, each of which is rounded up to 10-3. Let us assume that the resulting errors are random variables from a uniform distribution over the in terval (-10-3/2, 10-3/2). Find an interval (the narrowest possible), into which the total aggregate error will fall with probability equal to at least 0.95.

Answer 1

here for a random error, expected error =(a+b)/2 =(10^-3+(-10^-3))/2 =0

and std deviation for a random error=(b-a)/sqrt(12) =(10^-3-(-10^-3))/sqrt(12)=2*10^-3/sqrt(12)

therefore for sum of 1000 aggregate errors;

expected total error =0*1000=0

and std error =2*10^-3*sqrt(1000)/sqrt(12)=0.0183

hence interval in which 95% of aggregate error will fall =estimated mean -/+z*std error

=0 -/+1.96*0.0183 =-0.0358 to 0.0358