Question

45 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 probability that the resultant sum differs from the exact sum by more than 2?

Answer 1

SOLUTION:

From given data,

45 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 probability that the resultant sum differs from the exact sum by more than 2

Here each error has the distribution given as:

X U (-0.5, 0.5)

Therefore, for each trial, we get here:

E( ) = (-0.5 + 0.5) / 2 = 0

Var() = (0.5 - (-0.5))^2 / 12} =1/12

Therefore the distribution for 45 such numbers here will have the distribution as:

S N( =45*0, = 45/12 )

The probability here is computed as:

= 1 - P( -2 < X < 2 )

Converting this to a standard normal variable as:

= 1 - P( -2 / < Z < 2 / )

= 1 - P( - 1.03 < Z < 1.03 )

= 1 - P( Z < 1.03) + P(Z < - 1.03 )

Getting it from the standard normal tables, we get:

= 1 - 0.84849+ 0.15151 = 0.30302

Therefore 0.30302 is the required probability here.