Round-off error variance for uniform distribution over (-0.5, 0.5)

[xdrgnh]

http://kaharris.org/teaching/425/Lectures/lec37.pdf

"Fifty numbers are rounded-off to the nearest integer and the summed.
Suppose that the individual round-off errors are uniformly distributed
over (0:5; 0:5). What is the probability that the round-off error
exceeds the exact sum by more than 3?"I don't understand how they calculate the variance where they get the 1/12

This isn't homework nor am I taking a class in statistics. I will however be taking an exam in statistics so I can credit for a class in statistics in May. Any help will be appreciated.

 

[BvU]

The problem wording is rather bad and confusing. If I try to render what it should be, I get something like

Fifty numbers are rounded-off to the nearest integer and then summed.
Suppose that the individual round-off errors are uniformly distributed
over the interval [-0.5, 0.5]. What is the probability that the rounded-off sum exceeds the exact sum by more than 3?

Your question where the $\tfrac {1} {12}$ comes from, can be answered with a simple: from the definition of the variance: calculate E[x²] for a uniform distribution.

Spoiler

In your lecture suite: lecture 20.

 

[Mark44]

Suppose that the individual round-off errors are uniformly distributed
over (0:5; 0:5).

The interval should read "[-0.5, 0.5]".