Question

Forty-eight measurements are recorded to several decimal places. Each of these 48 numbers is rounded off to the nearest integer. The sum of the original 48 numbers is approximated by the sum of these integers. If we assume that the errors made by rounding off are iid and have a uniform distribution over the interval $(-\frac{1}{2},\frac{1}{2})$, compute approximately the probability that the sum of the integers is within two units of the true sum.

Short answer

There is approximately a 0.383 (or 38.3%) chance that the sum of the rounded integers is within two units of the true sum.

Step-by-step solution

Set Up The Variables

Let $x_i$ represent each original measurement. The rounding error is then $e_i=[x_i]-x_i$, where $[x_i]$ represents the integer rounded from $x_i$. These $e_i$ are independent and identically distributed (iid) random variables over the interval $(-\frac{1}{2},\frac{1}{2})$, a uniform distribution.

Compute Mean and Variance

For a uniform distribution over the interval $(-\frac{1}{2},\frac{1}{2})$, the mean is $E(e_i)=0$ and the variance is $Var(e_i)=\frac{1}{12}$. We are interested in the sum $S=\sum e_i=e_1+e_2+...+e_{48}$. The mean of $S$ is $E(S)=\sum E(e_i)=0$ and the variance of $S$ is $Var(S)=\sum Var(e_i)=48\cdot \frac{1}{12}=4$.

Apply the Central Limit Theorem

Since these are a sum of a large number of iid random variables, by the Central Limit Theorem, $S$ is approximately normally distributed. So, $S\sim N(0,4)$. We want the probability that $S$ is between -2 and 2, so we subtract the mean and divide by the standard deviation to normalize. That is $P(-2\le S\le 2)=P(-\frac{1}{2}\le \frac{S}{\sqrt{4}}\le \frac{1}{2})\sim \mathrm{\Phi }\left(\frac{1}{2}\right)-\mathrm{\Phi }(-\frac{1}{2})$, where $\mathrm{\Phi }$ is the cumulative distribution function of the standard normal distribution.

Use Standard Normal Distribution

The probabilities with standard normal distribution can be looked up in a standard normal table or calculated using software. So finally, $P(-\frac{1}{2}\le \frac{S}{\sqrt{4}}\le \frac{1}{2})=\mathrm{\Phi }\left(\frac{1}{2}\right)-\mathrm{\Phi }(-\frac{1}{2})=0.6915-0.3085=0.383$.