Question

Fifty numbers are rounded off to the nearest integer and then summed. If the individual round-off errors are uniformly distributed over (−.5, .5), approximate the probability that the resultant sum differs from the exact sum by more than 3.

Short answer

$P\left\{\left|\underset{1}{\overset{50}{\sum X_i}}\right|>3\right\}=0.1416$

Step-by-step solution

Step 1. Given information

Let X_irepresents the round-off error of i^thnumber. It is given that the random variables X_i, i = 1,...,50 are uniformly distributed over (-.5,.5).

Step 2. Calculating mean and variance 

Mean = $E\left(X_i\right)=\frac{0.5+(-0.5)}{2}=0$

Variance = $V\left(X_i\right)=\frac{{(-0.5-0.5)}^2}{12}=\frac{1}{12}$

Step 3. Using Central limit theorem

$$\begin{gathered} P\left\{\left|\underset{1}{\overset{50}{\sum X_i}}\right|>3\right\}=2\times P\left\{\underset{1}{\overset{50}{\sum X_i}}>3\right\} \\ P\left\{\left|\underset{1}{\overset{50}{\sum X_i}}\right|>3\right\}=2\times P\left\{\frac{{\displaystyle \underset{1}{\overset{50}{\sum X_i-50\left(0\right)}}}}{\sqrt{50\times {\displaystyle \frac{1}{12}}}}>\frac{3-50\left(0\right)}{\sqrt{50\times {\displaystyle \frac{1}{12}}}}\right\} \\ P\left\{\left|\underset{1}{\overset{50}{\sum X_i}}\right|>3\right\}=2\times P\{Z>1.47\} \\ P\left\{\left|\underset{1}{\overset{50}{\sum X_i}}\right|>3\right\}=2\times 0.0708=0.1416 \end{gathered}$$

Final answer

The probability that the resultant sum differs from the exact sum by more than 3 is 0.1416.