I need help with the following question:
A computer in adding numbers rounds each number to its nearest integer.
Suppose that all rounding errors are independent and uniformly distributed over (-0.5, 0.5).i) If 1500 numbers are added, what is the probability that magnitude of the total error exceeds 15?
ii) How many numbers may be added together so that magnitude of the total error is less than 10 with probability 0.9?
How I started:
Let X be a random variable denoting the rounding error.
Probability distribution function $f(x)$ of $X$ would be:
$1 \quad -0.5 \le X \le 0.5$
$0 \quad otherwise$
I can't figure out how to answer the first part. Someone please help.
Best Answer
Your question has been answered here: Connection to Normal distribution
Using Thomas Ahle's formula there gives the exact answer. \begin{align*} P(S<s) = \frac{1}{n!}\sum_{k=0}^{\lfloor s\rfloor}(-1)^k\binom{n}{k}(s-k)^n \end{align*}
We need to compute: $$\big(1-P(S<765)\big)\cdot 2 \approx 0.17972990949058$$
Using the normal approximation:
\begin{align*} 2\times \mathbb{P}\left(x>15\right) &= 2\times \mathbb{P}\left(z>\frac{15}{\sqrt{125}}\right) \\ &= 2\times \int_{15/\sqrt{125}}^{\infty} \, \frac{1}{\sqrt{2\, \pi}} e^{-z^2/2}\, dz \\ &\approx 0.17971249 \end{align*}
or use the table for normal distribution.