# using central limit theorem

I recently got a tute question which I don't know how to proceed with and I believe that the tutor won't provide solution... The question is

Pick a real number randomly (according to the uniform measure) in the interval $[0, 2]$. Do this one million times and let $S$ be the sum of all the numbers. What, approximately, is the probability that
a) $S\ge1,$
b) $S\ge0.001,$
c) $S\ge0$?
Express as a definite integral of the function $e^\frac{-x^2}{2}$.

Can anyone show me how to do it? It is in fact from a Fourier analysis course but I guess I need some basic result from statistcs which I am not familiar with at all..

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(c) is easy and has nothing to do with the function $e^\frac{−x^2}{2}$. – Henry Aug 19 '12 at 20:15
Since you are summing 1,000,000 random numbers with mean value 1, the sum will be usually about 1,000,000. If you are asking about the average value of the 1,000,000 numbers, it becomes more interesting. – marty cohen Aug 19 '12 at 21:24
@marty: not much: (a) would become $\frac12$, (c) would still be $1$ and (b) would still be very close to $1$ – Henry Aug 19 '12 at 22:58

You have $1\, 000\, 000$ independent random variables $X_j$ with uniform distribution on the interval $[0,2]$. Such a random variable has mean $\mu = 1$ and variance $\sigma^2 = 1/3$. The central limit theorem says that if $S_N$ is the sum of $N$ independent random variables with the same distribution, having mean $\mu$ and variance $\sigma^2$, then in the limit as $N \to \infty$, the distribution of $(S_N - N \mu)/(\sqrt{N} \sigma)$ approaches the standard normal distribution. Thus, for any $z$,

$$\text{Prob} \left( \frac{S_N - N\mu}{\sqrt{N} \sigma} \ge z \right) \to \int_z^\infty \frac{e^{-t^2/2}}{\sqrt{2\pi}} dt$$

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Do you think this approximation about $1732$ standard deviations from the mean is better in any sense than the value $1$? – Henry Aug 19 '12 at 20:25
Good point. I suspect the question was meant to be about the average rather than the sum. – Robert Israel Aug 20 '12 at 1:52
I don't think changing the question to the average would make the central limit theorem any more useful: the approximation might be correct for (a) but the exact answer is in any case obvious by symmetry, while (b) and (c) are still extreme. – Henry Aug 20 '12 at 7:13

Let's call $S_n$ the sum of the first $n$ terms. Then for $0 \le x \le 1$ it can be shown by induction that $\Pr(S_n \le x) = \dfrac{x^n}{2^n \; n!}$

So the exact answers are

a) $1 - \dfrac{1}{2^{1000000} \times 1000000!}$

b) $1 - \dfrac{1}{2000^{1000000} \times 1000000!}$

c) $1$

The first two are extremely close to 1; the third is 1. The central limit theorem will not produce helpful approximations here, so you may have misquoted the question.

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@André: Overall perhaps, but I think not for $0 \le x \le 1$ – Henry Aug 19 '12 at 21:24
Yes, you are absolutely right, I was thinking of the full range, which is irrelevant. – André Nicolas Aug 19 '12 at 21:27