# Thread: Approximating Uniform Dist by CLT

1. ## Approximating Uniform Dist by CLT

I'm given a question which asks the following:

In adding n real numbers, each is rounded to the nearest integer. Assume that the round-off errors, $X_{i}$, $i = 1,...,n,$ are independently distributed as $U(-0.5,+0.5)$

Obtain the approximate distribution of the total error $\sum X_{i}$ in the sum of the n numbers and hence find the probability that the absolute error in the sum is at most $\frac{1}{2}\sqrt{n}$.

Now, I assume it wants me to approx to Normal dist by CLT, which is:

$
\frac{\sqrt{n}(\overline{X} - \mu)}{\sigma}$
~ $N(0,1)$

From the Uniform dist, I know E(X) = 0, and Var(X) = $\frac{1}{12}$

So does this mean that the approximate dist of $\sum X_{i}$ is

$\overline{X_{i}}\sqrt{12n}$ ~ $N(0,1)$

or am I missing something regarding the fact that it's the dist of the sum rather than just X itself?

2. Originally Posted by Zenter
I'm given a question which asks the following:

In adding n real numbers, each is rounded to the nearest integer. Assume that the round-off errors, $X_{i}$, $i = 1,...,n,$ are independently distributed as $U(-0.5,+0.5)$

Obtain the approximate distribution of the total error $\sum X_{i}$ in the sum of the n numbers and hence find the probability that the absolute error in the sum is at most $\frac{1}{2}\sqrt{n}$.

Now, I assume it wants me to approx to Normal dist by CLT, which is:

$
\frac{\sqrt{n}(\overline{X} - \mu)}{\sigma}$
~ $N(0,1)$

From the Uniform dist, I know E(X) = 0, and Var(X) = $\frac{1}{12}$

So does this mean that the approximate dist of $\sum X_{i}$ is

$\overline{X_{i}}\sqrt{12n}$ ~ $N(0,1)$

or am I missing something regarding the fact that it's the dist of the sum rather than just X itself?
Total error $X=\sum X_i \sim N(0,n/12)$ (approx, by CLT)

(Note: $\overline{X_{i}}=0$)

CB

3. $\mu=0$

CB uses $\overline{X}$ as the population mean

The sample mean $\overline{X}$ is not zero, it's a random variable and there is no subscript i to it.

4. Originally Posted by matheagle
$\mu=0$

CB uses $\overline{X}$ as the population mean

The sample mean $\overline{X}$ is not zero, it's a random variable and there is no subscript i to it.
There appears to be some part of the conversation missing here, my comment about $\overline X_i$ is due to its appearence in the last expression in the original post, so the LHS of that expression is a numeric constant equal to zero and not a standard normal RV.

$X_i$ is the RV representing the $i$-th round off error and is $\sim U(-0.5,0.5)$ so by definition $\overline X_i =E( X_i)=0$

CB

5. Originally Posted by CaptainBlack
Total error $X=\sum X_i \sim N(0,n/12)$ (approx, by CLT)

(Note: $\overline{X_{i}}=0$)

CB
I understand what you did here, but then using this for the part where I work out the probability, I get

$P(Z \leq \frac{\sqrt{12n}}{2})$

Which, I don't think would be the right answer, since I'm not given a numerical figure for n, meaning I can't work that ^ out.

6. Originally Posted by Zenter
I understand what you did here, but then using this for the part where I work out the probability, I get

$P(Z \leq \frac{\sqrt{12n}}{2})$

Which, I don't think would be the right answer, since I'm not given a numerical figure for n, meaning I can't work that ^ out.
That's funny, 'cos I get $\mathrm P(|Z|<\surd3)$. Any other offers?

7. Originally Posted by halbard
That's funny, 'cos I get $\mathrm P(|Z|<\surd3)$. Any other offers?
Which is what I get as well, the $\sqrt{n}$ terms all cancel.

CB