Central Limit Theorem to determine approximate probability Ten numbers are rounded to the nearest integer and then summed.I want to determine the approximate probability that the sum of the unrounded numbers equals the rounded sum of the original numbers.
To solve this question, I have firstly assumed that the fractional part of each number is uniformly distributed.Then i started by applying the central limit theorem but couldn't proceed further.
Can someone give me hint on how to proceed further.
 A: The answer to the question as asked is $0$ since the distribution of the sum of the unrounded numbers is continuous and the probability that a variable with a continuous distribution equals an integer is $0$. I have therefore answered the question as I thought it might be intended: Find the probability that the rounded sum of the variables is equal to the sum of the rounded variables.
Computing the Probability
Let the probability density $f(x)=\left[|x|\le\frac12\right]$, where $[\dots]$ are Iverson Brackets. The variance of the distribution with the density $f$ is $\frac1{12}$. The probability that the rounded sum of the variables is equal to the sum of the rounded variables is the same as the probability that the sum of the rounding errors is between $-\frac12$ and $\frac12$. That is,
$$
\begin{align}
\int_{-1/2}^{1/2}\overbrace{f\ast f\ast\cdots\ast f}^{10\ f\text{s}}(x)\,\mathrm{d}x
&=\frac{381773117}{928972800}\\
&\doteq0.41096
\end{align}
$$

Approximation Using the Central Limit Theorem
The variance of $\overbrace{f\ast f\ast\cdots\ast f}^{10\ f\text{s}}$ is $10\cdot\frac1{12}=\frac56$. Thus, $\frac12$ is $\sqrt{\frac3{10}}$ standard deviations. Therefore, the Central Limit Theorem says the probability that the rounded sum of the variables is equal to the sum of the rounded variables is approximately
$$
\bbox[5px,border:2px solid #C0A000]{\frac1{\sqrt{2\pi}}\int_{-\sqrt{\frac3{10}}}^{\sqrt{\frac3{10}}}e^{-t^2/2}\,\mathrm{d}t\doteq0.41612}
$$

Evaluation Using the Fourier Transform
The Fourier Transform of $f$ is
$$
\begin{align}
\hat{f}(\xi)
&=\int_{-1/2}^{1/2}e^{-2\pi ix\xi}\,\mathrm{d}x\\
&=\frac{\sin(\pi\xi)}{\pi\xi}
\end{align}
$$
Since the Fourier Transform of a convolution is the product of the Fourier Transforms, and vice-versa, we have
$$
\begin{align}
\int_{-1/2}^{1/2}\overbrace{f\ast f\ast\cdots\ast f}^{n\ f\text{s}}(x)\,\mathrm{d}x
&=\int_{-\infty}^\infty\overbrace{f\ast f\ast\cdots\ast f}^{n\ f\text{s}}(x)\,f(x)\,\mathrm{d}x\\
&=\int_{-\infty}^\infty\hat{f}(\xi)^n\hat{f}(-\xi)\,\mathrm{d}\xi\\
&=\int_{-\infty}^\infty\left(\frac{\sin(\pi\xi)}{\pi\xi}\right)^{n+1}\mathrm{d}\xi
\end{align}
$$
which, using contour integration, evaluates to
$$
\frac1{2^nn!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n+1}{k}(n-2k+1)^n
$$
Setting $n=10$ yields $\frac{381773117}{928972800}$, as above.
