Confidence interval for parameter of discrete distribution for large sample size 
$X$ is a random variable that can take on $(-a, 0, a), a > 0$ values with corresponding probabilities $(\frac{1}{2}p, 1 - p, \frac{1}{2}p), 0 < p < 1$. Find confidence interval for $p$ with confidence level of $1 - \alpha$, assuming sample size, $n$, is large.

I'm stuck on this exercise. Because the sample size is large I assume that in the end I can approximate it with normal distribution to get the confidence intervals. However how to get the parameters of that normal distribution?
I can find some estimate of $p$ using method of moments. Then I get that $p = \frac{S^{2}_n}{a^2}$, can I use it as a mean for this normal distribution? What about variance? I'm not really sure how to move further and would appreciate some hints.
 A: We have an i.i.d set of $X_1,X_2,\ldots,X_n$ as described about.  Let $S^-$, $S^0$, and $S^+$ be sets of indices such that, for $i\in S^-$, $X_i=-a$, and likewise for the other two sets, and let corresponding $c^-$, $c^0$, and $c^+$, be the respective counts of the number of times $-a$, $0$, and $+a$, appear in the sample.
First, see that $c^0$ is a sufficient statistic for $p$.
$$
\begin{align}
f(x_1,\ldots,x_n) &= \prod_{i=1}^{n}f(x_i) \\
&= \prod_{i \in S^{-}} f(x_i) \times \prod_{i \in S^{0}} f(x_i) \times \prod_{i \in S^{+}} f(x_i) \\
&= \prod_{i \in S^{-}} \frac{1}{2}p \times \prod_{i \in S^{0}} (1-p) \times \prod_{i \in S^{+}} \frac{1}{2}p \\
&= \left(\frac{1}{2}p\right)^{c^-} \times \left(1-p\right)^{c^0} \times \left(\frac{1}{2}p\right)^{c^+} \\
&= \left(\frac{1}{2}p\right)^{c^-+c^+} \times \left(1-p\right)^{c^0} \\
&= \left(\frac{1}{2}p\right)^{n-c^0} \left(1-p\right)^{c^0} \\
\end{align}
$$
That fact that $c^0$ is a sufficient statistic tells us that we can estimate $p$ from the number of times zero appears in the experiment.  Each trial gives a zero with probability $1-p$ and we have $n$ trials, so
$$c^0 \sim Binom(1-p,n)$$
$c^0$ will have an expectation of $n(1-p)$ and a variance of $np(1-p)$, so we can estimate $p$ by
$$
\hat p = 1-\frac{c^0}{n}
$$
and we have the usual Normal approximation confidence interval as
$$
\hat p \pm Z_\alpha \sqrt{\hat p (1-\hat p)/n}
$$
A: Define a new random variable $Y=\frac{1}{a}|X|$. Then $Y$ is of Bernoulli distribution with the probability $p$ for $Y=1$ and the probability $(1-p)$ for $Y=0$. For Bernoulli distribution with a mean equal to $p$, the variance equals $p(1-p)$. Thus, when the sample size $n$ is sufficiently large, the variable $Z=\frac{n\bar{y}-np}{\sqrt{np(1-p)}}$ well approximates the standard normal distribution, where $\bar{y}=\frac{1}{n}\sum_{i=1}^ny_i$ is the sample mean of $y$. The rest calculation is straightforward to find the confidence interval of $p$ with multiple optional methods.
