Suppose that $U_i\sim \operatorname{iid} \operatorname{Unif} [-1,1]$ and consider $$\ X = \frac{U_1 + U_2 + \cdots + U_n} {\sqrt{U_1^2 + U_2^2 + \cdots + U_n^2}}$$ for large $n.$ What is the probability that $X > 1.$ Justify any approximation you use. How large $n$ should be?
2026-03-30 12:25:31.1774873531
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How do I solve this statistics question on approximation of distribution?
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More of a comment ... The $n=2$ case yields the symmetric Boobies distribution with pdf:
The exact cdf for the $n=2$ case is given below.
The following diagram illustrates convergence as $n$ increases...
Comparison to a standard Normal pdf when $n=7$:
and when $n = 20$:
Addendum: Given the joint pdf of $(U_1,U_2)$ as say $f(u_1,u_2)$, the cdf $P(X<x)$ is:





Note that $EU = 0$ and $Var(U) = EU^2 = 1/3$, then according to the comment by Nch, $$ X = \frac{\sqrt{3}\sum_{i=1}^n U_i/\sqrt{n}}{\sqrt{3}\sqrt{\sum_{i=1}^n U_i^2 / n}}. $$ Now, using WLLN and the continuous mapping theorem $$ \sqrt{\sum_{i=1}^n U_i^2 / n} \xrightarrow{p}\sqrt{E U_i^2}=1/\sqrt{3}, \,\, n \to \infty , $$ according to CLT $$ \sqrt{3}\sum_{i=1}^n U_i/\sqrt{n} \xrightarrow{D}Z, $$ where $Z \sim N(0,1)$, thus using Slutsky's theorem we have $$ X \xrightarrow{D} \frac{Z}{\sqrt{3}/\sqrt{3}} = Z. $$ Hence, for large enough $n$ $$ P(X>1) \approx P(\sqrt{3}Z>1)=1-\Phi(1/\sqrt{3}). $$