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ANSWER

Let Y_i ~ U(0,1) be IID.

where,

$X =\sum_{i=1}^{n} ln(Yi)$

now we can assume, Xi has

mean $\mu = E[Xi]$

and

variance $\sigma ^{^{2}} = V[Xi]$

using central limit theorem,

means,

$X \xrightarrow[n\rightarrow \infty]{d } N (n\mu ,n\sigma ^{2})$

Thus,

for large value of n, we get

where, X is approximately normally distributed with mean

$E[X] = n\mu$

and

standard deviation

$V[X] = \sigma{\sqrt{}}n$

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