Suppose that $U_{i},\dots,U_{n}$ are iid $U(0,1)$. Use the central limit theorem to find an approximation for:
$$P(U_{1}\times U_{2} \times\dots\times U_{25}\leq 6\times 10^{-6} )$$
Answer:
Using Wolfram http://mathworld.wolfram.com/UniformProductDistribution.html, the distribution function for the product of the standard uniform variates is:
$$\frac{-1^{n-1}}{(n-1)!}{(\ln u)}^{n-1} = \frac{-1^{24}}{(24)!}{(\ln u)}^{25-1} = \frac{1}{(24)!}{(\ln u)}^{24}$$
The $\mu$ for this distribution is $\frac{1}{(24)!}{(-1)}^{24} = \frac{1}{(24)!}1 = \frac{1}{(24)!} = 1.6 \times 10^{-24}$
The $\sigma ^{2}$ is $\frac{1}{(24)!}{(1)}^{24} = 1.6 \times 10^{-24}$
Therefore $\frac{\sigma }{\sqrt{n}}$ is $\frac{1.27\times 10^{-12}}{\sqrt{25}}= 2.54\times 10^{-13}$
Since n=25 is large enough, apply CLT:
$\bar{U_{i}}\sim N(1.6 \times 10^{-24},2.54\times 10^{-13})$, then
$$P(Z\leq \frac{6\times 10^{-6}-1.6\times 10^{-24}}{2.54\times 10^{-13}})$$ $$P(Z\leq 2.36\times 10^{7})$$
which doesn't make sense. Please help. Thanks.