Suppose that the error, in grams, of a balance has the density $$f(x)=\frac{1}{4}e^\frac{-|x|}{2}$$ for $-∞<x<∞$, and that 100 items are weighed, independently of each other. Use the Central Limit Theorem to approximate the probability that the absolute difference between the true total weight and the measured total weight is more than 50 grams.
I have calculated that $EX=0$ and $varX=8$ which I know is correct. As I understand it, I want to approximate $$P(X>50)=1-P(X\leqslant 0)$$But I am not sure I understand how this can be done using the theorem... Thanks for any help
The error on a single weigth has a Laplace distribution with mean $0$ and variance $8$.
By the CLT, when $100$ items are weigthed, the distribution of the cumulative error is very close to a normal distribution with mean $0$ and variance $800$, so a really close approximation of the probability that the cumulative error exceeds $50$ in absolute value is given by:
$$1-\frac{1}{40\sqrt{\pi}}\int_{-50}^{50}\exp\left(-\frac{x^2}{1600}\right)\,dx =1-\frac{1}{\sqrt{\pi}}\int_{-5/4}^{5/4}e^{-x^2}\,dx\approx \color{red}{7,71\%}.$$