Question:
Suppose that the error, in grammes, of a balance is a random variable with density $f(x) = \frac{1}{4}e^{−\frac{|x−1|}{2}}$, for $−\infty < x < \infty.$
Suppose that 100 items are weighed, and the errors of measurements are independent.
a) Find expectation and variance of the error in a single measurement.
b) Use the central limit theorem to find approximate value of the probability that absolute value of the total error is less than 50 grammes.
c) Use Chebyshev’s inequality to lower bound the probability that the total error is between 50 and 150 grammes.
My Answer:
(a) $\mathbb E(X)=\int_{-\infty}^{\infty}xf(x)dx=\frac{1}{4}\int_{-\infty}^{\infty}xe^{−\frac{|x−1|}{2}}dx=\frac{1}{4}\int_{-\infty}^{1}xe^{\frac{(x−1)}{2}}dx+\frac{1}{4}\int_{1}^{\infty}xe^{\frac{-(x−1)}{2}}dx=1$ (after integrating by parts a few times and using limits)
Similarly $Var(X)=\mathbb E(X^2)-(\mathbb E(x))^2=\int_{-\infty}^{\infty}\frac{x^2}{4}e^{\frac{-|x−1|}{2}}dx -1=9-1=8$ (after similar computations)
(b) I've used my pervious answers $\mu =\mathbb E(X)=1$ and $Var(X)=\sigma^2=8$. To give $\overline{X_n}\sim N(1,\frac{8}{100})$
$\Rightarrow \mathbb P(\overline{X_n}<50)=\Phi(\frac{50-1}{{\sqrt\frac{8}{100}}})=1$ Makes me think I've made a mistake as I dont think that makes sense :)
(c)
Think I need to be confident with (b) to tackle (c)
Comments
I'm unsure of all questions really, I dont think I fully understand how to use the central limit theorem. Any clarification, alternative solutions or even solution verification would be greatly appreciated
I'm doing a similar question for my homework, I found that E[X$^2$]=11, apologies, due to my reputation I am not able to comment.