Finding the probability of an absolute value being lower than, using the central limit theorem.

Question

So for example we're trying to find $P(|X -1| \le 0.6)$ with a mean $1$ and standard deviation $1$.

Using the teacher's advice, I've formatted the question to be

$P(|X -1| \le 0.6) = P(X -1 \le 0.6) + P(X -1 \ge -0.6)$

But my answer always seem to be wrong no matter where I go from here, any input about this?

Answer 1

Assuming @AndreNicholas has interpreted the problem correctly, an interpretation that makes sense to me, you want to find the $P(0.4 \le \bar X \le 1.6),$ where $\bar X$ is the mean of $n = 9$ iid $X_i \sim Exp(1).$

The precise distribution of the mean is given by $\bar X \sim Gamma(shape=9, rate=9).$ This distribution has mean 1, variance 1/9 and SD 1/3. From R statistical software, one finds the exact value of $P(0.4 \le \bar X \le 1.6) = 0.9374$ (to four places).

 diff(pgamma(c(.4,1.6), 9, 9))
 ## 0.9374498

I believe $n = 9$ is much too small for a good normal approximation via the Central Limit Theorem for many probabilities involving your $\bar X$. Nevertheless, that approximation happens to work pretty well for this particular case, yielding 0.9281 (to four places).

 diff(pnorm(c(.4,1.6), 1, 1/3))
 ## 0.9281394

The figure below illustrates these computations.