Using Central Limit Theorem to find a probability for the sample mean Let $X_1, X_2, \ldots , X_{10000}$ be i.i.d. Expo(1). Let $\bar X_{10000} = \sum X_i/10000$. Use CLT to
 $10000$ estimate $P(0.893 < \bar X_{10000} < 1.003)$
I know how to find probability using CLT given mean and variance, but I'm not familiar with this form.
 A: If $X_i,\; i = 1, 2, \dots, 10000$ are independently distributed as
$\mathsf{Exp}(1),$ then $E(\bar X) = 1,\, Var(\bar X) = 1/10000,$
and $SD(\bar X) = 1/\sqrt{10000} = 1/100.$
By the central limit theorem, $\bar X$ is approximately
$\mathsf{Norm}(\mu = 1,\, \sigma = 0.01).$
From R statistical software, $P(0.893 < \bar X < 1.003) = 0.61791.$
diff(pnorm(c(0.893, 1.003), 1, .01))
## 0.6179114

I suppose you are expected to evaluate the probability by standardizing
and using a printed normal table. Because approximations may be necessary in
using printed tables, you may get a slightly different answer. Here is
a start toward using a normal table:
$$P(0.893 < \bar X < 1.003) =
P\left(\frac{0.893 - 1}{0.01} < \frac{\bar X - \mu}{\sigma}
< \frac{1.003 - 1}{0.01}\right)
=P(-10.7 < Z < 0.3),$$
where $Z$ has a standard normal distribution. Because $-10.7$ is so far from 0
you won't find a useful value for the lower end of the interval in a printed table. But the probability
you want is essentially $P(Z < 0.3).$ There is almost no probability in a standard normal distribution below $-10.7.$ 
Note: The exact distribution of $\bar X$ is $\mathsf{Gamma}(shape=10000, rate=10000)$ so the exact probability to four places, not using the Central Limit Theorem, is 0.6191. The approach of an exponential mean to normal (according to the CLT)
is slow because the exponential distribution is extremely skewed, but
$n = 10000$ is a very large sample size, so the CLT gives a useful approximation.
diff(pgamma(c(.893,1.003), 10000, 10000))
## 0.6190677

