Central limit theorem, when do we know if n isn't large enough I'm working on a statistics question, and I'm stumped on how to answer it.
Here is the question
According to a survey conducted by the American Bar Association, 1 in every 410
Americans is a lawyer, but 1 in every 64 residents of Washington, D.C., is a lawyer.


(a) Use the Central Limit Theorem to approximate the probability that there is at
least one lawyer in a random sample of 1500 Americans. Is n = 1500 large enough
for the approximation to work well?

Using the CLT, I found that the answer was approximately .95, but I don't know how to determine if the sample size is large enough for the approximation? What do you look for in this case?
 A: You can calculate the exact probability and compare it to the CLT approximation. The number of lawyers in a sample of $1500$ has binomial distribution with $n=1500$ and $p=1/410$, so the probability that there are no lawyers in your sample is
$$(1-p)^n=\left(1-\frac1{410}\right)^{1500}.$$
Subtract this number from $1$ to get the exact prob of least one lawyer in your sample.
Aside: When you did the CLT approximation, did you apply the continuity correction? (Do you know about the continuity correction?)
A: $Exact\; Binomial$: As per the Answer by @grand_chat. Let the number of lawyers
in a random sample of 1500 Americans be $X \sim Binom(1500, 1/410)$.
We seek $P(X \ge 1) = 1  - P(X = 0).$ From R, we have:
 1 - (409/410)^1500
 ## 0.9743447
 1 - dbinom(0, 1500, 1/410)
 ## 0.9743447

$Poisson\; approximation\; to\; binomial$. We must have 
$\lambda = E(X) = 1500/410 = 3.658537,$ so that
$Y \sim Pois(\lambda).$ Again, we seek $P(Y \ge 1) = 1 - P(Y=0).$
 1 - exp(-lam)
 ## 0.9742298
 1 - dpois(0, lam)
 ## 0.9742298

$Normal\; approximation$. For the binomial distribution, $\mu = np = 1500/410,$
$\sigma^2 = np(1-p) = \mu(409/410).$ Now let $W \sim Norm(\mu, \sigma).$ We seek $P(W \ge 1) = P(W > .5) = 1 - P(W < .5).$
 mu = 1500*(1/410);  var = mu*(409/410); sg = sqrt(var)
 1 - pnorm(.5, mu, sg)
 ## 0.9508693

Of course $P(W > .5)$ can be found from printed normal tables
by standardizing: $P(W > .5) = P[Z = (W - \mu)/\sigma > (.5 - \mu)/\sigma)].$ This seems to be the result you obtained. [Note: Application of the 'continuity
correction' leads us to seek $P(W \ge .5)$ instead of $P(W \ge 1).$]
$Summary\; comments.$ The Poisson approximation is very good (see figure below). Although the 'usual rule' for using the the normal approximation as given in the
comment does not hold, the normal approximation is not very good here for two additional
reasons: (a) We are dealing with a probability in the tail of
the binomial distribution, (b) we are dealing with a
skewed binomial distribution ($p$ far from 1/2). In statistical practice, normal
approximations are less frequently used nowadays than before
software was available to do exact computations.
The figure below shows the binomial distribution (black bars), its
Poisson approximation (purple dots), and the normal curve that matches the binomial mean and variance.

