Markov's Inequality Summation Bound Let $X_1,\ldots,X_{20}$ be independent Poisson random variables with mean 1. Use central limit theorem to approximate the following equation. Use Markov's Inequality to obtain a bound:
$$\Pr\left[\sum_1^{20}X_i>15\right]$$
Since the mean is $1$, the distribution would be $\dfrac{1}{k!e}$. Markov's Inequality states that $\Pr\left[\sum_1^{20}X_i>15\right]\le 1/15$. I don't know how to apply the central limit theorem in this situation, though.
 A: The sum of the $20$ random variables has expected value $20$ since the expected value of each is $1$.
Have you heard that the sum of independent Poisson-distributed random variables also has a Poisson distribution?  Or that for the Poisson distribution, the expected value is the same as the variance?  So
$$
Z = \frac{\text{sum} - 20}{\sqrt{20}}
$$
has expected value $0$ and standard deviation $1$.
Notice that $\text{sum}>15$ if and only if $\text{sum}\ge 16$.  Should we seek $\Pr\left(Z > \dfrac{15-20}{\sqrt{20}} \right)$ or $\Pr\left(Z \ge \dfrac{16-20}{\sqrt{20}} \right)$?  When approximating this discrete distribution with the $N(0,1)$ distribution, just taking the average, $\dfrac{15+16} 2 = 15.5$, works quite well.
Markov's inequality doesn't say the probability that the sum exceeds $15$ is $\le 1/15$.  It does say that the probability that the sum exceeds $15$ times its expected value is $\le1/15$.  But its expected value is $20$, so it tells you the probability that the sum exceeds $15\times20=300$ is $\le 15$.  That's a very weak statement, since that probability is actually microscopic.
You're finding $\Pr(\text{sum}>15)$ and the expected value is $20$, so Markov's inequality just says the probability is $\le \dfrac{20}{16}$, which is more-or-less vacuous, since all probabilities of any kind are $\le$ that.
A: The "central limit theorem approximation" would replace a sum of $20$ iid Poisson variables with mean $1$ with a normal variable which have the appropriate mean and variance. Since the variance of a Poisson variable with mean $\lambda$ is also $\lambda$ and variances add for independent variables, the mean should be $20$ and the variance should be $20$. 
So this gives you a way to make an estimate. An interesting question is: is the estimate any good? A crude way to check is to use Markov's inequality or Chebyshev's inequality.
