CLT probalitity problem I have a probability problem with solution based on CLT.
I don't know how to solve it. Help me please.
Cleaner collects empty bottles on the street. Each bottle he finds with probability 0.2. Find the approximate value of the probability that he will collect at least 2200 bottles, if there are 10000 bottles in the street.
I know that I sould use the law of large numbers, but how?
Thank you.
 A: Let $X_i$, $1 \leq i \leq 10000$ be random variables which take the values $0$ or $1$. $X_i$ takes the values $1$ if the cleaner collects the $i$-th botte. The $X_i$ are thus identically distributed, with $\mathbb{P}(X_i=0) = 0.8$ and $\mathbb{P}(X_i=1)=0.2$.
The number of bottles the cleaner collects is $$
 Y = \sum_{i=1}^{10000} X_i,
$$ 
and you're interested in an approximate value for $$
  \mathbb{P}(Y \geq 2200) \approx 1 - \mathbb{P}(Y \leq 2200).
$$
Per the central limit theorem, $Y$ is approximately normally distributed, because it's the sum of independent and identically distributed random variables. It follows that $$
  \mathbb{P}(Y \leq 2200) \approx \Phi\left(\frac{2200 - \mu}{\sigma}\right),
$$
where $\Phi$ is the CDF of the standard normal distribution, $\mu$ is the expected value of $Y$ and $\sigma$ is the standard deviation of $Y$. All that remains is to figure out $\mu$ and $\sigma$ - can you do that using the CLT?
Hint: You'll need the mean $\mu_X$ and standard deviation $\sigma_X$ of the $X_i$ to find $\mu$ and $\sigma$.
A: The number of bottles found (denoted by $X$) follows a binomial distribution $\text{Binom}(N,p)$ where $N=10,000$ and $p=0.2$. Then, the probability $X\geq 2200$ can be computed precisely but since we're looking for an approximate value, let's look at the CLT.
We can write $X=\sum_{i=1}^NY_i$ where $Y_1,\ldots,Y_N$ are independent Bernoulli random variables each with probability $p$ of realizing $1$ . Then, by the Lindeberg-Levy CLT
$$
\frac{X}{\sqrt{N}}-\sqrt{N}p=\sqrt{N}\left(\frac{1}{N}\sum_{i=1}^NY_i-p\right)
$$
is approximately normal with mean $0$ and variance $\text{Var}(Y_i)=p(1-p)$. Thus,
$$
\Pr(X\geq 2200)=\Pr\left(\frac{1}{\sqrt{p(1-p)}}\left(\frac{X}{\sqrt{N}}-\sqrt{N}p\right)\geq\frac{1}{\sqrt{p(1-p)}}\left(\frac{2200}{\sqrt{N}}-\sqrt{N}p\right)\right)\\
\approx 1-\Phi(5)\approx 2.86652\times 10^{-7}.
$$
On the other hand, the exact probability (given by the binomial distribution) is
$$
\sum_{k=2200}^{10000}\binom{10000}{k}p^k(1-p)^{10000-k}\approx 4.09963\times 10^{-7}.
$$
A: Let $X_i$ be the random variable that takes the value $1$ if the bottle is empty and $0$ if the bottle isn't empty, with $i=1,2,\dots,10000$
You want $\mathbb P(\sum_i X_i\geq 2200)$. Note that $\mathbb P(\sum_iX_i=x)=\binom{10000}{x}(0.2)^x(0.8)^{10000-x}$. If $Y\stackrel{d}{=}\mathcal B(n,p)$, then you can approximate $Y\stackrel{d}{\approx}\mathcal N(np, np(1-p))$. Then you look for $1-\mathbb P(\sum_iX_i\le 2200)=1-\mathbb P\left(\dfrac{\sum_iX_i-2000}{40}\le 5\right) = 1-\Phi(5)\approx 0.$
