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\title{Stat 311-Lecture 11 Central Limit Theorem}
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%\author{\Large Probability}
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Oct 24, 2005

\begin{enumerate}
\item If the cdfs of $X$ and $Y$ are identical, two random variables
are identically distributed. This does not imply $X=Y$ which is
nonsense. To denote the equality of distribution, we will use
notation $X {\sim} Y$.
\item If $S_n = X_1 + X_2 + \cdots X_n$, where $X_i$ are identically
distributed random variables coming from independent events with
$\mathbb{E} X_i = \mu$ and $\mathbb{V} X_i = \sigma^2$. $X_1,
X_2,\cdots, X_n$ are usually called i.i.d. random variables. Let
$$Z_n = \frac{S_n -\mathbb{E}S_n}{\sqrt{\mathbb{V} S_n}} =\frac{S_n
-n\mu}{\sqrt{n}\sigma}.$$ Then for large $n$,
$$P(Z_n \leq x) \approx \Phi(x),$$
where $\Phi$ is the cdf for a standard normal distribution. Note
$$P(a \leq Z_n \leq b) \approx \Phi(b) -\Phi(a).$$

\item Problem. Let $X$ be the number of heads in 40 tossed coins.
Find the probability $X=20$. {\em Solution.} Note $X=X_1 + \X_2 +
\cdots X_{40}$ with $X_i \sim Bernoulli(0.5)$. Note $\mathbb{E} X =
40 \cdot 0.5, \mathbb{V} X = 40 \cdot 0.5^2$. Let $S= \frac{X -
20}{\sqrt{10}}$. \bq P(X=20) = P(19.5 \leq X \leq 20.5) \\=
\Phi(\frac{0.5}{\sqrt{10}}) - \Phi(-\frac{0.5}{\sqrt{10}})\\
= 2\Phi(\frac{0.5}{\sqrt{10}}) - 1 = 2 \codt 0.5636-1 = 0.1272.\eq
The exact result is $P(X=20) = {40 \choose 20}0.5^{40} = 0.1254$.


\item  Problem. A fair coin is thrown 1000 times. Find the approximate probability
that the total number of heads among 1000 tosses will lie between 400 and 600
using the Central Limit Theorem. \\
{\it Solution.} Let $X_i \sim \mbox{Bernoulli}(\frac{1}{2})$. With
notation $\bar X = \frac{X_1 + X_2 \cdots + X_n}{n}$, $P(400 \leq
\sum_{i=1}^{1000} X_i \leq 600) = P(-0.1 \leq \bar X-0.5 \leq 0.1) =
P\Big(\frac{-0.1}{1/20 \sqrt{10}} \leq \frac{\bar X - 0.5}{1/20
\sqrt{10}} \leq \frac{0.1}{1/20 \sqrt{10}}\Big) =
2\Phi(\frac{2}{\sqrt{10}})-1 =0.47$.

\item Problem. The expected service time for a customer coming through a checkout counter in a retail store is 2 minutes while its variance
is 1. (a) Approximate the probability that 100 customers can be served in less than 3 hours of total service time.
(b) Find the number of customers that can be served in less than 3 hours with probability 0.9.\\
{\em Solution.} (a) Let $X_i$ be the service time for the $i$-th
customers. $S= X_1 + \cdots + X_{100}$. $\mathbb{E} S = 200,
\mathbb{V} S = 100$. Let $Z= \frac{S - 200}{10}$. Then $$P(S \leq
180) = \Phi(\frac{180 - 200}{10})= 1-\Phi(2) = 0.0228$$. (b) For
$S_n = X_1 + \cdots +X_{n}$, $\mathbb{E} S = 2n, \mathbb{V} S = n$.
Let $Z= \frac{S_n - 2n}{\sqrt{n}}$. Then we need $P(Z \leq
\frac{180-2n}{\sqrt{n}}) = 0.9$. From the table $\Phi(1.28)=0.9$. So
we need to solve $180-2n = 1.28\sqrt{n}$.

\end{enumerate}
{\bf HW 3} Due Nov. 10. 8:00AM. Solve the following 8 problems in
Chapter 5. Problems 1, 10, 11, 16, 26, 37 Theoretical Exercises 26,
29.\\ {\bf Exam schedule:} Nov. 17 (Thursday). Second midterm exam
will cover chapters 4-5 and parts of chapter 6 up to November 10th
lecture. We will have 4-5 problems.


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