        \problemdata       % Takes 5 *mandatory* arguments
        {variance-6}             % latex-friendly label for the prob.
        {variance}               % The topic of the problem content.
        {Velleman}               % Source (if known)
        {S98 PS11-7; F97 PS11-8} % Usage (list ones you are aware of).
        {Theory Pig, S02}        % Last revision info (author, date).
% FA 2001, tutorial 13
\begin{problem} 
\textbf{Central Limit Theorem.} Let $B_n$ be a random variable
with binomial distribution $f_{n,p}$.

\begin{problemparts}

\problempart Write a formula defining the random variable, $B^*_n$, which is the
``normalized'' version of $B_n$ (normalized to have mean 0 and standard
deviation 1).

\solution{
The normalized version of any variable, $R$, is $(R - \mu_R)/\sigma_R$.
But $\expect{B_n} = np$ and $\variance{B_n} = np(1-p)$, so
\[
B^*_n = \frac{B_n-np}{\sqrt{np(1-p)}}.
\]
}

\problempart Explain why
\[
\lim_{n \rightarrow \infty}\pr{B^*_n < \beta} = N(\beta),
\]
where $N(\beta)$ is the normal distribution from the Central Limit Theorem
(see Appendix).

\solution{

Note that $B_n$ has the same distribution as the sum, $S_n$, of $n$
mutually independent Bernoulli random variables, $G_i$, with mean $p$ and
variance $p(1-p)$.  So the normalized version, $B^*_n$ of $B_n$ has the same
distribution as the normalized version, $T_n$, of $S_n$.  The Central
Limit Theorem says that
\begin{equation}\label{T}
\lim_{n \rightarrow \infty}\pr{T_n < \beta} = N(\beta),
\end{equation}
and since the distributions of $T_n$ and $B^*_n$ are the same,
equation~(\ref{T}) holds for $B^*_n$ as well.
}

\problempart Suppose you flip a fair coin 100 times.  The coin flips are all
mutually independent.  Use the standard normal distribution to approximate
the probability that the number of heads is between $30$ and $70$.  Can you
give a lower bound on this probability using the generalized Chernoff
bound?  (You can leave $N(\beta)$ and $e$ in your answer.)

\solution{ We have an unbiased binomial distribution for this case.  The
expected number of heads is $50$ and the variance of the number of heads
is $100/4 = 25$.  Let $S_{100}$ be the number of heads.  We have
\begin{align*}
\pr{ 30 < S_{100} < 70} & = \pr{-4 < T_{100} < 4}\\
& \approx N(4)-N(-4)\\
& = 2N(4)-1=0.9999.
\end{align*}
Note that to make this approximation practical $n$ needs to be
``sufficiently large''.  A rule often stated is that $n$ is ``sufficiently
large'' if $np\leq 5$ and $n(1-p)\leq 5$.

Since the distribution is symmetric, $\pr{30 < S_{100} <
70}=1-2\pr{S_{100} \geq 70}$. Applying the generalized Chernoff bound
derived in the lecture note, we have
$$\pr{S_{100} \geq 70} \leq e^{-4^2/2}$$
where $4=\frac{20}{5}$. Therefore, 
$$\pr{30 < S_{100} < 70}=1-2\pr{S_{100} \geq 70} \geq 1-2e^{-4^2/2}=0.9993$$ 
which gives us a good lower bound on $\pr{ S_{100} \geq 70}$.
}

\end{problemparts}
\end{problem}

