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{\bf Question}

If 16 digits are chosen from a table of random digits, what is the
probability that their average will lie between 4 and 6?




\vspace{.25in}

{\bf Answer}

Here $n=16$

Let $X_i$ be the chosen digit in the $i$th draw.

$X_i = 0,1,2,3,4,5,6,7,8,9$ with probability $\ds\frac{1}{10}$

Therefore $E(X_i)=\mu=\ds\frac{0+1+2+...+9}{10}=\frac{9 \times
10}{2 \times 10} = 4.5$

$E(X_i^2)=\ds\frac{0^2+1^2+2^2+...+9^2}{10}=\ds\frac{9(10)(19)}{6
\times 10}=\ds\frac{57}{2}=28.5$

[Remember $1^2+2^2+...+n^2=\ds\frac{n(n+1)(2n+1)}{6}$]

Now $\sigma^2=E(X_i^2)-\{E(X_i)\}^2=28.5-(4.5)^2=8.25$

We want $P(4<\bar X_n<6)$ where $\ds\frac{\sqrt{n}(\bar
X_n-\mu)}{\sigma} \sim N(0,1)$ approximately.

\begin{eqnarray*} & = & P\left\{\ds\frac{\sqrt{16}}{\sqrt{8.25}}(4-4.5)<\ds\frac{\sqrt{n}(\bar
X_n-\mu)}{\sigma}<\ds\frac{\sqrt{16}}{\sqrt{8.25}}(6-4.5)\right\}\\
& = & P\{-0.69<Z<2.09\}\\ & = & 0.7366 \end{eqnarray*}



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