91Ó°ÊÓ

Here, 16 digits are chosen at random with replacement from a set \(\left\{ {0,...,9} \right\}\)

The mean of a random digit X is,

\(\begin{array}{c}E\left( X \right) = \frac{1}{{10}}\left( {0 + 1 + ... + 9} \right)\\ = \frac{{45}}{{10}}\\ = 4.5\end{array}\)

Also,

\(\begin{array}{c}E\left( {{X^2}} \right) = \frac{1}{{10}}\left( {{0^2} + {1^2} + ... + {9^2}} \right)\\ = \frac{{285}}{{10}}\\ = 28.5\end{array}\)

The variance of random digit X is,

\(\begin{array}{c}Var\left( X \right) = E\left( {{X^2}} \right) - {\left( {E\left( X \right)} \right)^2}\\ = 28.5 - {\left( {4.5} \right)^2}\\ = 28.5 - 20.25\\ = 8.25\end{array}\)

The distribution of the average\({\bar X_n}\)of 16 random digits will be approximately the normal distribution, with the mean being,

\(\mu = 4.5\)

The variance is,

\(\begin{array}{c}{\sigma ^2} = \frac{{8.25}}{{16}}\\ = 0.5156\end{array}\)

The standard deviation is,

\(\begin{array}{c}\sigma = \sqrt {0.5156} \\ = 0.7181\end{array}\)

The distribution is,

\(\begin{array}{c}Z = \frac{{{{\bar X}_n} - \mu }}{\sigma }\\ = \frac{{{{\bar X}_n} - 4.5}}{{0.7181}}\end{array}\)

Z will be approximately standard normal distribution

\({\bar X_n} = 6\)

\(\begin{array}{c}Z = \frac{{{{\bar X}_n} - \mu }}{\sigma }\\ = \frac{{{{\bar X}_n} - 4.5}}{{0.7181}}\\ = \frac{{6 - 4.5}}{{0.7181}}\\ = 2.0888\end{array}\)

Again,

\({\bar X_n} = 4\)

\(\begin{array}{c}Z = \frac{{{{\bar X}_n} - \mu }}{\sigma }\\ = \frac{{{{\bar X}_n} - 4.5}}{{0.7181}}\\ = \frac{{4 - 4.5}}{{0.7181}}\\ = - 0.6963\end{array}\)

The probability is,

\(\begin{array}{c}{\mathop{\rm P}\nolimits} \left( {4 \le {{\bar X}_n} \le 4} \right) = {\mathop{\rm P}\nolimits} \left( { - 0.6963 \le Z \le 2.0888} \right)\\ = P\left( {Z \le 2.0888} \right) - \left( {1 - P\left( {Z \le 0.6963} \right)} \right)\\ = 0.9817 - \left( {1 - 0.7580} \right)\\ = 0.7397\end{array}\)

Therefore, the probability that their average will lie between 4 and 6 is 0.7397