91Ó°ÊÓ

Suppose 16 digits are chosen at random with replacement from the set \(\left\{ {0,1,......9} \right\}\)

Let X be the number chosen at random.

The mean of random variable X is

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

also

\(\)\(\begin{array}{c}E\left( {{X^2}} \right) = \frac{1}{{10}}\left( {{0^2} + {1^2} + ..... + {9^2}} \right)\\ = \frac{1}{{10}} \times \frac{{\left( 9 \right)\left( {10} \right)\left( 9 \right)}}{6}\\ = 28.5\end{array}\)

Variance of random variable X is:

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

Let S denote the sum of the 16 digits, then

\(\begin{array}{c}E\left( s \right) = 16\left( {4.5} \right)\\ = 72\end{array}\)

\(\begin{array}{c}{\sigma _X} = \sqrt {\left[ {16\left( {8.25} \right)} \right]} \\ = 11.49\end{array}\)

Hence,

\(\begin{array}{c}{\rm P}\left( {4 \le \overline {{X_n}} \le 6} \right) = {\rm P}\left( {64 \le S \le 96} \right)\\ = {\rm P}\left( {63.5 \le S \le 96.5} \right)\\ = {\rm P}\left( {\frac{{63.5 - 72}}{{11.49}} \le Z \le \frac{{96.5 - 72}}{{11.49}}} \right)\end{array}\)

\(\begin{array}{c}{\rm P}\left( {4 \le \overline {{X_n}} \le 6} \right) \approx \phi \left( {2.132} \right) - \phi \left( { - 7.40} \right)\\ \approx 0.9835 - 0.2296\end{array}\)

\({\rm P}\left( {4 \le \overline {{X_n}} \le 6} \right) = 0.7539\)

Hence, probability of average of 16 digits will lies between 4 and 6 is \(0.7539\)