91Ó°ÊÓ

Random samples of 25 observations are taken from the normal distribution with mean μ and standard deviation 2.

We know that\(E\left( {\overline {{X_n}} } \right) = \mu \)and

\(\begin{array}{c}Var\left( {\overline {{X_n}} } \right) = \frac{{{\sigma ^2}}}{n}\\ = \frac{4}{{25}}\end{array}\)

\(Z = \frac{{\left( {\overline {{X_n}} - \mu } \right)}}{{\left( {\frac{2}{5}} \right)}} = \left( {\frac{5}{2}} \right)\left( {\overline {{X_n}} - \mu } \right)\)

Thus, Z will have the standard normal distribution.

Hence,

\(\begin{array}{c}{\rm P}\left( {\left| {\overline {{X_n}} - \mu } \right| \le 1} \right) = {\rm P}\left( {\left| Z \right| \le 2.5} \right)\\ = 2\phi \left( {2.5} \right) - 1\\ = 0.9876\end{array}\)

Therefore , the probability that the sample mean will lie within one unit of \(\mu \) is 0.9876.