91Ó°ÊÓ

The given information is \(E\left( {|{{\bar X}_n} - \theta |} \right) \le 0.1\)

Here \({\bar X_n}\) has the normal distribution with mean \(\theta \) and variance \(4/n\) . Hence, the random variable \(Z = \left( {{{\bar X}_n} - \theta } \right)/\left( {2/\sqrt n } \right)\)will has standard normal distribution.

Therefore,

\(\begin{align}E\left( {|{{\bar X}_n} - \theta |} \right) &= \frac{2}{{\sqrt n }}{E_\theta }\left( {|Z|} \right)\\ &= \frac{2}{{\sqrt n }}\int_{ - \infty }^\infty {|z|} \frac{1}{{\sqrt {2\pi } }}\exp \left( { - {z^2}/2} \right)dz\\ &= 2\sqrt {\frac{2}{{n\pi }}} \int_0^\infty {z\exp \left( { - {z^2}/2} \right)dz} \\ &= 2\sqrt {\frac{2}{{n\pi }}} \end{align}\)

Here given that,

\(\begin{align}E\left( {|{{\bar X}_n} - \theta |} \right) \le 0.1\\2\sqrt {\frac{2}{{n\pi }}} \le 0.1\\4 \times \frac{2}{{n\pi }} \le 0.01\\n \ge \frac{8}{{0.01 \times \pi }}\\n \ge \frac{8}{{0.01 \times 3.141593}}\\n \ge 254.6479\end{align}\)

Hence, n must be \( \ge 255\)