Question

Suppose that a random sample of size n is to be taken from a distribution for which the mean is μ, and the standard deviation is 3. Use the central limit theorem to determine approximately the smallest value of n for which the following relation will be satisfied:

\({\bf{Pr}}\left( {\left| {{{{\bf{\bar X}}}_{\bf{n}}}{\bf{ - \mu }}} \right|{\bf{ < 0}}{\bf{.3}}} \right) \ge {\bf{0}}{\bf{.95}}\).

Short answer

The smallest possible value of n is 385

Step-by-step solution

Given information

A random sample of size n taken from the distribution with a mean of \(\mu \) and standard deviation \(\sigma = 3\)

Finding the smallest value of n

The distribution,

\(\begin{array}{c}Z = \frac{{\sqrt n \left( {{{\bar X}_n} - \mu } \right)}}{\sigma }\\ = \frac{{\sqrt n \left( {{{\bar X}_n} - \mu } \right)}}{3}\end{array}\)

Then the distribution of Z will be an approximately standard normal distribution.

Therefore,

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| < 0.3} \right) = \Pr \left( {\left| Z \right| < 0.1\sqrt n } \right)\\ \approx 2\Phi \left( {0.1\sqrt n } \right) - 1\end{array}\)

But,

\(\begin{array}{c}2\Phi \left( {0.1\sqrt n } \right) - 1 \ge 0.95\\2\Phi \left( {0.1\sqrt n } \right) \ge 1 + 0.95\\\Phi \left( {0.1\sqrt n } \right) \ge \frac{{1.95}}{2}\\0.1\sqrt n \ge {\Phi ^{ - 1}}\left( {0.975} \right)\end{array}\)

\(\begin{array}{c}0.1\sqrt n \ge 1.96\\\sqrt n \ge 19.6\\n \ge 384.16\end{array}\)

Therefore, the smallest possible value of n is 385