Question

Suppose that \({X_1},...,{X_n}\) form a random sample from the normal distribution with mean μ and variance \({\sigma ^2}\) . Assuming that the sample size n is 16, determine the values of the following probabilities:

\(\begin{align}a.\,\,\,\,P\left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \mu } \right)}^2} \le 2{\sigma ^2}} } \right)\\b.\,\,\,\,P\left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2} \le 2{\sigma ^2}} } \right)\end{align}\)

Short answer

1. The required probability is 0.94.
2. The required probability is 0.91.

Step-by-step solution

Given information

Suppose that \({X_1},...,{X_n}\) form a random sample from the normal distribution with mean μ and variance \({\sigma ^2}\).and the sample size is 16.

Computation of the probability

a.

\(\begin{align}\Pr \left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \mu } \right)}^2} \le 2{\sigma ^2}} } \right) &= \Pr \left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{{16}}\sum\limits_{i = 1}^{16} {{{\left( {{X_i} - \mu } \right)}^2} \le 2{\sigma ^2}} } \right)\\ &= \Pr \left( {8 \le \sum\limits_{i = 1}^{16} {\frac{{{{\left( {{X_i} - \mu } \right)}^2}}}{{{\sigma ^2}}} \le 32} } \right)\\ &= \Pr \left( {8 \le V \le 32} \right)\\ &= 0.94\end{align}\)

Where V has chi-square distribution with 16 degrees of freedom and the values are obtained from chi-square table.

Thus the required probability is .94.

Computation of the probability

b.

\(\begin{align}\Pr \left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2} \le 2{\sigma ^2}} } \right) &= \Pr \left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{{16}}\sum\limits_{i = 1}^{16} {{{\left( {{X_i} - {{\bar X}_{16}}} \right)}^2} \le 2{\sigma ^2}} } \right)\\ &= P\left( {8 \le \sum\limits_{i = 1}^{16} {\frac{{{{\left( {{X_i} - {{\bar X}_{16}}} \right)}^2}}}{{{\sigma ^2}}} \le 32} } \right)\\ &= P\left( {8 \le V \le 32} \right)\\ &= 0.91\end{align}\)

Where V has chi-square distribution with 16-1=15 degrees of freedom and the values are obtained from chi-square table.

Thus the required probability is 0.91.