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The lower bounded for \(\lambda \) with confidence level \(95\% \) is

\(\frac{{{X^2}_{0.95,2n}}}{{2\sum\limits_{i = 1}^n {{X_i}} }}\)

The \({e^{ - \lambda }}\) needs to be obtained, therefore from the inequality,

\({e^{ - \lambda t}} < \exp \left\{ {\frac{{ - t{X^2}_{0.05,2n}}}{{2\sum\limits_{i = 1}^n {{X_i}} }}} \right\}\)

From the mentioned example, random variable

\(2\lambda \sum\limits_{i = 1}^n {{X_i}} \)has a chi squared distribution with \(2n\) degrees of freedom. The area to the right of value \({X^2}_{0.95,2n}\) is \(0.95\). The following is true

\(P\left( {{X^2}_{0.95,2n} < 2\lambda \sum\limits_{i = 1}^n {{X_i}} } \right) = 0.95\)

From this, the lower bounded for \(\lambda \) with confidence level \(95\% \) is

\(\frac{{{X^2}_{0.95,2n}}}{{2\sum\limits_{i = 1}^n {{X_i}} }}\)

In the example, \(n = 10\), therefore the chi squared distribution has \(2n = 20\) degrees of freedom. From the given data the lower bound is

\(\begin{aligned}{}\frac{{{X^2}_{0.95,2n}}}{{2\sum\limits_{i = 1}^n {{X_i}} }} & = \frac{{10.851}}{{2(550.87)}}\\\frac{{{X^2}_{0.95,2n}}}{{2\sum\limits_{i = 1}^n {{X_i}} }} & = 0.0098\end{aligned}\)

Hence, \({X^2}_{0.95,2n} = 10.851\) was obtained from the table A.7. that given the chi squared critical value for \(20\) degrees of freedom.

Similarly as in part a), the following stands,鈥� From the mentioned example, random variable

\(P\left( {2\lambda \sum\limits_{i = 1}^n {{X_i}} < {X^2}_{0.05,2n}} \right) = 0.95\)

From this, \(\lambda \) can be separated in the parentheses as,

\(\lambda < \frac{{{X^2}_{0.05,2n}}}{{2\sum\limits_{i = 1}^n {{X_i}} }}\)

The \({e^{ - \lambda }}\) needs to be obtained, therefore from the inequality above

\({e^{ - \lambda t}} < \exp \left\{ {\frac{{ - t{X^2}_{0.05,2n}}}{{2\sum\limits_{i = 1}^n {{X_i}} }}} \right\}\)

Finally, the lower confidence bound is,

\(\begin{aligned}{}\exp \left\{ {\frac{{ - t{X^2}_{0.05,2n}}}{{2\sum\limits_{i = 1}^n {{X_i}} }}} \right\} & = \exp \left\{ {\frac{{ - 100(31.41)}}{{2(550.87}}} \right\}\\ & = 0.058\end{aligned}\)

Hence, The lower bound is \(0.058\).