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At the end of Example 8.5.11, compute the probability that \(\left| {{{\bar X}_2} - \theta } \right| < 0.3\) given Z = 0.9. Why is it so large?

Short Answer

Expert verified

\(P\left( {\left| {{{\bar X}_2} - \theta } \right| < 0.3|Z = 0.9} \right) = 1\)

Step by step solution

01

Step 1:Given information

It is given that

\(\begin{align}P\left( {\left| {{{\bar X}_2} - \theta } \right| < c|Z = z} \right) &= \frac{{2c}}{{1 - z}}\,\,\,\,\,c \le \left( {1 - z} \right)\\ &= 1\,\,\,\,\,\,\,\,,c > \left( {1 - z} \right)\end{align}\)

We need to compute the probability \(P\left( {\left| {{{\bar X}_2} - \theta } \right| < 0.3|Z = 0.9} \right)\)

02

Computation of \(P\left( {\left| {{{\bar X}_2} - \theta } \right| < 0.3|Z = 0.9} \right)\) 

\(\begin{align}P\left( {\left| {{{\bar X}_2} - \theta } \right| < c|Z = z} \right) &= \frac{{2c}}{{1 - z}}\,\,\,\,\,c \le \left( {1 - z} \right)\\ &= 1\,\,\,\,\,\,\,\,,c > \left( {1 - z} \right)\end{align}\)

This implies that

\(\begin{align}c &= 0.3 > \frac{{1 - 0.9}}{2}\\ &= 0.05\end{align}\)

This implies \(P\left( {\left| {{{\bar X}_2} - \theta } \right| < 0.3|Z = 0.9} \right) = 1\)

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