7.
\includegraphics[max width=\textwidth, alt={}, center]{65369843-222f-48b2-b8cd-a1c304eac3d9-6_707_718_347_660}
The diagram above shows a cyclic quadrilateral \(A B C D\), where \(\widehat { B A D } = \alpha , \widehat { B C D } = \beta\) and \(\alpha + \beta = 180 ^ { \circ }\). These angles are measured.
The random variables \(X\) and \(Y\) denote the measured values, in degrees, of \(\widehat { B A D }\) and \(\widehat { B C D }\) respectively. You are given that \(X\) and \(Y\) are independently normally distributed with standard deviation \(\sigma\) and means \(\alpha\) and \(\beta\) respectively.
- Calculate, correct to two decimal places, the probability that \(X + Y\) will differ from \(180 ^ { \circ }\) by less than \(\sigma\).
- Show that \(T _ { 1 } = 45 ^ { \circ } + \frac { 1 } { 4 } ( 3 X - Y )\) is an unbiased estimator for \(\alpha\) and verify that it is a better estimator than \(X\) for \(\alpha\).
- Now consider \(T _ { 2 } = \lambda X + ( 1 - \lambda ) \left( 180 ^ { \circ } - Y \right)\).
- Show that \(T _ { 2 }\) is an unbiased estimator for \(\alpha\) for all values of \(\lambda\).
- Find \(\operatorname { Var } \left( T _ { 2 } \right)\) in terms of \(\lambda\) and \(\sigma\).
- Hence determine the value of \(\lambda\) which gives the best unbiased estimator for \(\alpha\).