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.

1. Calculate, correct to two decimal places, the probability that \(X + Y\) will differ from \(180 ^ { \circ }\) by less than \(\sigma\).
2. 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 \(\alpha\).
3. Now consider \(T _ { 2 } = \lambda X + ( 1 - \lambda ) \left( 180 ^ { \circ } - Y \right)\).
   1. Show that \(T _ { 2 }\) is an unbiased estimator for \(\alpha\) for all values of \(\lambda\).
   2. Find \(\operatorname { Var } \left( T _ { 2 } \right)\) in terms of \(\lambda\) and \(\sigma\).
   3. Hence determine the value of \(\lambda\) which gives the best unbiased estimator for \(\alpha\).