91Ó°ÊÓ

The Mean Deviation form of any \(p \times N\)is given by:

\(B = \left( {\begin{array}{*{20}{c}}{{{\hat X}_1}}&{{{\hat X}_2}}&{........}&{{{\hat X}_N}}\end{array}} \right)\)

Whose \(p \times p\)covariance matrix is:

\(S = \frac{1}{{N - 1}}B{B^T}\)

From exercise 1, the maximum eigenvalue is:

\({\lambda _1} = 95.2041\)

The respective unit vector is:

\({u_1} = \left( {\begin{array}{*{20}{c}}{0.946515}\\{ - 0.322659}\end{array}} \right)\)

The new variable will be:

\({y_1} = 0.946515{x_1} - 0.322659{x_2}\)

Now, the percentage of change in variance can be obtained as:

\(\begin{array}{c}\Delta = \frac{{{\lambda _1}}}{{tr\left( S \right)}} \times 100\\ = \frac{{95.2041}}{{86 + 16}} \times 100\\ = 93.3374\% \end{array}\)

Hence, this is the required answer.