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For equal prior probabilities, $b=\frac{1}{2}{a}^T({\mu }_1+{\mu }_2)$. The projected score is normal under either class because it is a linear image of a multivariate normal vector, and its variance isMoreover $a^T{\mu }_1-a^T{\mu }_2={\delta }^2$, so the midpoint threshold lies ${\delta }^2/2$ from either projected mean. HenceStandardizing the two normal distributions, with $\Phi$ the standard normal distribution function, givesThus the equal-covariance Gaussian classification error is $p(\delta )=\Phi (-\delta /2)$. The Mahalanobis distance $\delta$ between the class means is their separation in units of within-class noise. It yields $p\to 1/2$ as $\delta \downarrow 0$ and $p\to 0$ as $\delta \to \infty$. The stipulated $\delta >0$ means the two class means are distinct.

For common positive-definite covariance, subtracting the two Gaussian log densities cancels the quadratic term. Equal priors leave the linear discriminant analysis scoreHereThusAssign class two if the expression is negative; a tie has zero probability under either continuous Gaussian law.

To compute the error rather than just specify the boundary, put $D^2={\Delta }^T{\Sigma }^{-1}\Delta =27/2$. Within class one the score has mean $D^2/2=27/4$, and within class two mean $-27/4$. In both classes its variance is ${\Delta }^T{\Sigma }^{-1}\Sigma {\Sigma }^{-1}\Delta =D^2$. Hence both conditional error probabilities are $\Phi (-D/2)$, where $\Phi$ is the standard normal distribution function. Equal priors give the equal-covariance Gaussian classification errorFor the specified observation, the score is $\delta (2,1)=27/4>0$, so assign it to class one.