I have a bivariate normal distribution and I want to determine the axes of the ellipse that will contain 60% probability. According to my textbok, It follows from the spectral decomposition of the covariance matrix (and the fact that the mahalanobis distances follow the chi square distribution) that the formula for the axes will be given by $\sqrt{\lambda(i)} \cdot$ eigen vector associated with $\lambda(i) \cdot c$. Here, $c^2 = \chi^2(\alpha)$, where $\chi^2(\alpha)$ is the upper 100th percentile of a chi-square distribution with $p$ degrees of freedom, leading to the contours that contain $(1 - a) \cdot 100%$ of the probability. So if I want to plot the 60% probability contour for bivariate normal, to calculate the value of $c^2$, I should look into a right-tailed chi square table and take the value associated with 2 degrees and freedom and $0.4$, right?
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