Consider the random variable X with cumulative distribution function F(x) = (1/2) + (1/pi)tan^-1 (x), -infinity < x < infinity. Find the 25%, 50%, and 75% quantiles of this distribution Solution As F(x) = 1/2 + (1/pi) tan^-1(x) Let y = F(x) = 1/2 + (1/pi)tan^-1(x) Then, (y - 1/2) = (1/pi) tan^-1(x), so pi(y-1/2) = tan^-1(x), so tan(pi(y-1/2) ) = x Thus, the 25th percentile is tan(pi(1/4-1/2) ) = x tan(-pi/4) = x x = -1 The 50th percentile is tan(pi(1/2-1/2) ) = x tan(0) = x x = 0 The 75th percentile is tan(pi(3/4-1/2) ) = x tan(pi/4) = x x = 1 .