In this article we shall discuss the geometry of the curves which Edgeworth obtains by this transformation and derive a method for an approximate solution of the two equations, one of the fourth and the other of the sixth degree, which arise in the fitting of a curve of this class. In order that the final curve may be written in terms of the co-ordinates x and y the equation of the base or generating normal probability curve is written: 1 z = —————— e^(—(t^2/2)) √(2π) where t denotes abscissas and z ordinates. Let the abscissas of the transformed curve be functions of the corresponding abscissas of the base curve. Then it may be assumed that x can be developed in powers of t, and hence we may write on omitting fourth and higher powers, x = a(t + κt^2 + λt^3), where a, κ and λ are constants to be determined in “fitting” the curve. Since x denotes the value of a measurement and y the frequency of x, that is, the number of individuals possessing that value of x, the magnitude of an element of area denotes the number of individuals between two values of x. Obviously, therefore, if the transformation is to be of concrete value the magnitude of an element of area must not be altered, though of course the shape will be changed. Hence y dx = z dt, and y = z dt/dx 1 1 = —————— e^(—(t^2/2)) · —————————————————— √(2π) a(1 + 2κt + 3λt^2) The formulas of transformation are thus: x = a(t + κt^2 + λt^3), 1 1 y = —————— e^(—(t^2/2)) · —————————————————— √(2π) a(1 + 2κt + 3λt^2) =Maximum and Minimum Points.= Since only curves with one maximum point or mode are practically useful it is desirable to determine what values of the constants a, κ and λ give unimodal curves. We have dy dy dt ——— = ——— · ——— dx dt dx 1 (3λt^3 + 2κt^2 + (1 + 6λ)t + 2κ) = —————— e^(—(t^2/2)) · ———————————————————————————————— √(2π) a(1 + 2κt + 3λt^2) From the vanishing of the numerator of dy/dx there must result either one or three real modes for each pair of values for λ and κ, that is, for each translated curve. To determine what values of λ and κ give uni-modal curves and what tri-modal it is convenient to consider the plane of λ and κ. The discriminant of the equation 3λt^3 + 2κt^2 + (1 + 6λ)t + 2κ = 0 is 16κ^4 - κ^2(1 + 66λ + 117λ^2) + 3λ(1 + 6λ)^3 = 0