Page:Biometrika - Volume 6, Issue 1.djvu/10

10 Substituting this value of $$D$$ we get

Consequently the value of the standard deviation of a standard deviation which we have found $$\left(\frac{\sigma}{\sqrt{2n}\sqrt{1-\frac{1}{4n}}}\right)$$ becomes the same as that found for the normal curve by Professor Pearson ($$\sigma/\sqrt{2n}$$) when $$n$$ is large enough to neglect the $$1/4n$$ in comparison with $$1$$.

Neglecting terms of lower order than $$\frac{1}{n}$$ we find

Consequently as $$n$$ increases $$\beta_2$$ very soon approaches the value $$3$$ of the normal curve, but $$\beta_1$$ vanishes more slowly, so that the curve remains slightly skew.

Diagram I shows the theoretical distribution of the found from samples of 10.

Writing $$z=\textrm{tan}\,\theta$$ the equation becomes $$y=\frac{n-2}{n-3}\cdot\frac{n-4}{n-5}\,\textrm{...}\;\textrm{etc.}\times \textrm{cos}^n \theta$$, which affords an easy way of drawing the curve. Also $$dz=d\theta/\textrm{cos}^2 \theta$$.