338 Life and Letters of Francis Galton

by Galton to have the median value of the character (m). The two men with

75 °/o and 25 °/o of the population above them are said to have the lower and

upper quartile values (q, and q,). If the distribution be symmetrical about the median then m. - q, and q, - m will be equal; if the distribution obeys the so-called normal curve of deviations, then all the constants of the distribution can be found by measuring the intensity of the character in the median and in the quartile individuals. Thus Galton would place a hundred and one savages in a row, the. curves formed by the apices of their heads would be his "ogivel" for their stature, and by measuring only the 25th, the 51st and 76th men he would obtain a reasonable distribution for the stature of adult men in that tribe.

Theoretically there are difficulties about Galton's "ogive," if we suppose it to correspond to a normal curve of deviations, in particular at the terminals. Galton endeavoured to get over these difficulties by replacing the normal curve by a symmetrical binomial, which has a finite range. He treats of this matter in a paper on "Statistics by Intercomparison with Remarks on the Law of Frequency of Error 2" In this paper after mentioning that Quetelet had shown that a binomial to the 999th power was practically a normal curve of deviations, Galton goes on to indicate that the same holds very closely for symmetrical binomials of quite low powers. Thus he plots (p. 39) the Binomial Ogive of 17 elements against the Binomial Ogive of 999 equal elements, which is practically identical with the Exponential Ogive, and argues therefrom to the binomial of the 17th power being very close, indeed (which is a fact), to the normal curves. Galton then passes to some suggestive remarks on the origin of the distribution of deviations according to the normal law. He rejects any idea of its source in a very large number of small and independent contributory causes. He supposes the exponential curve to arise because it nearly resembles the curve based upon a binomial of moderate power, i. e. he supposes that in nature the contributory causegroups are relatively few; but he has to suppose in this case that nature works all her processes by equal additions or subtractions, i.e. prefers the mathematics of coin-tossing to those of the dice-box.

"I shall show," he writes, "by quite a different line of argument that the exponential view contains inherent contradictions when nature is appealed to, that the binomial of a moderate power is the truer one and that we have means of ascertaining a limit which the number of elements [independent cause-groups as the individual coins of a combined toss]

cannot exceed." (p. 40.)

Galton takes the mean (m) and divides it by the quartile deviation (q, - m or m - q,) and computes the ratio m/(q, - m). In the case of the binomial (a+ b)" this will be

nb/•67449 /nab =1.48257 , / nb '~ a

i Philosophical Magazine, January, 1875, Vol. XLIX, pp. 33-46.

2 If X be the abscissa, i.e. the rank, and Y the ordinate or value of the variate, m the mean, a the standard deviation of the population =1.4825 7 (q, - m), and N the total population, then

the equation to Galton's "ogive" will be X=.-+ (Y N e 1(Y-m)2/o2 dP.

s I have not been able to agree wit# the values given in the Table on p. 42.