The difficulty felt by most students is in applying the formula to actual statistics, in other words in putting in the correct units. To take an actual numerical example, suppose that 1460 men have been measured in regard to their height “true to the nearest inch,” and let it be known that the modulus here is 3.6 inches. Then dx = 1 (inch); h−1 = 3.6 inches. Now ∑h/√πe−h2x2 dx = 1; that is, the sum of all the consecutive possible values is equal to unity. When therefore we want the sum, as here, to be 1460, we must express the formula thus;— y = 1460/√π × 3.6 e−(x/3.6)2, or y = 228e−(x/3.6)2.
Here x stands for the number of inches measured from the central or mean height, and y stands for the number of men referred to that height in our statistical table. (The values of e−t2 for successive values of t are given in the handbooks.)
For illustration I give the calculated numbers by this formula for values of x from 0 to 8 inches, with the actual numbers observed in the Cambridge measurements recently set on foot by Mr Galton.
| inches | calculated | observed |
|---|---|---|
| x = 0 | y = 228 | = 231 |
| x = 1 | y = 212 | = 218 |
| x = 2 | y = 166 | = 170 |
| x = 3 | y = 111 | = 110 |
| x = 4 | y = 82 | = 66 |
| x = 5 | y = 32 | = 31 |
| x = 6 | y = 11 | = 10 |
| x = 7 | y = 4 | = 6 |
| x = 8 | y = 1 | = 3 |
Here the average height was 69 inches: dx, as stated, = 1 inch. By saying, ‘put x = 0,’ we mean, calculate the number of men who are assigned to 69 inches; i.e.[ TN: space] who fall between 68.5 and 69.5. By saying, ‘put x = 4,’ we mean, calculate the number who are assigned to 65 or to 73; i.e.[ TN: space] who lie between 64.5 and 65.5, or between 72.5 and 73.5. The observed results, it will be seen, keep pretty close to the calculated: in the case of the former the means of equal and opposite divergences from the mean have been taken, the actual results not being always the same in opposite directions.
(2) The other point concerns the interpretation of the familiar probability integral, 2/√π ∫0te−t2 dt. Every one who has calculated the chance of an event, by the help of the tables of this integral given in so many handbooks, knows that if we assign any numerical value to t, the corresponding value of the above expression assigns the chance that an
error taken at random shall lie within that same limit, viz. t. Thus put t = 1.5, and we have the result 0.96; that is, only 4 per cent.[ TN: space] of the errors will exceed ‘one and a half.’ But when we ask, ‘one and a half’ what?[ TN: space] the answer would not always be very ready. As usual, the main difficulty of the beginner is not to manipulate the formulæ, but to be quite clear about his units.
It will be seen at once that this case differs from the preceding in that we cannot now choose our unit as we please. Where, as here, there is only one variable (t), if we were allowed to select our own unit, the inch, foot, or