14 SUMMARIZATION OF DATA Ch. 2 



X 4 = 185, X 5 = 201, Z 6 = 174, X 7 = 191, and X 8 = 210, all in 

 pounds. Therefore, 



8 



J^Xi = Xj + X 2 +---+X 8 = 152+ 170 H f-210 = 1448 



i 



so that jj. = 1448/8 = 181 pounds, which is the arithmetic mean of 

 the weights. 



Although the number 181 pounds gives a useful impression of the 

 general weight size of the players of problem 2.11, it is obvious that 

 the same mean weight could have been obtained for many other 

 groups of eight weights, some of which might be considered to be 

 quite different from those above. For example, each of the following 

 sets of eight weights (in pounds) has /x = 181: 



Set 1. 185, 180, 181, 184, 182, 179, 177, and 180. 

 Set 2. 190, 190, 190, 182, 184, 183, 190, and 139. 

 Set 3. 172, 180, 165, 1G0, 175, 168, 180, and 248. 



In Set 1, the extreme weights are but 8 pounds apart; three weights 

 are higher than the mean, four are lower than the mean, and one is 

 the same as the mean weight. This set differs from that of problem 

 2.11 chiefly by being more uniform. In Set 2, the difference between 

 the extremes is 51 pounds, and every weight but one is higher than 

 the mean. This set differs from that in problem 2.11 chiefly in the 

 fact that the mean actually is not very representative of the weights 

 of the squad members. Similar remarks hold for Set 3 except that 

 seven of the eight weights are below the mean. To summarize : Sets 

 1, 2, and 3 differed from the example of Problem 2.11 in the amount 

 of dispersion, or non-uniformity, among the numbers and in the 

 manner in which it occurred. All sets of data had the same arith- 

 metic mean. 



A measure of the dispersion, or variation, of the measurements, 

 X if about their arithmetic mean, fi, logically would be based upon 

 the amounts by which the Xi are greater than or less than that mean. 

 It is customary to symbolize those amounts by Xi — Xi — fx, and to 

 call the Xi the deviations from the mean. It is observed that when 

 an X is smaller than the mean, the x is negative ; when the X is larger 

 than the mean, the corresponding deviation, x, is positive. 



In the first numerical example of this chapter, X\ = — 1, rr 2 = +2, 

 x s = —2, Xi = 0, and x 5 = +1. For problem 2.11, Xi = —29, x 2 — 

 — 11, x s = —16, Xi = +4, x 5 = +20, x & — — 7, x 7 = +10, and 

 x 8 = +29. It is observed that, at least in these instances, %x = 0. 



