16 SUMMARIZATION OF DATA Ch. 2 



the x's are difficult to compute, the following results are useful. In 

 view of the fact that x 2 = (X - M ) 2 = (X 2 - 2pX + it 2 ), it should 

 be clear that 2x 2 = 2(X 2 - 2iiX + it 2 ) = 2X 2 - 2iiSX + 2/j 2 . But 

 2/i. 2 = Nfj 2 , as explained earlier, and /X = 2X/iV; therefore, — 2ii2A r = 

 -2(2X) 2 /A r and V = + 1(2X) 2 /A'. It follows that 2x 2 = 2X 2 - 

 (2A r ) 2 /A r . If this substitution is made into formula 2.12, the follow- 

 ing alternative method for computing the standard deviation, o-, is 

 obtained: 



/2Z 2 - CEX) 2 /N 

 (2.13) , = J i-LZ-. 



For a numerical example considered earlier in this chapter, formula 



/55 - (15) 2 /5 /- 



2.13 becomes a = */ = V2 . The variance is <r = 2. 



\ 5 



To further illustrate the uses to which a 2 and o- can be put, con- 

 sider again the ACE scores of Table 2.01. The arithmetic mean is 

 95.7. The standard deviation is calculated to be 26.1 (see problem 

 11 at the end of this section), with the extreme scores being 23 and 

 183. It is noted that 95.7 is a bit less than midway between the 

 lowest and highest scores, but in general it is quite centrally located 

 in that respect. To obtain a clearer picture of the dispersion of the 

 scores between the extreme scores, and about the mean, the standard 

 deviation will be found to be very useful in subsequent discussions. 

 It is entirely possible for different sets of data to have essentially 

 the same extremes but very different distributions with respect to the 

 mean. The <r 2 and o- will help to describe these differences. This 

 use of the variance and standard deviation is illustrated, in part, by 

 the following discussion. 



As the student can verify, approximately 67.1 per cent of the ACE 

 scores in Table 2.01 lie within a distance (on the ACE scale) of lo- 

 below or above the mean, /*; that is, approximately 67.1 per cent of 

 the 1290 scores are among the numbers from 70 to 121, inclusive. 

 This fact can be put in the following brief form: /x ± lo- includes 

 67.1 per cent of the scores. In a strictly normal population the cor- 

 responding percentage is 68.3. Such information sometimes is con- 

 sidered useful in the summarization of sets of numbers like Table 

 2.01. 



Likewise the interval fx ± 2o- (which includes scores from 44 to 147 r 

 inclusive) contains 95.2 per cent of the 1290 scores in the table. If 



