158 SAMPLING NORMAL POPULATIONS Ch. 6 



ard deviation, a, and if a very large number of random samples of 

 n observations each is drawn from that population: 



(a) The population of Xi thus formed will have a normal fre- 

 quency distribution. 



(6) The mean of the X{ will be \i also. 



(c) The standard deviation of the X{ will be a/\/n. 



Note from this theorem that, if the X's are normally distributed we 

 automatically know the form of the distribution of the sample means 

 and hence can write down specifically the formula for their distribu- 

 tion curve, namely; 



1 



(x-m) 2 



(6.21) Vl = -y= T=r- • e 2aVn 



Formula 6.21 can be transformed into the standard normal formula of 

 Chapter 4 by means of the substitutions 



V = l/i-ir/Vw and X = (x — n)/(<r/-\/n) 



whereupon Table III can be employed as shown earlier. 



If ten measurements are taken per sample and the parent population 

 has a mean of 60 and a standard deviation of 10, as above, the mean 

 of the x's is also 60, and the standard deviation is 10/\/i0 = \/l0 

 = 3.16. Table 6.21 and Figure 6.21 furnish approximate verification 

 of these statements from actual experience. 



If n = 15 and the samples are drawn from a normal population with 

 ix = 60 and a = 10, the mean of the resulting population of X{ also 

 will be 60, and the standard deviation will be 10/\/Y5. On 200 such 

 samples, their mean was 60.22 and their standard deviation was 2.53 

 instead of the expected 60 and 2.58, respectively. Two hundred is a 

 relatively small number of samples from which to seek empirical verifi- 

 cation of mathematical theory, but these results do agree quite well 

 with the theorem given above. 



Problem 6.21. If chemical determinations of the percentage of protein in 

 samples of a certain variety of wheat are known to have a normal frequency 

 distribution with /j. = 14 and a = 2, what is the probability that five random 

 samples will have a mean per cent protein above 16? 



In the following discussion, a £ will be used to denote the standard 

 deviation of the population of sampling means. In this problem, n 

 = 5, m = 14, and <r £ = 2/^5 = 0.90. Therefore, X = (16 - 14)/0.90 

 = 2.22; and P(X > 2.22) = .013, approximately. In other words, 

 only about 13 times in 1000 sets of 5 observations like these would you 



