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The Mean and Standard Deviation of a Finite Distribution I. The idea We want to have a measure of the center and the spread of a finite random variable whose probabilities are given by a probability mass function. Since we can think of the distribution as giving the probabilities for an enormous number of repetitions of the experiment which provides the random variable, this is equivalent to finding the population mean and standard deviation. The symbols for these are µ and σ, respectively, and if there are more than one random variable wandering about, subscripts are used, i.e. µX and σY . Note that these are not the X n and s which come from samples. II. The mean The formal definition is n X xi · P (X = xi ). i=1 The most convenient way to mechanize this is in a table which displays the probability model and organizes the calculation of the sum. i xi 1 2 2 4 3 6 Sum P (X = xi ) xi · P (X = xi ) .40 .8 .30 1.2 .30 1.8 1.00 3.8 Thus in the case of this random variable, µ = 0.8 + 1.2 + 1.8 = 3.8. This is the long-term average outcome if I run the experiment very many times. III. Variance and Standard Deviation As with samples, there are two measures of spread for a random variable. The first is called distribution variance VX or VAR(X). The idea is similar to the calculations from a sample, but just enough different to cause problems. We have n VAR(X) = X (xi − µ)2 · P (X = xi ) i=1 and σ= q VAR(X) We can calculate these things by extending the above table in a fairly obvious way, and I will use the same example. i 1 2 3 Sums xi 2 4 6 P (X = xi ) xi · P (X = xi ) xi − µ .40 .8 -1.8 .30 1.2 .2 .30 1.8 2.2 1.00 3.8 1 (xi − µ)2 3.24 0.04 4.84 (xi − µ)2 · P (X = xi ) 1.2960 0.0120 1.4520 2.7600 Thus √ the variance of this distribution is 2.7600, and its standard deviation is 2.7600 = 1.6613. IV. Binomials A calculation using the rules of sigma notation yields the following results for a Binomial Distribution with n repetitions and probabilty of success p: µ VAR σ where q = = = = np npq √ npq 1−p V. Continuous Random Variables Similar calculations for continuous random variables require calculus, and are therefore not a part of this course. The results of such calculations will be supplied as we need them. 2