OVERVIEW: If X is a
discrete random variable with possible valuesx_{i }havingprobabilities p_{i}, then
...mean of xvalues = m= x_{1} p_{1 } + x_{2} p_{2}+ ... + x_{k} p_{k}= sum(x_{i} p_{i} ), and
...variance of xvalues = s= (x_{1} m )^{2} p_{1}+ (x_{2} m )^{2} p_{2}+ ... + (x_{k} m )^{2} p_{k}= sum[(x_{i} m )^{2} p_{i} ]
This section provides illustrations using these formulas.
Example: Consider a random variable x that assumesthe values 1,2,3 with respective probabilities 60%, 30%, 10%. Thefollowing table illustrates use of the formulas in theOVERVIEW.
x_{i} p_{i} x_{i}p_{i} x_{i}  m (x_{i}  m)^{2} (x_{i}  m)^{2}p_{i} 1 .6
.6 .5 .25 .15 2 .3
.6 .5 .25 .075 3 .1
.3 1.5 2.25 .225 SUMS
1.5 = m
.45 = s^{2}
The formulas shown in the OVERVIEW are consistentwith previously established formulas for mean, standard deviation,and variance. To illustrate this, note that in the example above, onewould "expect" this distribution of the random variable in tentrials: {1,1,1,1,1,1,2,2,2,3}. If you place this set in listL_{1} andcalculate 1Var Stats, you will find that the mean is 1.5 and thestandard deviation is 0.6708203932. Squaring the standard deviationto obtain variance yields 0.4499999999, or 0.45.
Law of Large Numbers
Example: When you flip a coin numeroustimes, you "expect" heads 50% of the time.
...Flip ten coins and count the number ofheads. Do this many times. It is highly likely that in some ofthe trials, the percentage of heads will be 30% or less.
...Flip one hundred coins and count the number of heads. Dothis many times. It is highly unlikely that any of the trialswill yield a percentage of heads that is 30% or less.Advanced topic, but related to the above...On TI83,binomcdf(10,.5,3) = .171875 and binomcdf(100,.5,30) =.00003925. Basically, if you flip 10 coins, the probability of 3 orfewer heads is about 17%. If you flip 100 coins, the probability of30 or fewer heads is very close to 0%.
RULES FOR MEANS:
The purpose here is to illustrate important rules(page 396) relating to the mean statistic.
Let S_{x}= {1,3}. The mean of S_{x} is 2.
Now, multiply each element by 7, and then add 5, producing the setS_{7x+5} ={12,26}.
The mean of S_{7x+5}is 19 = 9(2) + 5.
This illustrates Rule 1 (page 396). In this situation, mean(S_{7x+5 }) =7mean(S_{x }) +5Now, let S_{y} = {5,9,22}, which has mean=12. Construct the set S_{x+y} ={1+5,1+9,1+22,3+5,3+9,3+22} = {6,8,10,12,23,25}.
The mean of S_{x+y}is 14 = 2 + 12.
This illustrates Rule 2 (page 396). In this situation,mean(S_{x+y}) =mean(S_{x}) +mean(S_{y}).
RULES FOR VARIANCES:
The purpose here is to illustrateimportant rules (page 400) relating to the variance statistic.
Let T_{x} = {10,14}. The mean,standard deviation, and variance of T_{x} are, respectively, 12, 2,and 4.
Now, multiply each element by 3 and add 5, obtaining the setT_{3x+5} ={35,47}. The mean, standard deviation, and variance ofT_{3x+5} are,respectively, 41, 6, and 36. Note that 36 = 3^{2}(4). In other words,var(T_{3x+5}) =3^{2}var(T_{x}). Note also thatStDev(T_{3x+5})= 3StDev(T_{x} ). This illustrates Rule 1 (page 400). Now, let T_{y} = {6,9,12}. The mean,standard deviation, and variance of T_{y} are, respectively 9,2.4494897, and 6. Assume that T_{x}and T_{y} are
independent sets. Construct theset T_{xy} ={106,109,1012,146,149,1412} = {2,1,2,4,5,8}. The mean,standard deviation, and variance for T are, respectively, 3,3.16227766, and 10 respectively. Note that 10 = 4 + 6, orvar(T_{xy}) =var(T_{x}) +var(T_{y}). Ifyou construct the set T_{x+y}, you will find thatvar(T_{x+y}) = var(T_{xy}) = var(T_{x}) + var(T_{y})
This illustrates Rule 2 on page 400. In anutshell, with independence, variances add when setsT_{x+y} andT_{xy} areconstructed as indicated above. Note carefully that thestandard deviations
do not add. (Check this out with the examples above.) You shouldalso note mean(T_{xy} ) = 3 = 12  9 = mean(T_{x})  mean(T_{y}). If you constructT_{x+y}, youwill find that mean(T_{x+y}) = mean(T_{x}) + mean(T_{y}).

These bright Algebra II students are playing roulette. The only reason they are smiling is because they are not playing the game with real money. This brilliant group know that you can bet on...
Using math power, they have determined that on every bet except the fivenumber bet, the casino will take in $.0526 for every $1 that is wagered on the game. On the five number bet, the casino will take in $.0789 for every dollar wagered. As an example, there are 18 black slots, 18 red slots, and 2 green slots in American roulette. If you bet $1 on one number (such as black 17, the favorite of James Bond), you have a probability of 1/38 of winning $35. The probability you will lose your dollar is 37/38. Your expectation is ($35)(1/38)  ($1)(37/38) = $.0526. Roulette is a fun game, and it is certainly possible for you to win some money if you play for a short period of time. However, math power easily illustrates that you can't expect to make a living playing this game in a casino. Of course, it has to admitted that James Bond always seems to do unusually well in casinos. A good gambling strategy would be to follow him around and put your money on whatever he bets on! RETURN TO TEXTBOOK HOME PAGE / Back to the top of this page
