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Rules for Means and Variances. Target Goal: I can find the mean and standard deviation of the sum or difference of independent random variables. I can determine if two random variables are independent. I can probabilities of independent Normal random variables.
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Rules for Means and Variances Target Goal: I can find the mean and standard deviation of the sum or difference of independent random variables. I can determine if two random variables are independent. I can probabilities of independent Normal random variables. 6.2b h.w: pg 378: 49, 51, 57 – 59, 63
Rules for Means • Rule #1 • If X is a random variable and a and b are fixed numbers, then • If X and Y are random variables, then
Warm up: • Suppose the equation Y = 20 + 10X converts a PSAT math score, X, into an SAT math score, Y. Suppose the average PSAT math score is 48. What is the average SAT math score?
Example: 5 years later • Let represent the average SAT math score. • Let represent the average SAT verbal score.
represents the average combined SAT score. Then is the average combined total SAT score.
Rules for Variances • (Addition rule not always true for variances.) • Take X to be the % of a family’s after tax income that is spent. • Take Y to be the % that is saved.
If X goes down , Y goes up but the sum of X + Y always equals 100% and does not vary at all. • It is the association between X and Y that prevents their variances from adding.
Independent • If random variables are independent, the association between them is ruled out and their variances will add.
Rules for Variances If X is a random variable and a and b are fixed numbers, then • Rule #1
Rule #2 • If X and Y are independent random variables, then
Note: • The variances of independent variables add but their standard deviations do not! σX+Y = sqrt (σ2X + σ2Y),not σX+Y = σX + σY • Also, the variance of the differenceis the sum of the variances. • (b/c the square of -1 is 1, pg. 421).
Example: • Suppose the equation Y = 20 + 10X converts a PSAT math score, X, into an SAT math score, Y. Suppose the standard deviation for the PSAT math score is 1.5 points.
Standard Deviation: σX • We prefer σXas a measure of variability. • Use the rules for variance and then take the square root of.
Example: Winning the Lottery The payoff X of a $1 ticket in the Tri-State Pick 3 game is $500 with probability 1/1000 and $0 the rest of the time. Here is the combined calculation of mean and variance.
Calculation of mean and variance • So, the standard deviation σx = sqrt 249.75 = $15.80 • Games of chance have high σx.
Winnings: W = X - $1 μW= μX – 1, “payoff – cost” • 0.5– 1 = -$0.50 • Players lose money on average. The standard deviation σWof W = X -1 is the same as σW of X. • Subtracting a fixed number affects the mean not the variance.
Buy tickets two days in a row: Payoff: X + Y Find the mean and standard deviation. • Mean • μX+Y = μX + μY = $0.50 + $0.50 = $1.00
Standard Deviation X and Y are independent so, σ2X+Y = σ2X + σ2Y = 249.75 + 249.75 = 499.50 σX+Y = sqrt 499.5 Standard dev. of the total payoff = $22.35
What does this mean? • Your mean payoff for a year is 0.50 x 365 = $182.50 • Your cost to play is $365.00 • The state mean winnings is 365 – 182.50 = $182.50
Combining Normal Random Variables Linear combinations of independent random variables are also normally distributed. • If X and Y are independent and, • a and b are fixed numbers, • aX + bY is normally distributed • Find μ and σ using the rules.
Example: A round of Golf Tom and George are playing in the club golf tournament. Their scores vary as they play the repeatedly. • Tom’s score is N(110,10) • George’s score is N(100,8) If they play independently, what is the probability Tom will score lower than George?
The difference in their scores X – Y is normally distributed with: μX-Y = μX – μY = 110 – 100 = 10 σ2X-Y = σ2X + σ2Y = 102 + 82 = 164 σX-Y = 12.8
So, X – Y has the N(10, 12.8) distribution. Standardize to compute the probability. • P(X<Y) = P(X – Y < 0) • = P (X-Y) – 10 < 0 - 10 12.8 12.8 • = P(Z < -0.78) • Use Table A or calculator;
2ndVARS:normalcdf(-E99,-.78)= 0.2177 Conclusion • Although George’s score is 10 strokes lower on the average, Tom will have the lower score in about one of every five matches.