Combining Two Random Variables: Means and Variances Lesson 7.2.3

Starter Suppose I measure the heights of a class of fourth-graders and find the distribution of heights to be N(100, 2 cm). Then I have them all stand on top of a 10 cm high step and I measure again. What change would you expect in the mean? What change would you expect in the standard deviation?

Objectives Find the mean and variance of a random variable formed from a linear transformation of a given random variable with known mean and variance. Find the mean and variance of a random variable formed from a linear combination of two given random variables with known mean and variance. California Standards 5.0 Students know the definition of the mean of a discrete random variable and can determine the mean for a particular discrete random variable. 6.0 Students know the definition of the variance of a discrete random variable and can determine the variance for a particular discrete random variable.

Activity Write the PDF for an ordinary fair die where X is the number of spots that show. Calculate the mean and variance of X. X P(X)1/6

Mean & Variance on the TI To do the same problem on the TI, treat it as a one-variable distribution with the probabilities being shown as frequencies in L 2 for each outcome in L 1. –seq(x, x, 1, 6) → L 1 –seq(1/6, x, 1, 6) → L 2 –1-Var Stats(L 1, L 2 ) Note that the calculator gives you σ = You would have to square σ to get the variance we previously calculated

Practice on Another Variable Toss 4 coins and let Y be the number of heads that show. Enter the PDF in L 1 and L 2 and find the mean and variance of Y. –Recall that we have done this PDF before; it’s in your notes. You should have gotten μ Y = 2 and σ Y 2 = 1 Save these results; we are now ready to combine two random variables.

Linear Variations of a Random Variable Consider the fair die (Variable X) with mean 3.5 and standard deviation Suppose we re-paint the die so that its spots are 4, 5, 6, 7, 8, 9 –Now the mean is 6.5(verify on the TI) Why did the mean increase from 3.5 to 6.5? –Note that the standard deviation is still Why didn’t it change? Now let’s use a die with faces 8, 10, 12, 14, 16, 18 –Use the TI to find the mean and standard deviation Explain the new mean of 13 Explain the new standard deviation of 3.416

Generalize What We Just Saw For a given random variable, adding a uniform constant to all outcomes increases the mean by that constant. –The data shift but do not spread more, so the standard deviation and variance stay the same. Multiplying by a constant multiplies both the mean and standard deviation. –So the variance would increase by the square of the constant. Expressing this as a general idea: –For a given random variable X with known μ x and σ x, adding a constant A and multiplying by a constant B gives a new random variable A+BX with: mean μ A+BX = A+Bμ X and variance σ 2 A+BX = B 2 σ 2 X –Note that σ A+BX = Bσ X

Example We know that a fair die has a mean of 3.5 and standard deviation of If I multiply each face by three and add five to each result, write the values of the six faces. –8, 11, 14, 17, 20, 23 Based on the rules we just discussed, find the mean and standard deviation of this new die, then verify your results on the TI. μ A+BX = A+Bμ X and σ 2 A+BX = B 2 σ 2 X

Answer We expect μ new = 5 + 3μ old –So μ new = x 3.5 = 15.5 We expect σ new = 3σ old –So σ new = 3 x = To confirm, put the outcomes in L 1 and the probabilities (all 1/6) in L 2 ; run 1-Var Stats

Combining Two Random Variables If two random variables are added (or subtracted), the mean of the new random variable will be the sum (or difference) of the two original means. –μ X+Y = μ X + μ Y –μ X-Y = μ X - μ Y If two independent random variables are added (or subtracted), the variance of the new random variable will be the sum of the variances of the two original variables. –σ 2 X+Y = σ 2 X-Y = σ 2 X + σ 2 Y –Note that we do not add standard deviations

Example Return to the first two variables we considered: –X is the outcome of a fair die μ X = 3.5σ X = 1.708σ 2 X = –Y is the number of heads in 4 tossed coins μ Y = 2σ Y = 1σ 2 Y = 1 Define a new variable Z which is the sum of the die and the number of heads. –Use the approach on the previous slide to find the mean, variance and standard deviation of Z

Answer μ Z = μ X + μ Y = = 5.5 σ 2 Z = σ 2 X + σ 2 Y = = σ Z = √3.917 = –Note that σ X + σ Y = = ≠ σ Z Challenge: Write the PDF of Z and use the TI (or formulas) to verify the results we just found. –Hint: What are the greatest and least values Z can take on? –Another hint: 6 die faces x 16 ways for the coins to come up gives 96 ways to do the trial. –The answer is posted as a link on the assignments page.

Objectives Find the mean and variance of a random variable formed from a linear transformation of a given random variable with known mean and variance. Find the mean and variance of a random variable formed from a linear combination of two given random variables with known mean and variance. California Standards 5.0 Students know the definition of the mean of a discrete random variable and can determine the mean for a particular discrete random variable. 6.0 Students know the definition of the variance of a discrete random variable and can determine the variance for a particular discrete random variable.

Homework Read pages 395 – 396, 399 – 402 Do problems 24, 25, 26