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Mean & Std. Dev. of the Sum of Random Variables Page 1 of 3 1. Let X be a random variable representing the number on the top of a die after a roll. Fill in X's probability distribution in the chart below: X P(X) 1 2 3 4 5 6 • By inspection determine the expected value of X, E(X). E(X) = • Write the expression which you would have to compute by hand to calculate the standard deviation of X. See formula sheet? Do not compute it by hand. Have this formula checked by your teacher. • Use your calculator to compute E(X), (X), and the Var(X). (X) = E(X) = Var(X) = 2. Now consider rolling two die and adding up their top pips. We could create a new random variable Z which is the sum of two random variable just like X above. We could think of the first die as X and the second as Y. So Z = X + Y. If the first die showed 3 and the second showed 5 the new random variable would have the value 3 + 5 or 8. The question we wish to consider is what is E(Z) and (Z). In other words, how do we describe the new distribution Z. • Fill in the table below with all the possible outcomes of adding the top pips after rolling two dice. X+Y X=1 X=2 X=3 X=4 X=5 X=6 Y=1 2 Y=2 Y=3 Y=4 Y=5 Y=6 3. Now complete the probability distribution for Z: Z P(Z) 2 1/36 3 4 5 6 - over - 7 8 9 10 11 12 Mean & Std. Dev. of the Sum of Random Variables Page 2 of 3 4. Use your calculator to compute E(Z), (Z), and the Var(Z). You will have to enter the new data. E(Z) = (Z) = Var(Z) = 5. Which of the following is true (at least in this case)? a. E(X + Y) = E(X) + E(Y) b. (X + Y) = (X) + (Y) c. Var(X + Y) = Var(X) + Var(Y) 6. You should have concluded that a and c in the previous question are true. Write two sentences which contain NO symbols which state these two major conclusions of this exercise. Conclusions: • • Mean & Std. Dev. of the Sum of Random Variables Page 3 of 3 You do not need to know or study the material on this page. Topic: Why standard deviation of the sum of two INDEPENDENT random variables is less than the sum of their standard deviations. BVD 3rd, p 373 Here is a proof from Floyd Bullard, an AP Stats guru and one of the best teachers I know: Var(X)+Var(Y)=Var(X+Y) (due to independence) Sigma^2(X) + sigma^2(Y) = sigma^2(X+Y) (another notation for the same thing) Sigma^2(X) + sigma^2(Y) + 2*sigma(X)*sigma(Y) > sigma^2(X+Y) (add a positive number the left side only, it must become greater than the right) (sigma(X)+sigma(Y))^2 > sigma^2(X+Y) (replace binomial-squared expansion on LHS with the square of the binomial) Sigma(X)+sigma(Y)>sigma(X+Y) (take square root of both sides; legal in inequality because sqrt function strictly increases, monotonically)