## Presentation on theme: "Copyright © 2010 Pearson Education, Inc. Slide 16 - 1."— Presentation transcript:

Copyright © 2010 Pearson Education, Inc. Slide 16 - 2 Solution: C

Copyright © 2010 Pearson Education, Inc. Slide 16 - 4 Expected Value: Center A random variable assumes a value based on the outcome of a random event. We use a capital letter, like X, to denote a random variable. A particular value of a random variable will be denoted with the corresponding lower case letter, in this case x.

Copyright © 2010 Pearson Education, Inc. Slide 16 - 5 Expected Value: Center (cont.) There are two types of random variables: Discrete random variables can take one of a countable number of distinct outcomes. Example: Number of credit hours Continuous random variables can take any numeric value within a range of values. Example: Cost of books this term

Copyright © 2010 Pearson Education, Inc. Slide 16 - 6 Expected Value: Center (cont.) A probability model for a random variable consists of: The collection of all possible values of a random variable, and the probabilities that the values occur. Of particular interest is the value we expect a random variable to take on, notated μ (for population mean) or E(X) for expected value.

Copyright © 2010 Pearson Education, Inc. Slide 16 - 7 Expected Value: Center (cont.) The expected value of a (discrete) random variable can be found by summing the products of each possible value by the probability that it occurs: Note: Be sure that every possible outcome is included in the sum and verify that you have a valid probability model to start with.

Copyright © 2010 Pearson Education, Inc. Slide 16 - 8 Example: On Valentines Day a restaurant offers a Lucky Lovers Special (I warned you!) that could save couples money on their romantic dinners. When the waiter brings the check, hell also bring four aces from a deck of cards. Hell shuffle them and lay them out face down on the table. The couple will then get to turn one card over. If its a black ace, theyll owe the full amount, but if its the ace of hearts, the waiter will give them a \$20 discount. If they first turn over the ace of diamonds (its red), theyll then get another turn to turn over another card, earning a \$10 discount if it is the ace of hearts. a. Draw a probability model for x, the dollar value of the discount. b. Based on the probability model, what is the expected discount for a couple?

Copyright © 2010 Pearson Education, Inc. Slide 16 - 9 First Center, Now Spread… For data, we calculated the standard deviation by first computing the deviation from the mean and squaring it. We do that with discrete random variables as well. The variance for a random variable is: The standard deviation for a random variable is:

Copyright © 2010 Pearson Education, Inc. Slide 16 - 10 TI – Tips Finding the mean and SD of a random variable Enter the values of the variable is L1 Enter the probability model in L2. STAT CALC L1, L2 Note: The mean is called xbar, but it is really mu. The standard deviation is lower case sigma.

Copyright © 2010 Pearson Education, Inc. Slide 16 - 11 Example: Using the probability model from the previous example. Find the mean (expected value) and the standard deviation on your calculator.

Copyright © 2010 Pearson Education, Inc. Slide 16 - 12 What happened to the mean and standard deviation when you added a number to each value? The same is true for expected values: If I add a constant to each independent random variable then I add the constant to the mean (expected value). E(X ± c) = E(X) ± c If I add a constant to each independent random variable then the standard deviation and variance doesnt change. Var(X ± c) = Var(X) SD(X ± c) = SD(X)

Copyright © 2010 Pearson Education, Inc. Slide 16 - 13 Example: The couples dining with the Lucky Lovers discounts averaging \$5.83 with a standard deviation of \$8.62. There was also a coupon in the paper for \$5 off any one meal (one discount per table). If every couple dining there also brings in a coupon, what will be the mean and standard deviation of the new discount they will receive?

Copyright © 2010 Pearson Education, Inc. Slide 16 - 14 What happened to the mean and standard deviation when you multiplied by a number? The same is true for expected values: If I multiply each independent random variable by a constant then I multiply the mean (expected value) by the constant. E(aX) = aE(X) If I multiply each independent random variable by a constant then the standard deviation is multiplied by the constant. SD(aX) = aSD(X)

Copyright © 2010 Pearson Education, Inc. Slide 16 - 15 Example: The couples dining with the Lucky Lovers discount averages \$5.83 with a standard deviation of \$8.62. When two couples dine together on a single check, the restaurant doubles the discount offer -- \$40 for the ace of hearts on the first card and \$20 on the second. What are the mean and standard deviation of discounts for such foursomes?

Copyright © 2010 Pearson Education, Inc. Slide 16 - 16 But there is a new spin … If two independent random variables are added together: the means of each expected value are added (or subtracted) together. E(X ± Y) = E(X) ± E(Y) the standard deviation is the square root of the sum of the standard deviations squared. SD(X ± Y) = SQRT(X 2 + Y 2 )

Copyright © 2010 Pearson Education, Inc. Slide 16 - 17 Example: The couples dining with the Lucky Lovers discounts averaging \$5.83 with a standard deviation of \$8.62. You go with another couple to the restaurant, but they dont like the idea of the double discount. So they sit at the table next to you, so they can have their own discount. Does this change the expected value or the standard deviation? (Calculate E(X 1 +X 2 ) and SD(X 1 +X 2 ))

Copyright © 2010 Pearson Education, Inc. Slide 16 - 18 Example: The couples dining with the Lucky Lovers discounts averaging \$5.83 with a standard deviation of \$8.62. Down the road, another restaurant is having a competing Lottery of Love promotion. A couple can select a chocolate from a bowl and unwrap it to learn the size of their discount. The manager (another math geek) says the discounts vary with an average of \$10 and a standard deviation of \$15. How much more can you expect to save at the Lottery of Love promotion? With what standard deviation?

Copyright © 2010 Pearson Education, Inc. Slide 16 - 19 One more rule! SD(X + X) does NOT equal SD(2X) SD(X + X) = SQRT(X 2 + X 2 ) Heres a short cut … SD(X + X + X + …+ X) = SQRT(n(SD) 2 )

Copyright © 2010 Pearson Education, Inc. Slide 16 - 20 Example: The couples dining with the Lucky Lovers discounts averaging \$5.83 with a standard deviation of \$8.62. The restaurant owner is planning to serve 40 couples on Valentines Day. What is the expected total of the discounts the owner will give? With what standard deviation?

Copyright © 2010 Pearson Education, Inc. Slide 16 - 21 One Last Example: Given independent random variables with means and standard deviations as shown, find the mean and standard deviation of: X – 20 0.5Y X + Y X – Y Y 1 + Y 2 MeanSD X8012 Y 3

Copyright © 2010 Pearson Education, Inc. Slide 16 - 22 Homework: Pg. 383 1-41 odd (we will do 23a together)