Copyright © Cengage Learning. All rights reserved. 8 PROBABILITY DISTRIBUTIONS AND STATISTICS

Copyright © Cengage Learning. All rights reserved. 8.2 Expected Value

3 Mean

4 The average value of a set of numbers is a familiar notion to most people. For example, to compute the average of the four numbers 12, 16, 23, 37 we simply add these numbers and divide the resulting sum by 4, giving the required average as

5 Mean In general, we have the following definition:

6 Applied Example 1 – Waiting Times Find the average number of cars waiting in line at the bank’s drive-in teller at the beginning of each 2-minute interval during the period in question. Solution: The number of cars, together with its corresponding frequency of occurrence, are reproduced in Table 7. Table 7

7 Applied Example 1 – Solution Observe that the number 0 (of cars) occurs twice, the number 1 occurs 9 times, and so on. There are altogether = 60 numbers to be averaged. Therefore, the required average is given by or approximately 3.1 cars. cont’d (1)

8 Expected Value

9 Let’s reconsider the expression on the left-hand side of Equation (1), which gives the average of the frequency distribution shown in Table 7. Table 7

10 Expected Value Dividing each term by the denominator, we may rewrite the expression in the form Observe that each term in the sum is a product of two factors; the first factor is the value assumed by the random variable X, where X denotes the number of cars waiting in line, and the second factor is just the probability associated with that value of the random variable.

11 Expected Value This observation suggests the following general method for calculating the expected value (that is, the average or mean) of a random variable X that assumes a finite number of values from the knowledge of its probability distribution.

12 Expected Value Note: The numbers x 1, x 2,..., x n may be positive, zero, or negative. For example, such a number might be positive if it represents a profit and negative if it represents a loss.

13 Expected Value The expected value of a random variable X is a measure of the central tendency of the probability distribution associated with X. In repeated trials of an experiment with random variable X, the average of the observed values of X gets closer and closer to the expected value of X as the number of trials gets larger and larger.

14 Expected Value Geometrically, the expected value of a random variable X has the following simple interpretation: If a laminate is made of the histogram of a probability distribution associated with a random variable X, then the expected value of X corresponds to the point on the base of the laminate at which the laminate will balance perfectly when the point is directly over a fulcrum (Figure 6). Figure 6 Expected value of a random variable X

15 Applied Example 4 – Expected Profit A private equity group intends to purchase one of two motels currently being offered for sale in a certain city. The terms of sale of the two motels are similar, although the Regina Inn has 52 rooms and is in a slightly better location than the Merlin Motor Lodge, which has 60 rooms. cont’d

16 Applied Example 4 – Expected Profit Records obtained for each motel reveal that the occupancy rates, with corresponding probabilities, during the May–September tourist season are as shown in the following tables. cont’d

17 Applied Example 4 – Expected Profit The average profit per day for each occupied room at the Regina Inn is $40, whereas the average profit per day for each occupied room at the Merlin Motor Lodge is $36. a. Find the average number of rooms occupied per day at each motel. b. If the investors’ objective is to purchase the motel that generates the higher daily profit, which motel should they purchase? (Compare the expected daily profit of the two motels.) cont’d

18 Applied Example 4(a) – Solution Let X denote the occupancy rate at the Regina Inn. Then the average daily occupancy rate at the Regina Inn is given by the expected value of X—that is, by E (X) = (.80)(.19) + (.85)(.22) + (.90)(.31) + (.95)(.23) + (1.00)(.05) =.8865 The average number of rooms occupied per day at the Regina Inn is (.8865)(52)  46.1 or approximately 46.1 rooms.

19 Applied Example 4(a) – Solution Similarly, letting Y denote the occupancy rate at the Merlin Motor Lodge, we have E(Y) = (.75)(.35) + (.80)(.21) + (.85)(.18) + (.90)(.15) + (.95)(.09) + (1.00)(.02) =.8240 The average number of rooms occupied per day at the Merlin Motor Lodge is (.8240)(60)  49.4 or approximately 49.4 rooms. cont’d

20 Applied Example 4(b) – Solution The expected daily profit at the Regina Inn is given by (46.1)(40) = 1844 or $1844. The expected daily profit at the Merlin Motor Lodge is given by (49.4)(36)  1778 or approximately $1778. From these results, we conclude that the private equity group should purchase the Regina Inn, which is expected to yield a higher daily profit. cont’d

21 Expected Value The next example illustrates the Games that is “fair.” In a fair game, neither party has an advantage, a condition that translates into the condition that E(X) = 0, where X takes on the values of a player’s winnings.

22 Applied Example 7 – Fair Games Mike and Bill play a card game with a standard deck of 52 cards. Mike selects a card from a well-shuffled deck and receives A dollars from Bill if the card selected is a diamond; otherwise, Mike pays Bill a dollar. Determine the value of A if the game is to be fair. Solution: Let X denote a random variable whose values are associated with Mike’s winnings. Then X takes on the value A with probability P(X = A) = (since there are 13 diamonds in the deck) if Mike wins and takes on the value –1 with probability P(X = –1) = if Mike loses.

23 Applied Example 7 – Solution Since the game is to be a fair one, the expected value E(X) of Mike’s winnings must be equal to zero; that is, Solving this equation for A gives A = 3. Thus, the card game will be fair if Bill makes a $3 payoff when a diamond is drawn. cont’d

24 Practice p. 448 Self-Check Exercises #1 & 2

25 Assignment p. 448 Exercises #1-15 odd, 16, odd