Expected Values and Standard Deviation of Random Variables Chapter 4-2 Expected Values and Standard Deviation of Random Variables

Looking back at 4-1 Remember Example 2: Rolling a fair die 2, 3, 4, or 5 you would win $1.50, and if it rolled a 1, or 6 you would pay $1 or $6 respectively. Well you could write that out in an equation: Expected gain = (−6) 1 6 + (−1) 1 6 + 1.5 4 6 =− 1 6 In general: Pr 𝑋 = 𝑥 1 𝑝 1 + 𝑥 2 𝑝 2 +. . . .+ 𝑥 𝑘 𝑝 𝑘 The x’s are the random variables (possible outcomes) and p’s are the probability for each of them.

Example 1 VideoMax has a promotion featuring 3 premium label DVDs and 5 discount specials. Each premium label DVD sells for $24, and each discount special sells for $16. Two DVDs are selected at random, and a random variable X is defined as the total cost of the DVDs selected. Find the expected value of the random variable X. Value of selecting 2 Discount DVDs is $32 (2 x $16) Value of selecting 1 Discount and 1 Premium is $40 Value of selecting 2 Premium DVDs is $48 Probability of spending $32: 𝐶(5,2)∙𝐶(3,0) 𝐶(8,2) = 20 56 Probability of spending $40: 𝐶(5,1)∙𝐶(3,1) 𝐶(8,2) = 30 56 Probability of spending $48: 𝐶(5,0)∙𝐶(3,2) 𝐶(8,2) = 6 56 Expected value: 32 20 56 +40 30 56 +48 6 56 =38 Basically, $38 is the average amount of money someone choosing 2 CDs would spend

Example 2 A friend invites you to play the following game: 2 coins are selected at random from 6 coins: 2 nickels, 3 dimes and 1 quarter. If the sum of the values of the coins is 10, 20, or 30 cents, your friend pays you 25 cents. Otherwise, you must pay your friend the value of the coins. 2 nickels: 10 cents (+25) 1 nickel/1 dime: 15 cents (-15) 2 dimes: 20 cents (+25) 1 nickel/1 quarter: 30 cents (+25) 1 dime/1 quarter: 35 cents (-35) −35: 𝐶 3,1 ∙𝐶 1,1 𝐶 6,2 = 3 15 −15: 𝐶 2,1 ∙𝐶 3,1 𝐶 6,2 = 6 15 +25: 𝐶 2,2 𝐶 6,2 + 𝐶 3,2 𝐶 6,2 + 𝐶 2,1 ∙𝐶(1,1) 𝐶 6,2 = 6 15 Expected: −35 3 15 + −15 6 15 + 25 6 15 =−3 You should expect to lose 3 cents to your friend each time you play

Example 3 A sales representative makes sales to 20% of the customers he calls. His commissions are $100 each for the first 3 sales each week, $200 each for every sale above 3. In a certain week the sales representative calls on 8 customers. Find the expected value of the total income of the sales representative for that week. Bernoulli! Each call is either a success or failure 0 sales: $0, 𝐶 8,0 (0.2) 0 (0.8) 8 =0.1678 x0 =0 1 sale: $100, 𝐶 8,1 (0.2) 1 (0.8) 7 =0.3355 x100 =33.55 2 sales: $200, 𝐶 8,2 (0.2) 2 (0.8) 6 =0.2936 x200 =58.72 3 sales: $300, 𝐶 8,3 (0.2) 3 (0.8) 5 =0.1468 x300 =44.04 4 sales: $500, 𝐶 8,4 (0.2) 4 (0.8) 4 =0.0459 x500 =22.95 5 sales: $700, 𝐶 8,5 (0.2) 5 (0.8) 3 =0.0092 x700 =6.44 6 sales: $900, 𝐶 8,6 (0.2) 6 (0.8) 2 =0.0011 x900 =.99 7 sales: $1100, 𝐶 8,7 (0.2) 7 (0.8) 1 =0.0001 x1100 =.11 8 sales: $1300, 𝐶 8,8 (0.2) 8 (0.8) 0 =0.0000 x1300 =0 All adds up to 166.80. The sales person can expect $166.80

Expected value of a binomial random variable If X is a binomial random variable for a Bernoulli process consisting of n trials with probability of success p, then Expected value = np

Example 4 A political poll is to be taken by mailing 1000 questionnaires. It is known that the probability that any given questionnaire will be returned is .14. Find the expected number of questionnaires to be returned. n = 1000 p = 0.14 Expected return: 1000∙0.14=140

Example 5 An NBA basketball player makes 30 percent of his 3-point shot attempts and 45 percent of his 2-point shot attempts. If each year he attempts 300 three-point shots and 800 two-point shots, what is the average number of points he makes per year on field goals? n = 300 p = 0.3 Expected 3-pointers made: 300∙0.3=90 (270 points) n = 800 p = 0.45 Expected 2-pointers made: 800∙0.45=360 (720 points) Total points in a year: 270+720=990

Standard Deviation The graphs of the density functions in table 4-10 is show in figure 4-2. Notice the means of each: 𝐸 𝑋 𝐴 = 5 0.5 + −5 0.5 =0 𝐸 𝑋 𝐴 = 1 0.25 + 0 0.5 +(−1)(0.25)=0 𝐸 𝑋 𝐴 = 5 0.1 + 0 0.8 + −5 0.1 =0 All the means are zero, even though each table/graph is very different.

Standard Deviation Standard deviation doesn’t measure average, it measures dispersion of data (how spread out it is) Variance: 𝑉𝑎𝑟 𝑋 = ( 𝑥 1 −𝜇) 2 𝑝 1 + ( 𝑥 2 −𝜇) 2 𝑝 2 +…+ ( 𝑥 𝑘 −𝜇) 2 𝑝 𝑘 x’s are the random variables, 𝜇 is the mean, p’s are probabilities of each Standard deviation: 𝜎= 𝑉𝑎𝑟[𝑋]

Example 7 Determine variance and standard deviation from the table/graph earlier. Variance: 𝑉𝑎𝑟 𝑋 𝐴 = 5−0 2 0.5 + −5−0 2 0.5 =25 𝑉𝑎𝑟 𝑋 𝐵 = 1−0 2 0.25 + 0−0 2 0.5 + −1−0 2 0.25 =0.5 𝑉𝑎𝑟 𝑋 𝐶 = 5−0 2 0.1 + 0−0 2 0.8 + −5−0 2 0.1 =5 Standard deviation: 𝑋 𝐴 : 𝜎= 25 =5 𝑋 𝐵 : 𝜎= 0.5 =0.71 𝑋 𝐶 : 𝜎= 5 =2.2

Variance of a Binomial Random Variable If X is a binomial random variable for a Bernoulli process consisting of n trials and with probability of success p, then the variance of X, Var[X], is 𝑉𝑎𝑟 𝑋 =𝑛𝑝 1−𝑝 And the standard deviation of X, 𝜎 𝑋 = 𝑛𝑝(1−𝑝)

Example 8 Sales information provided by O.A. Rowe Seed Company asserts that the germination rate for its best selling Robust Red Tomato seeds is 98%. Suppose 2000 seeds are planted. Let X denote a random variable which assigns to each outcome, for the “experiment” of planting 2000 seeds, the number of seeds which germinate. Assume that this is a Bernoulli process and that the assertion is correct. Find the expected value and variance of X. 𝐸 𝑋 =2000 0.98 =1960 𝑉𝑎𝑟 𝑋 =2000 0.98 0.02 =39.2