Rules for Random Variables EQ What is the

Rules for Random Variables EQ: What is the expected value of a random variable and how much will it vary?

Expected Value You draw a card from a deck. If you get a red card, you win nothing. If you get a spade, you win $5. For any club, you win $10 pus an extra $20 for the ace of clubs. X $0 $5 $10 $30 P(X) . 50 . 25 . 23 . 02 Expected Value: Find the expected amount you will win from playing this game E(X) = (0)(. 50) + (5)(. 25) + (10)(. 23) + (30)(. 02) E(X) = $4. 15 How much should be charged to play the game?

Variance Def: How much can the data vary from the expected value on average Standard deviation:

Example X $0 $5 $10 $30 P(X) . 50 . 25 . 23 . 02

Example A couple plans to have children until they get a girl, but they agree that they will not have more than three children even if all are boys (assume boys and girls are equally likely. ) X = number of children they have X 1 2 3 P(X) 0. 5 0. 25 How many children can they expect to have? E(X) = 1(0. 5) + 2(. 25) + 3(. 25) E(X) = 1. 75 What is the standard deviation?

Combining Random Variables An insurance policy costs $100 and will pay policyholders $10, 000 if they suffer a major injury (resulting in hospitalization) or $3000 if they suffer a minor injury (resulting in lost time from work). The company estimates that each year 1 in every 2000 policy holders may have a major injury, and 1 in 500 a minor injury only. a. Create a probability model for the PROFIT on a policy b. what’s the company’s expected profit on this policy? c. What’s the standard deviation?

Insurance X = the profit on a policy X P(X) $100. 9975 $-9900. 0005 E(X) = 100(. 9975) + (-9900)(. 0005) + (-2900)(. 002) E(X) = $89 $-2900. 002

Rules for Random Variables

Example W. J. Youden (Australian, 1900 -1971) weighed many new pennies and found the distribution of weights to be approximately N(3. 11 g, 0. 43 g). What are the reasonably likely mean and standard deviation of weights of rolls of 50 pennies? X = A single penny 50 pennies = 50 (X) X ~ N(3. 11 g, 0. 42 g) 50 X ~ N(155. 5 g, 21. 5 g)

Putting it all together Suppose the amount of propane needed to fill a customer’s tank is a random variable with a mean of 318 gallons and a standard deviation of 42 gallons. Hank Hill is considering two pricing plans for propane. Plan A would charge $2 per gallon. Plan B would charge a flat rate of $50 plus $1. 80 per gallon. Calculate the mean and standard deviation of the distributions of money earned under each plan. X = amount of propane needed to fill tank X ~ N (318, 42) Plan A: $2 per gallon Plan B: $1. 80 per gallon + $50

X ~ N (318, 42) Plan A: $2 per gallon E(2 X) = 2(318) = $636 SD(2 X) = $54 Plan A ~ N ($636, $54)

X ~ N (318, 42) Plan B: $1. 80 per gallon + $50 E(1. 80 X + 50) = 1. 80(318) + 50 = $622. 40 Var(1. 80 X + 50) = Std Dev = $75. 6 Plan B ~ N (622. 40, 75. 6)

Example Suppose the average SAT-Math score is 625 and the standard deviation of SAT-Math is 90. The average SATVerbal score is 590 and the standard deviation of SATVerbal is 100. The composite SAT score is determine by adding the Math and Verbal portions. What is the mean and standard deviation of the composite SAT score? X = SAT Math Y = SAT Verbal E(X) = 625 E(Y) = 590 E(X + Y) = 625 + 590 = 1215 Var(X + Y) = 8100+ 1000 = 9100 Std Dev (X + Y) = 95. 39