The mean of a set of observations is their ordinary average, whereas the mean of a random variable X is an average of the possible values of X The mean of a random variable X is often called the expected value of X, and describes the long-run average outcome. It is a weighted average.

Example 7.6, p. 483 X Prob 1/9 V Prob Mean of X (μ x ) = Mean of V (μ V ) =

Locating the Mean in a discrete distribution = the center of the distribution

Example (Mean of a prob. distribution = Expected Value) Two dice are rolled simultaneously. If both show a 6, then the player wins $20, otherwise the player loses the game. It costs $2.00 to play the game. What is the expected gain or loss of the game?

Ex. 7.7 Linda sells cars and motivates herself by using probability estimates of her sales. She estimates her car sales as follows: Find the mean and variance of X. Cars sold 0123 Prob

Ex. 7.8: The distribution of the heights of women is close to normal with N(64.5, 2.5). The graph plots the values as we add women to our sample. Eventually the mean of the observations gets close to the population mean and settles down at that value.

More on Law of Large #’s  The law says that the average results of many independent observations/decisions are stable and predictable. Insurance companies, grocery stores, and other industries can predict demand even though their many customers make independent decisions.  A grocery store deciding on how many gallons of milk to stock  A fast food restaurant deciding how many beef patties to prepare  Gambling casinos (games of chance must be variable in order to hold the interest of gamblers). Even a long evening in a casino is unpredictable; the house plays often enough to rely on the law of large numbers, but you don’t. The average winnings of the house on tens of thousands of bets will be very close to the mean of the distribution of winnings (guaranteeing a profit).

How large is a large number?  Can’t write a rule for how many trials are needed to guarantee a mean outcome close to ; this depends on the variability of the random outcomes. The more variable the outcomes, the more trials are needed to ensure that the mean outcome is close to the distribution mean.

You! Emergency Evacuation A panel of meteorological and civil engineers studying emergency evacuation plans for Florida’s Gulf Coast in the event of a hurricane has estimated that it would take between 13 and 18 hours to evacuate people living in a low- lying land, with the probabilities shown here. Find the mean, variance, and standard deviation of the distribution. Time to Evacuate (nearest hr) Probability

To find the sum (or difference) of the means of 2 random variables, add the individual means of the random variables X and Y together. If 2 random variables are independent, the variance of the sum (or difference) of the 2 random variables is equal to the sum of the 2 individual variances. DOES NOT WORK for standard deviations!

Example The following data comes from a normally distributed population. Given that X={2, 9, 11, 22} and Y={5, 7, 15, 21}, illustrate the rules for means and variances. Rule 1 for means: Find 3 + 2X and the new mean Rule 2 for means: Find X + Y and mean of X + Y Rule 1 for variances: Find 3 + 2X and the new variance

Example #1 Gain Communication sells units to both the military and civilian markets. Next year’s sales depend on market conditions that are unpredictable. Given the military and civilian division estimates and the fact that Gain makes a profit of $2000 on each military unit sold and $3500 on each civilian unit sold, find: a) The mean and the variance of the number X of communication units. b) The best estimate of next year’s profit. Military Units sold: ,000 Probability: Civilian Units sold: Probability:.4.5.1

Example #2 The probabilities that a randomly selected customer purchases 1, 2, 3, 4, or 5 items at a convenience store are.32,.12,.23,.18, and.15, respectively. a) Construct a probability distribution (table) for the data and verify that this is a legitimate probability distribution. b) Calculate the mean of the random variable. Interpret this value in the context of this problem. c) Find the standard deviation of X. d) Suppose 2 customers (A and B) are selected at random. Find the mean and the standard deviation of the difference in the number of items purchased by A and by B. Show your work.

Example #3 Any linear combination of independent Normal random variables is also Normally Distributed. Suppose that the mean height of policemen is 70 inches w/a standard deviation of 3 inches. And suppose that the mean height for policewomen is 65 inches with a standard deviation of 2.5 inches. If heights of policemen and policewomen are Normally distributed, find the probability that a randomly selected policewoman is taller than a randomly selected policeman.

Example #4 Here’s a game: If a player rolls two dice and gets a sum of 2 or 12, he wins $20. If the person gets a 7, he wins $5. The cost to play the game is $3. Find the expected payout for the game.

Example #5 The random variable X takes the two values and, each with probability 0.5. Use the definition of mean and variance for discrete random variables to show that X has mean and standard deviation.