# High School: Statistics and Probability

### Using Probability to Make Decisions HSS-MD.A.2

2. Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.

You know how there are about a zillion names for the same drug? Or all the different names, nicknames, and scientific names for the same animal? Or that there are over twenty different synonyms for "buttocks" in the English language?

Well, students should know that mathematicians are just as good at making up words (though none as funny as "derriere" or "hiney.") So when they say expected value, they mean weighted average. But the expected value is the true mean, whereas the actual mean may vary. You know what we mean?

No? Well let's get to the "bottom" of this.

Students can think about it this way. If we conduct a short experiment with ten random coin tosses, we expect to get 5 tails. That's our expected value. However, our sample mean may actually be 3 because of the random nature of the flips combined with the small sample size. That's the real difference. Over a much longer time frame with a much large sample size, our sample mean should approach the expected value: 5 tails for every 10 tosses.

Students should know that in order to calculate our expected value, we use the formula E(x) = x1p1 + x2p2 + … + xipi, where E(x) is our expected value and p is the probability of event x occurring. Remember that x is a number associated with the event. (If you were picking coins out of a hat, the event of picking up a penny would be x = 1, while the event of picking up a dime would be x = 10.)

Bring in a deck of cards to class. Assign values for randomly drawing a card. (For example, each numbered card can equal that payoff.) Calculate the expected value of the payoff and then conduct an actual experiment to see if the actual value matches the expected value. Repeat this with increasing sample amounts to help your students understand how a sample mean may differ from the actual expected value depending on sample size.

#### Drills

1. What is the expected value of a situation whereby one of 34 events can occur, each one with equal probabilities, most like?

Weighted average

An expected value takes into account probability as well as the value of the random variable's outcome, which is therefore a weighted average. A simple average of only probabilities wouldn't yield an expected value. The sum of probabilities should give 1, and the probabilities all multiplied together would give the probability of all of those events happening one after another rather than averaged.

2. When you're over playing darts in your friend's basement, you have a 50% chance of hitting the outer most rings, a 40% chance of hitting the middle set of rings, and a 10% probability of hitting the bull's eye. If you get 1, 5, and 10 points respectively for hitting those areas, what is the expected value of each throw?

3.5

If we use the equation E(x) = x1p1 + x2p2 + … + xipi, we get 0.5 × 1 + 0.4 × 5 + 0.1 × 10 = 3.5. Even though there can't be a throw that earns you 3.5 points, it represents the expected value. As you throw more and more darts, each throw should earn you an approximate value of 3.5 points.

3. Explain why the expected value is not (10 + 5 + 1) ÷ 3 for the question above.

The probabilities for each event are different

Since each event has a different probability of occurring, we cannot treat them as equal parts in the weighted average. This means we should incorporate the probabilities into the average since they'll "weigh" the likely outcome of each event appropriately.

4. An insurance agent claims that if you buy a specific health insurance policy, you're covered for 80% of all health related costs when you break an arm or a leg. She also claims it is about 25% likely that you will break your leg and 50% likely you'll break your arm in the next year. Google (in its almighty power) says that a broken leg costs an average of \$5,000 and a broken arm costs \$2,000, and that's after the 80% coverage has been put into effect. What is the difference in expected value if you do not buy the insurance policy over buying it?

\$9,000

If \$2,000 and \$5,000 are only 20% of the total cost you'd be paying, we can calculate the total price of breaking an arm and a leg: \$10,000 and \$25,000, respectively. Then again, there's only a chance these will happen, so we can use E(x) = x1p1 + x2p2 + … + xipi and find out that E(x) = 25,000 • 0.25 + 10,000 • 0.5 = 11,250. Calculating the expected value with insurance yields 5,000 • 0.25 + 2,000 • 0.5 = 2,250. The answer is the difference between the two, or \$9,000.

5. You have compiled a list of the ten richest people in the world. Their worth is listed in billions below. If you were to randomly pick someone on the list, what would you expect his or her worth would be?

 Billionaire Worth Adam Moneybuttons 50 Shirley Z. Cash 40 Glitter E. Diamondshoes 45 Stanley Greenback 32 Leslie A. Blank-Czech 34 Gordon Goldfarb 23 Audrey Platinumy 24 Richard Wealthyboots 25 Sterling Silver 18 William Bigbucks 20

31

Calculating the E(x) value just means multiplying the billionaire's money (x) and the probability of randomly selecting that individual (p = 0.1). If we add up those values, we'll get E(x) = 31.1, or (C).

6. You decide to pick a random person's age from a pool of 22 people. (Literally, they're swimming in the community pool.) Their ages are given below. What is the expected value of the age you'll get if you do this over and over?

 Person # Age 1 75 2 28 3 92 4 75 5 83 6 10 7 48 8 72 9 10 10 48 11 32 12 38 13 19 14 8 15 49 16 2 17 4 18 9 19 75 20 29 21 48 22 27

40

Since you're randomly picking a people, the probability of choosing any one person is 1 in 22, or about 0.045455. Our E(x) value would equal the sum of the people's ages times 0.045455. If we calculate the expected value, we should get E(x) = 40. That means our answer is (B).

7. The Department of Transportation Safety believes that people do not know what their average tire pressures should be. They randomly test the air pressure inside car tires as they come in for their scheduled emissions test. The data is found below. What is the expected value of the tire pressure?

 Car Average PSI 1 12 2 42 3 65 4 32 5 33 6 34 7 33 8 12 9 18 10 19 11 24 12 27

30

The cars that come in are random, so the probability associated with each one is 1 in 12, or 0.0833. If we multiply that by the average tire pressure (in psi) and add up the values, we'll get an average tire pressure of E(x) = 29.7. That's about 30, so we can say our answer is (C).

8. A health insurance company has calculated that a critical heart surgery has a 0.25 chance of being a failure, which will lead to medical claims damages awarded to the patient in the amount of \$400,000 per case. Given 10 operations; what is the expected value of claims being paid out?

1,000,000

Each surgery has a p = 0.25 chance of being a failure, and the claim for a failed surgery is x = \$400,000. That means that on average, every surgery costs the hospital \$100,000 (by multiplying x and p). If ten surgeries are performed, the expected amount paid in damages would be 10 × \$100,000 = \$1,000,000.

9. An unfair die is being used to gamble. It has a 0.4 probability of landing on the number 3. If it lands on this number, you stand to lose \$50. (You don't know the die is unfair.) What is the expected value of your earnings when the die is rolled three times? (Assume you'll win nothing on any other number and you don't have to pay to play. Sounds like a rotten deal, but just go with it.)

-\$60

Each time the dice is rolled, you stand to lose \$50, but only if it lands on 3, which has a 0.4 chance of happening. That means your earnings are 0.4 • -50 = -20 for each roll of the dice. When rolled three times, your earnings will amount to an expected value of -\$60. (Yes, the negative sign matters.)

10. What is the probability that must be used in calculating the expected value of throwing a fair six-sided die, regardless of payoff for each throw?