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POD #50 12/2/2011 2007B #6b Is this model appropriate? Write the equation of the LSRL Stats: Modeling the World Chapter 16 Random Variables Greedy Pig game What is a Random Variable? A random variable assumes a value based on the outcome of a random event. •We use a capital letter, like X, to denote a random variable. •A particular value of a random variable will be denoted with the corresponding lower case letter, in this case x. Two Types of Random Variables: Discrete random variables can take one of a countable number of distinct outcomes. Example: Number of credit hours Continuous random variables can take any numeric value within a range of values. Example: Cost of books this term Measure of Center: Expected Value E ( X ) xP( x) Measure of Center: Expected Value Policyholder Outcome Death Disability Neither Payout x 10,000 5,000 0 Probability P(X=x) 1/1000 2/1000 997/1000 Find the expected value of the insurance company payout E(X): Measure of Spread: Standard Deviation Var ( X ) ( x ) P( x) 2 SD( X ) Var ( X ) 2 Measure of Spread: Standard Deviation Policyholder Outcome Death Disability Neither Payout x 10,000 5,000 0 Probability P(X=x) 1/1000 2/1000 997/1000 Deviation (X-µ) (10,000-20)=9980 (5000-20)=4980 (0-20)=-20 Find the standard deviation of the insurance company payout SD(X) or σ: Example: A carnival game offers a $100 cash prize for anyone who can break a balloon by throwing a dart at it. It costs $5 to play, and you’re willing to spend up to $20 trying to win. You estimate that you have about a 10% chance of hitting the balloon on any throw. a) Create a probability model for this carnival game. b) Find the expected number of darts you will throw. c) Find your expected winnings. d) Find the standard deviation of your winnings. . Example: . A carnival game offers a $100 cash prize for anyone who can break a balloon by throwing a dart at it. It costs $5 to play, and you’re willing to spend up to $20 trying to win. You estimate that you have about a 10% chance of hitting the balloon on any throw. a) Create a probability model for this carnival game. Net Winnings $95 $90 Number of Darts 1 dart 2 darts P(Amount won) =0.1 (0.9)(0.1) =0.09 $85 3 darts (0.9)(0.9)(0.1) =0.081 $80 4 darts (win) (0.9)3(0.1) =0.0729 -$20 4 darts (lose) (0.9)4 =0.6561 Net Winnings $95 $90 Number of Darts 1 dart 2 darts P(Amount won) =0.1 (0.9)(0.1) =0.09 $85 3 darts (0.9)(0.9)(0.1) =0.081 $80 4 darts (win) (0.9)3(0.1) =0.0729 -$20 4 darts (lose) (0.9)4 =0.6561 A carnival game offers a $100 cash prize for anyone who can break a balloon by throwing a dart at it. It costs $5 to play, and you’re willing to spend up to $20 trying to win. You estimate that you have about a 10% chance of hitting the balloon on any throw. b) Find the expected number of darts you will throw. Net Winnings $95 $90 Number of Darts 1 dart 2 darts P(Amount won) =0.1 (0.9)(0.1) =0.09 $85 3 darts (0.9)(0.9)(0.1) =0.081 $80 4 darts (win) (0.9)3(0.1) =0.0729 -$20 4 darts (lose) (0.9)4 =0.6561 A carnival game offers a $100 cash prize for anyone who can break a balloon by throwing a dart at it. It costs $5 to play, and you’re willing to spend up to $20 trying to win. You estimate that you have about a 10% chance of hitting the balloon on any throw. c) Find your expected winnings. Net Winnings $95 $90 Number of Darts 1 dart 2 darts P(Amount won) =0.1 (0.9)(0.1) =0.09 $85 3 darts (0.9)(0.9)(0.1) =0.081 $80 4 darts (win) (0.9)3(0.1) =0.0729 -$20 4 darts (lose) (0.9)4 =0.6561 A carnival game offers a $100 cash prize for anyone who can break a balloon by throwing a dart at it. It costs $5 to play, and you’re willing to spend up to $20 trying to win. You estimate that you have about a 10% chance of hitting the balloon on any throw. d) Find the standard deviation of your winnings. 2 Var (Winnings) (95 17.20)2 (0.1) (90 17.20)2 (0.09) (85 17.20)2 (0.081) (80 17.20)2 (0.0729) (20 17.20)2 (0.6561) 2650.057 SD(Winnings) Var (Winnings) 2650.057 $51.48 Try This: . A small software company bids an two contracts. It anticipates a profit of $50,000 if it gets the larger contract and a profit of $20,000 on the smaller contract. The company estimates there’s a 30% chance it will get the larger contract and a 60% chance it will get the smaller contract. Assuming the contracts will be awarded independently. a) Organize the events in a Venn diagram. b) Create a probability model for the company. c) What is the expected profit? d) What is the standard deviation of the profit? Try This: . A small software company bids an two contracts. It anticipates a profit of $50,000 if it gets the larger contract and a profit of $20,000 on the smaller contract. The company estimates there’s a 30% chance it will get the larger contract and a 60% chance it will get the smaller contract. Assuming the contracts will be awarded independently. a) Organize the events in a Venn diagram. Larger 0.12 0.18 Smaller 0.42 0.28 A small software company bids an two contracts. It anticipates a profit of $50,000 if it gets the larger contract and a profit of $20,000 on the smaller contract. The company estimates there’s a 30% chance it will get the larger contract and a 60% chance it will get the smaller contract. Assuming the contracts will be awarded independently. b) Create a probability model for the company. Profit P(Profit) Larger Only Smaller Only Both $50,000 $20,000 $70,000 0.12 0.42 0.18 Neither $0 0.28 Profit P(Profit) Larger Only Smaller Only Both $50,000 $20,000 $70,000 0.12 0.42 0.18 Neither $0 0.28 A small software company bids an two contracts. It anticipates a profit of $50,000 if it gets the larger contract and a profit of $20,000 on the smaller contract. The company estimates there’s a 30% chance it will get the larger contract and a 60% chance it will get the smaller contract. Assuming the contracts will be awarded independently. c) What is the expected profit? E (Profit)=$50,000(0.12)+$20,000(0.42)+$70,000(0.18)+$0(0.28) =$27,000 d) What is the standard deviation of the profit? 2 Var (Profit)=($50,000-$27,000) 2 (0.12) ($20,000-$27,000)2 (0.42) ($70,000-$27,000) 2 (0.18) ($0-$27,000)2 (0.28) 621,000,000 Var (Profit) 621, 000, 000 $24,919.87 More about Means and Variances Adding or subtracting a constant from data shifts the mean but doesn’t change the variance or standard deviation: E(X ± c) = E(X) ± c Var(X ± c) = Var(X) Example: Consider everyone in a company receiving a $5000 increase in salary. More about Means and Variances In general, multiplying each value of a random variable by a constant multiplies the mean by that constant and the variance by the square of the constant: E(aX) = aE(X) Var(aX) = a2Var(X) Example: Consider everyone in a company receiving a 10% increase in salary. More about Means and Variances In general, The mean of the sum of two random variables is the sum of the means. The mean of the difference of two random variables is the difference of the means. E(X ± Y) = E(X) ± E(Y) If the random variables are independent, the variance of their sum or difference is always the sum of the variances. Var(X ± Y) = Var(X) + Var(Y) *Standard Deviations DO NOT add!!! Example: . Couples dining at the Quiet Nook can expect Lucky Lovers discounts averaging $5.83 with a standard deviation of $8.62. a) Suppose that for several weeks the restaurant has also been distributing coupons worth $5 off any one meal (one discount per table). If every couple dining there on Valentine’s Day bring a coupon, what will the mean and standard deviation of the total discounts they’ll receive? Example: . Couples dining at the Quiet Nook can expect Lucky Lovers discounts averaging $5.83 with a standard deviation of $8.62. b) When two couples dine together on a single check the restaurant doubles the discount offer. What are the mean and standard deviation of discounts for the foursome? Example: . Couples dining at the Quiet Nook can expect Lucky Lovers discounts averaging $5.83 with a standard deviation of $8.62. c. Now two couple are dining together but decide to get separate checks. What is the mean and standard deviation for this situation? Example: . Couples dining at the Quiet Nook can expect Lucky Lovers discounts averaging $5.83 with a standard deviation of $8.62. d. Now there is another restaurant called the Wise Fool that has a competing discount. The discount at the wise fool has an average of $10 and a standard deviation of $15. How much more can you expect to save at the Wise Fool and with what standard deviation? Try This: . A casino knows that people play the slot machines in hopes of hitting the jackpot but that most of them lose their dollar. Suppose a certain machine pays out an average of $0.92, with a standard deviation of $120. a) Why is the standard deviation so large? b) If you play 5 times, what are the mean and standard deviation of the casino’s payout? c) If gamblers play this machine 1000 times in a day, what are the mean and standard deviation of the casino’s payout? Try This: . The American Veterinarian Association claims that the annual cost of medical care for dogs averages $100, with a standard deviation of $30, and for cats averages $120, with a standard deviation of $35. a) What’s the expected difference in the cost of medical care for dogs and cats? b) What’s the standard deviation of the difference? Example: . A company manufacturer small stereo systems. At the end of the production line, the stereos are packaged and prepared for shipping. Stage 1 is called “packing” can be described by the Normal model with a mean of 9 minutes and a standard deviation of 1.5 minutes. The second stage is called “boxing” which can also be modeled as Normal with a mean of 6 minutes and standard deviation of 1 minute. a) What is the probability that packing two consecutive systems takes over 20 minutes? b) What percentage of stereo systems take longer to pack than to box? Example: . A company manufacturer small stereo systems. At the end of the production line, the stereos are packaged and prepared for shipping. Stage 1 is called “packing” can be described by the Normal model with a mean of 9 minutes and a standard deviation of 1.5 minutes. The second stage is called “boxing” which can also be modeled as Normal with a mean of 6 minutes and standard deviation of 1 minute. a) What is the probability that packing two consecutive systems takes over 20 minutes? b) What percentage of stereo systems take longer to pack than to box?