Section 5.2 Random Variables PDF

Section 5.2 Random Variables  Key Concept This section introduces the important concept of a probability distribution, which gives the probability for each value of a variable that is determined by chance. Give consideration to distinguishing between outcomes that are likely to occur...

Section 5.2 Random Variables  Key Concept This section introduces the important concept of a probability distribution, which gives the probability for each value of a variable that is determined by chance. Give consideration to distinguishing between outcomes that are likely to occur by chance and outcomes that are “unusual” in the sense they are not likely to occur by chance.  Key Concept The concept of random variables and how they relate to probability distributions Distinguish between discrete random variables and continuous random variables Develop formulas for finding the mean, variance, and standard deviation for a probability distribution Determine whether outcomes are likely to occur by chance or they are unusual (in the sense that they are not likely to occur by chance)  Why Is This Important??? There is a 0.9986 probability that a randomly selected 30 year-old male lives through the year (based on data from the U.S Department of Health and Human Services). A Fidelity life insurance company charges $161 for insuring that the male will live through the year. If the male does not survive the year, the policy pays out $100,000 as a death benefit. Is life insurance worth the cost?  Let’s Get Started … The Basics A random variable is a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure. A probability distribution is a description that gives the probability for each value of the random variable. It is often expressed in the format of a graph, table, or formula.  For Example … The Secret Garden A Probability Distribution Mr. Llorens has a strange x fascination with pea pods. (Number of Every summer, Mr. Llorens peas with P(x) grows peas in his garden and green pods) keeps track of how many 0 0.01 green pea pods and yellow pea pods each plant produces. 1 0.015 Here is a sample of some of 2 0.088 the results in his creeper 3 0.264 journal. 4 0.396 5 0.228 Characteristics of a Probability Distribution 1. The sum of all probabilities must be 1, but values such as 0.999 or 1.001 are acceptable because they may result from rounding errors. 2. Each probability value must be between 0 and 1 inclusive. CHECK YOURSELF!!! x P(x) 0 0.250 1 0.382 2 0.119 3 0.481 4 0.260 5 0.127 No ! Σ P(x) = 1.619 > 1 Probability Distribution? x P(x) 0 0.250 1 0.382 2 0.119 3 0.481 4 0.487 5 -0.010  Graphs The probability histogram is very similar to a relative frequency histogram, but the vertical scale shows probabilities.  CAUTION!!! Section 5.2 only deals with discrete random variables.  REFRESH YOURSELF!!! A discrete random variable has either a finite number of values or a countable number of values, where “countable” refers to the fact that there might be infinitely many values, but they can be associated with a counting process so that the number of values is 0, 1 or 2 or 3, etc. A continuous random variable has infinitely many values, and those values can be associated with measurements on a continuous scale without gaps or interruptions. Examples??? Temperature of the day  Lightning Practice Round The number of tennis balls Brendan misses every time he goes to the tennis courts after school. Discrete Random Variable The number of awesome hairstyles Tori can rock. Discrete Random Variable The amount of coffee (in ounces) Sarah can drink every morning. Continuous Random Variable  Lightning Practice Round The number of times Ryan says his name is Chris, not Ryan. Discrete Random Variable The amount of time Casey spends googling T. Swizzle’s supposed relationship with One Direction band members. Continuous Random Variable The amount of money Olivia spends on her badminton swagg. Discrete Random Variable What is Expected Value? We can think of the mean as the expected value in the sense that it is the average value that we would expect to get if the trials could continue indefinitely. How do I calculate Expected Value? Example 1 You are playing a game that requires you to flip a coin and roll a die. If you flip a heads and roll an even, you earn 2 points, if you flip a tails and roll an odd you earn 1 point, any other outcome will lose 1 point. What is the expected value of a single turn? Outcome Points P(X) E(X) = 2(0.25) + 1(0.25) + (-1)(0.5) (x) = 0.5 + 0.25 - 0.5 Heads, Even 2 (.5)(.5) =.25 = 0.25 Tails, Odd 1 (.5)(.5) =.25 Other -1 1 – (.25 +.25) =.5 Interpreting the Meaning of E(X) What does a positive E(X) mean? You are earning points on average. If you do enough trials, you can expect to earn E(x). What does a negative E(X) mean? You are losing points on average. If you do enough trials, you can expect to lose E(x). What does a zero E(X) mean? You are neither earning or losing points on average. If you do enough trials, you can expect to break even.  Example 2 The Winning Wheel has 7 slots that are all equally likely to occur. The blue slot wins $20, the orange slot wins $10, the red slot wins $5, the two yellow slots win $1, and the remaining purple slots lose $10. Is the game worth playing? Outcome Net $ (x) P(X) Blue +20 1/7 E(X) = 20(1/7)+10(1/7)+5(1/7)+1(2/7)+(-10)(2/7) Orange +10 1/7 = 17/7 Red +5 1/7 = 2.43 Yellow +1 2/7 Purple -10 2/7 Example 3 A new gambling game called crazy 4 has come out in Las Vegas. The player must bet $5, if the dealers rolls a 4, the player is given $15, otherwise they get nothing back. Who has the advantage in this game? Outcome Net $ (x) P(X) Roll a 4 (Win) +10 1/6 Roll a 1, 2, 3, 5, or 6 (Lose) -5 5/6 E(X) = 10(1/6)+(-5)(5/6) = -15/6 = -2.5 Example 4: There is a 0.9986 probability that a randomly selected 30 year-old male lives through the year (based on data from the U.S Department of Health and Human Services). A Fidelity life insurance company charges $161 for insuring that the male will live through the year. If the male does not survive the year, the policy pays out $100,000 as a death benefit. Is life insurance worth the cost? Outcome Net $ (x) P(X) He lives -161 0.9986 He dies +99839 0.0014 E(X) = -161(0.9986) + 99839(0.0014) = -21 The New Found about the Mean x 1 2 4 5 7 8 9 P(x) 6/20 3/20 2/20 4/20 1/20 3/20 1/20  “Freaking Finally, Ms. P!!!” – Direct Quote by You Now that you have the background info … on to the MAIN EVENT Formula 5-1 Mean for a probability distribution Formula 5-2 Variance for a probability distribution (easier to understand) Formula 5-3 Variance for a probability distribution (easier to compute) Formula 5-4 Standard deviation for a probability distribution Roundoff Rule for µ, σ, and σ2 Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of x are integers, round µ, σ, and σ 2 to two decimal place.  How to use the Calculator TI-83 / TI-84 Computing Mean and Standard Deviation Enter the values of the random variable x in the list L1. Enter the corresponding probabilities in the list L2. Press STAT, select CALC option and choose 1-Var Stats For TI-83, “1-Var Stats L1,L2" and Press the ENTER key. For TI-84, “List: L1” and “FreqList: L2” and select “Calculate”  So, How Do We Use These? Remember Mr. Llorens and his creepy plant obsession? Let’s Calculate! x Find the mean, variance, (Number of and standard deviation for peas with P(x) the probability green pods) distribution. 0 0.01 What does this mean if 1 0.015 these results are the 2 0.088 number of green pods out 3 0.264 of 5 possible pods? 4 0.396 5 0.228  Solution for µ, σ 2, and σ 1. Save values of variable x to L1 2. Save values of P(x) to L2 3. Calculate µ = ΣxP(x) = sum(L1*L2) = 3.707 🡪 A 4. Calculate σ 2 = Σx2P(x) – µ2 = sum((L1) 2*L2) – A2 = 1.0372 5. Calculate σ = 1.0184 6. The value µ = 3.707 is the mean number of green pods out of 5 pods. HOMEWORK Pg. 214-215 #5 -10 Section 5.2 Random Variables Homework Answers 5. a) discrete 7. yes; μ = 1.5; σ =.9 b) Continuous c) Continuous 8. No d) Discrete e) continuous 9. no 6. a) continuous b) Discrete 10. yes; μ = 0.6; σ = 0.7 c) Discrete d) Continuous e) discrete  Building Onto Previous Knowledge Important Definition The expected value of a discrete random variable is denoted by E, and it represents the mean value of the outcomes. It is obtained by finding the value of Where have we seen this before?  A Little (LOT) Clarification We can think of the mean as the expected value in the sense that it is the average value that we would expect to get if the trials could continue indefinitely. For example, an expected value of 3.2 girls is not meant to be interpreted as the number of girls in one trial will be equal to 3.2, but rather it means that among many such trials, the mean number of girls is 3.2  How to use the Calculator TI-83 / TI-84 Computing Mean and Standard Deviation Enter the values of the random variable x in the list L1. Enter the corresponding probabilities in the list L2. Press STAT, select CALC option and choose 1-Var Stats Enter "L1,L2" and Press the ENTER key.  Identifying Unusual Results Range Rule of Thumb According to the range rule of thumb, most values should lie within 2 standard deviations of the mean. We can therefore identify “unusual” values by determining if they lie outside these limits: Maximum usual value = μ + 2σ Minimum usual value = μ – 2σ  Identifying Unusual Results Probabilities Rare Event Rule for Inferential Statistics If, under a given assumption (such as the assumption that a coin is fair), the probability of a particular observed event (such as 992 heads in 1000 tosses of a coin) is extremely small, we conclude that the assumption is probably not correct.  Identifying Unusual Results Probabilities Using Probabilities to Determine When Results Are Unusual ❖ Unusually high: x successes among n trials is an unusually high number of successes if P(x or more) ≤ 0.05. ❖ Unusually low: x successes among n trials is an unusually low number of successes if P(x or fewer) ≤ 0.05.  The Exciting Stuff So how do we know what to expect? Rare Event Rule x successes among n trials is an unusually high result when P(x or more) ≤ 0.05 x successes among n trials is an unusually low results when P(x or fewer) ≤ 0.05 Now Apply It! Based on data from CarMax. Com when a car is randomly selected the number of bumper stickers and the corresponding probabilities are 0 (0.823), 1(0.092), 2 (0.039), 3 (0.014), 4 (0.012), 5 (0.006), 6 (0.006), 7 (0.004), 8 (0.002), and 9 (0.002). a) Does the given information describe a probability distribution? 0.823 + 0.092 + 0.039 +0.014 +0.012 + 0.006 * 2+ 0.004 + 0.002 * 2 = 1.000 Yes. b) Is it unusual for a car to have more than one bumper sticker? P(X > 1) = P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) = 0.039 + 0.014 + 0.012 + 0.006 *2 + 0.004 + 0.002*2 = 0.085 > 0.05 OR P(X >1) = 1 - P(0) - P(1) = 1 – 0.823 – 0.092 = 0.085 No, it is usual. Now Apply It! In the Illinois Pick 3 lottery game, you pay $0.50 to select a sequence of three digits, such as 233. If you select the same sequence of three digits that are drawn, you win and collect $250. a) How many different selections are possible? 10 x 10 x 10 = 1000 b) What is the probability of winning? 1/1000 c) Find your expected value of winning. x P(x) -0.5 999/1000 249.5 1/1000 - 0.5 * 999/1000 + 249.5 * 1/1000 = - 0.25  Now Apply It! There is a 0.9968 probability that a randomly selected 50-year-old female lives through the year (based on data from the U.S Department of Health and Human Services). A Fidelity life insurance company charges $226 for insuring that the female will live through the year. If she does not survive the year, the policy pays out $50,000 as a death benefit. a) If she purchases this policy, what is her expected value? x P(x) -226 * 0.9968 + 49774 * 0.0032 = -66 -226 0.9968 49774 0.0032 b) Can the insurance company expect to make a profit from many such policies? Yes, averagely speaking, insurance company would make $66 for each one who bought life insurance.  Recap In this section we have discussed: ❖ Combining methods of descriptive statistics with probability. ❖ Random variables and probability distributions. ❖ Probability histograms. ❖ Requirements for a probability distribution. ❖ Mean, variance and standard deviation of a probability distribution. ❖ Identifying unusual results. ❖ Expected value.  HOMEWORK Pg. 215-217 #16, 17, 18, 28, 29