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PROBABILITY & STATISTICS FOR P-8 TEACHERS Chapter 5 Probability Distributions PROBABILITY DISTRIBUTIONS We will now combine the methods of descriptive statistics (Chapter 2 and 3) and those of probability (Chapter 4) to describe and analyze probability distributions. Probability Distributions describe what will probably happen instead of what actually did happen, and they are often given in the format of a graph, table, or formula. COMBINING DESCRIPTIVE METHODS AND PROBABILITIES In this chapter we will construct probability distributions by presenting possible outcomes along with the relative frequencies we expect. PROBABILITY DISTRIBUTIONS A random variable is a variable whose values are determined by chance. A discrete probability distribution consists of the values a random variable can assume and the corresponding probabilities of the values. The sum of the probabilities of all events in a sample space add up to 1. Each probability is between 0 and 1, inclusively. RANDOM VARIABLES Two types of random variables: A discrete random variable can assume a countable number of values. Number of steps to the top of the Eiffel Tower A continuous random variable can assume any value along a given interval of a number line. The time a tourist stays at the top once s/he gets there RANDOM VARIABLES Discrete random variables Number of sales Number of calls Shares of stock People in line Mistakes per page Continuous random variables Length Depth Volume Time Weight PROBABILITY DISTRIBUTIONS The probability distribution of a discrete random variable is a graph, table or formula that specifies the probability associated with each possible outcome the random variable can assume. 0 ≤ P(x) ≤ 1 for all values of x P(x) = 1 PROBABILITY DISTRIBUTIONS Is the following a probability distribution? Not a probability distribution x 0 1 P(x) 2 3 4 0.22 0.10 0.30 5 -0.01 0.16 0.18 0.95 P(x) < 0 ∑P(x) ≠ 1 PROBABILITY DISTRIBUTION Construct a probability distribution for tossing a coin twice and recording the number of heads # of Heads Probability x P(x) 0 0.25 1 0.50 2 0.25 P(0 heads) = 1/4 P(1 head) = 2/4 P(2 heads) = 1/4 HH HT TH TT PROBABILITY HISTOGRAM A probability histogram is a histogram in which the horizontal axis corresponds to the value of the random variable and the vertical axis represents the probability of that value of the random variable. PROBABILITY HISTOGRAM Draw a probability histogram of the probability distribution to the right, which represents the number of DVDs a person rents from a video store during a single visit. Probability DVDs Rented at a Video Store 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 Number of DVDs Rented 4 5 x P(x) 0 0.06 1 0.58 2 0.22 3 0.10 4 0.03 5 0.01 PROBABILITY DISTRIBUTION A survey was completed regarding how many siblings are in your family. Let X denote the number of siblings of a randomly selected student. a. Determine the probability distribution of the random variable x. b. Construct a probability histogram for the random variable x. PROBABILITY DISTRIBUTION Solution Probability Distribution Probability Histogram PROBABILITY DISTRIBUTION Just like with any distribution, we would like to analyze the data. The mean and standard deviation are the most common measurements. To find these measurements, we will treat the probability distribution just like a frequency distribution. Since probability distributions represent theoretical data, we will treat the results as a true population. MEAN AND STANDARD DEVIATION Mean: µ = ∑ x P(x) Variance: σ2 = ∑ (x – µ)2 P(x) or σ2 = ∑ x2 P(x) – µ2 MEAN & STANDARD DEVIATION Find the mean and standard deviaton of the number of spots that appear when a die is tossed. Outcome x Probability P(x) x P(x) 1 1/6 1/6 2 1/6 2/6 3 1/6 3/6 4 1/6 4/6 5 1/6 5/6 6 1/6 6/6 21/6 µ = ∑ x P(x) = 21/6 = 3.5 MEAN & STANDARD DEVIATION Find the mean and standard deviaton of the number of spots that appear when a die is tossed. Outcome x Probability P(x) x2 P(x) 1 1/6 1/6 σ2 = ∑ x2 P(x) – µ2 2 1/6 4/6 3 1/6 9/6 4 1/6 16/6 5 1/6 25/6 6 1/6 36/6 σ = √ 105/36 91/6 = 1.708 = 91/6 – (21/6)2 = 105/36 MEAN & STANDARD DEVIATION The probability distribution shown represents the number of trips of five nights or more that American adults take per year. (That is, 6% do not take any trips lasting five nights or more, 70% take one trip lasting five nights or more per year, etc.) Find the mean. MEAN & STANDARD DEVIATION # Trips x Probability P(x) x P(x) x2 P(x) 0 0.06 0.00 0.00 1 0.70 0.70 0.70 2 0.20 0.40 0.80 3 0.03 0.09 0.27 4 0.01 0.04 0.16 1.23 1.93 µ = ∑ x P(x) = 1.23 σ2 = ∑ x2 P(x) – µ2 = 1.93 – (1.23)2 = 0.4171 σ = √0.4171 = 0.6458 EXPECTATION The expected value, or expectation, of a discrete random variable of a probability distribution is the theoretical average of the variable. The expected value is, by definition, the mean of the probability distribution. E(x) = µ = ∑ x P(x) WINNING TICKETS One thousand tickets are sold at $1 each for four prizes of $100, $50, $25, and $10. After each prize drawing, the winning ticket is then returned to the pool of tickets. What is the expected value if you purchase a ticket? Winnings = prize amount – $1 ticket price Winnings x Probability P(x) x P(x) 99 1/1000 99/1000 49 1/1000 49/1000 24 1/1000 24/1000 9 1/1000 9/1000 -1 996/1000 -996/1000 -815/1000 E(x) = ∑ x P(x) = –0.815 On average, you will lose 82 cents for every dollar you spend WINNING TICKETS (ALTERNATE APPROACH) One thousand tickets are sold at $1 each for four prizes of $100, $50, $25, and $10. After each prize drawing, the winning ticket is then returned to the pool of tickets. What is the expected value if you purchase a ticket? Winnings x Probability P(x) x P(x) 100 1/1000 100/1000 50 1/1000 50/1000 25 1/1000 25/1000 10 1/1000 10/1000 0 996/1000 0/1000 185/1000 E(x) = ∑ x P(x) – initial cost = 0.185 – $1 = –0.815 THE BINOMIAL DISTRIBUTION Many types of probability problems have only two possible outcomes or they can be reduced to two outcomes. Examples include: when a coin is tossed it can land on heads or tails, when a baby is born it is either a boy or girl, etc. THE BINOMIAL DISTRIBUTION The binomial experiment is a probability experiment that satisfies these requirements: 1. Each trial can have only two possible outcomes—success or failure. 2. There must be a fixed number of trials. 3. The outcomes of each trial must be independent of each other. 4. The probability of success must remain the same for each trial. NOTATION FOR THE BINOMIAL DISTRIBUTION p The numerical probability of success q The numerical probability of failure n The number of trials x The number of successes Note that x = 0, 1, 2, 3,...,n THE BINOMIAL DISTRIBUTION In a binomial experiment, the probability of exactly X successes in n trials is n! X n X P X p q n - X ! X ! or P X n Cx number of possible desired outcomes p X q n X probability of a desired outcome THE BINOMIAL DISTRIBUTION The Binomial Probability Distribution the probability of exactly x successes in n trials is P(x) = nCx x p n-x q THE BINOMIAL DISTRIBUTION If 40% of the class is female, what is the probability that 1 of the next 2 students walking in will be female? P(x) = nCx px qn-x P(1) = 2C1 p1 q2-1 = (2) (.4)1 (.6)1 = 0.48 n = 2 (# trials) x = 1 (# successes) p = .4 (probability of success) q = .6 (probability of failure) THE BINOMIAL DISTRIBUTION A survey found that one out of five Americans say he or she has visited a doctor in any given month. If 10 people are selected at random, find the probability that exactly 3 will have visited a doctor last month. n = 10 x=3 p = 1/5 q = 4/5 P(x) = nCx px qn-x P(3) = 10C3 p3 q10-3 3 10! 1 P 3 7!3! 5 7 4 0.201 5 THE BINOMIAL DISTRIBUTION Individual baseball cards, chosen at random from a set of 20, are given away inside cereal boxes. Stan needs one more card to complete his set so he buys five boxes of cereal. What is the probability that he will complete his set? P(x) = nCx px qn-x n=5 x=1 p = 1/20 q = 19/20 P(1) = 5C1 p1 q5-1 = (5) (.05)1 (.95)4 = 0.204 The probability of Stan completing his set is 20%. THE BINOMIAL DISTRIBUTION A test consists of 10 multiple choice questions, each with four possible answers. To pass the test, one must answer at least nine questions correctly. Find the probability of passing, if one were to guess the answer for each question. P(x successes) = nCx px qn - x P(x ≥ 9 successes) = P(9 successes) + P(10 successes) n = 10 9 1 x = 9, 10 1 3 P(x ≥ 9) 10 C9 4 4 p = 1/4 q = 3/4 = 0.000028610 + = 0.000029564 10 0 1 3 10 C10 4 4 0.000000954 The probability of passing is 0.003%. THE BINOMIAL DISTRIBUTION A survey from Teenage Research Unlimited (Northbrook, Illinois) found that 30% of teenage consumers receive their spending money from part-time jobs. If 5 teenagers are selected at random, find the probability that at least 3 of them will have part-time jobs. n=5 p = 0.3 q = 0.7 x = 3, 4, 5 5! 3 2 P 3 0.30 0.70 0.132 2!3! 5! 4 1 P 4 0.30 0.70 0.028 1!4! 5! 5 0 P 5 0.30 0.70 0.002 0!5! P(x ≥ 3) = 0.132 + 0.028 + 0.002 = 0.162 THE BINOMIAL DISTRIBUTION A family has nine children. What is the probability that there is at least one girl? This can be best solved using the compliment, that is, the probability of zero girls: n=9 p = 0.5 q = 0.5 x=0 P(x successes) = nCx px qn - x P (0) = 9C0 p0 q9 - 0 = (1) (.5)0 (.5)9 = 0.001953 P(x ≥ 1) = 1 – P(0) = 1 – 0.001953 = 0.998 BINOMIAL DISTRIBUTION Create a probability distribution table for tossing a coin 3 times P(x) = nCx px qn-x # Heads x P(x) 0 0.125 1 0.375 2 0.375 3 0.125 P(0) P(1) P(2) P(3) = 3C0 p0 q3-0 = (1) (.5)0 (.5)3 = (3) (.5)1 (.5)2 = (3) (.5)2 (.5)1 = (1) (.5)3 (.5)0 = 0.125 = 0.375 = 0.375 = 0.125 TOSSING COINS A coin is tossed 3 times. Find the probability of getting exactly two heads, using Table B. n 3, p 12 0.5, X 2 P 2 0.375 BINOMIAL DISTRIBUTION (a) Construct a binomial probability histogram with n = 8 and p = 0.15. (b) Construct a binomial probability histogram with n = 8 and p = 0. 5. (c) Construct a binomial probability histogram with n = 8 and p = 0.85. For each histogram, comment on the shape of the distribution. THE BINOMIAL DISTRIBUTION The mean, variance, and standard deviation of a variable that has the binomial distribution can be found by using the following formulas. Mean: np Variance: npq 2 Standard Deviation: npq THE BINOMIAL DISTRIBUTION The Statistical Bulletin published by Metropolitan Life Insurance Co. reported that 2% of all American births result in twins. If a random sample of 8000 births is taken, find the mean, variance, and standard deviation of the number of births that would result in twins. np 8000 0.02 160 npq 8000 0.02 0.98 156.8 157 2 npq 8000 0.02 0.98 12.5 13 THE BINOMIAL DISTRIBUTION According to the Experian Automotive, 35% of all carowning households have three or more cars. Find the mean and standard deviation for a random sample of 400 homes. µ = np σ = √npq = (400)(0.35) = √(400)(0.35)(0.65) = 140 = 9.54