Chapter 4: Probability s 4. Probability s 4. Binomial s Section 4. Objectives Distinguish between discrete random variables and continuous random variables Construct a discrete probability distribution and its graph Determine if a distribution is a probability distribution Find the mean, variance, and standard deviation of a discrete probability distribution Find the expected value of a discrete probability distribution Random Variables Random Variables Random Variable Represents a numerical value associated with each outcome of a probability distribution. Denoted by x Examples x = Number of sales calls a salesperson makes in one day. x = Hours spent on sales calls in one day. Discrete Random Variable Has a finite or countable number of possible outcomes that can be listed. Example, x = Number of sales calls a salesperson makes in one day. x 0 3 4 5 Continuous Random Variable Has an uncountable number of possible outcomes, represented by an interval on the number line. Example, x = Hours spent on sales calls in one day. x 0 3 4
Example: Random Variables Example: Random Variables Decide whether the random variable x is discrete or continuous.. x = The number of Fortune 500 companies that lost money in the previous year. Discrete random variable (The number of companies that lost money in the previous year can be counted.) {0,,, 3,, 500} Decide whether the random variable x is discrete or continuous.. x = The volume of gasoline in a -gallon tank. Continuous random variable (The amount of gasoline in the tank can be any volume between 0 gallons and gallons.) Example: Random Variables Example: Random Variables Decide whether the random variable x is discrete or continuous. 3. x = The distance your car travels on a tank of gas Solutions The distance your car travels at is a continuous random variable because it is a measurement that cannot be counted. Decide whether the random variable x is discrete or continuous. 4. x = The number of students in a statistics class. The number of students is a discrete random variable because it can be counted.
More Examples of Random Variables Discrete Random Variable:. The number of people in the waiting line in a bank. More Examples of Random Variables Continuous Random Variable:. The height and weight of a person.. The number of kids lining up for the ice cream.. The amount of sugar in an apple. 3. The number of homes in a area. 3. The length of a light bulb. 4. The number of defective light bulbs. 4. The length of time of a phone call. 5. The number of phone calls. 5. The length of waiting time in a line at a super market. Discrete Probability s Discrete probability distribution Lists each possible value the random variable can assume, together with its probability. Must satisfy the following conditions: In Words. The probability of each value of the discrete random variable is between 0 and, inclusive.. The sum of all the probabilities is. In Symbols 0 P (x) ΣP (x) = Constructing a Discrete Probability Let x be a discrete random variable with possible outcomes x, x,, x n.. Make a frequency distribution for the possible outcomes.. Find the sum of the frequencies. 3. Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies. 4. Check that each probability is between 0 and, inclusive, and that the sum of all probabilities is. 3
Example: Constructing a Discrete Probability An industrial psychologist administered a personality inventory test for passive-aggressive traits to 50 employees. Individuals were given a score from to 5, where was extremely passive and 5 extremely aggressive. A score of 3 indicated neither trait. Construct a probability distribution for the random variable x. Then graph the distribution using a histogram. Score, x Frequency, f 4 33 3 4 4 30 5 Constructing a Discrete Probability Divide the frequency of each score by the total number of individuals in the study to find the probability for each value of the random variable. 4 P () = = 0.6 50 33 P () = = 0. 50 30 P (4) = = 0.0 P (5) = = 0.4 50 50 Discrete probability distribution: 4 P (3) = = 0.8 50 x 3 4 5 P(x) 0.6 0. 0.8 0.0 0.4 Constructing a Discrete Probability x 3 4 5 P(x 0.6 0. 0.8 0.0 0.4 ) This is a valid discrete probability distribution since. Each probability is between 0 and, inclusive, 0 P(x).. The sum of the probabilities equals, ΣP(x) = 0.6 + 0. + 0.8 + 0.0 + 0.4 =. Constructing a Discrete Probability Histogram Probability, P(x) 0.3 0.5 0. 0.5 0. 0.05 0 Passive-Aggressive Traits 3 4 5 Score, x Because the width of each bar is one, the area of each bar is equal to the probability of a particular outcome. 4
Example: Constructing a Discrete Probability Example: The spinner below is divided into two sections. The probability of landing on the is 0.5. The probability of landing on the is 0.75. Let x be the number the spinner lands on. Construct a probability distribution for the random variable x. x P (x) 0.5 0.75 Each probability is between 0 and. The sum of the probabilities is. Constructing a Discrete Probability Example: The spinner below is spun two times. The probability of landing on the is 0.5. The probability of landing on the is 0.75. Let x be the sum of the two spins. Construct a probability distribution for the random variable x. The possible sums are, 3, and 4. P (sum of ) = 0.5 0.5 = 0.065 Spin a on the first spin. and Spin a on the second spin. Continued. Constructing a Discrete Probability Constructing a Discrete Probability Example continued: Example continued: P (sum of 3) = 0.5 0.75 = 0.875 Spin a on the first spin. and or Spin a on the second spin. P (sum of 4) = 0.75 0.75 = 0.565 Spin a on the first spin. and Spin a on the second spin. P (sum of 3) = 0.75 0.5 = 0.875 Sum of P (x) spins, x 0.065 3 0.375 4 Spin a on the first spin. 0.875 + 0.875 and Spin a on the second spin. Continued. Sum of P (x) spins, x 0.065 3 0.375 4 0.565 Each probability is between 0 and, and the sum of the probabilities is. Continued. 5
Graphing a Discrete Probability Example: Graph the following probability distribution using a histogram. P(x) Sum of P (x) spins, x 0.065 3 0.375 4 0.565 Probability 0.6 0.5 0.4 0.3 0. 0. 0 Sum of Two Spins 3 4 Sum x More Examples Example. Find the probability distribution of number of Heads (random variable: X) when you flip three different coins. Solution Random Var. X: number of heads, Sample Space S: {TTT, TTH, THT, HTT, HHT, HTH, THH, HHH} Possible values of X: 0,,, and 3 Probability Dist. of X in Table form: X 0 3 P(x) /8 3/8 3/8 /8 Note:. each P(x) is between 0 and. p(x) = A Probability Histogram and a Line graph can also be made to show the probability distribution graphically. More Examples More Examples Example. Check whether the following function can be a probability distribution function of the random variable x. f(x) = (x + 3) / 5 for x =,, and 3 = 0 otherwise f(x) = (x + 3) / 5 for x =,, and 3 f() = ( + 3)/5 = 4/5 f() = 5/5 and f(3) = 6/5 Note:. each f(x) is between 0 and. and. f(x) = Therefore, f(x) can represent a p.d.f. Example 3. Some one claims that the following is the probability distribution function (p.d.f.) for a random variable Y. Give two reasons why it is not a valid p.d.f. of the random variable Y. Y 0 3 P(Y) /0 /0 5/0 3/0. p(0) = /0 violates 0 P(x). p(y) = 7/0 6
Mean Example: Finding the Mean Mean of a discrete probability distribution μ = ΣxP(x) The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean score. Each value of x is multiplied by its corresponding probability and the products are added. x P(x) xp(x) 0.6 (0.6) = 0.6 0. (0.) = 0.44 3 0.8 3(0.8) = 0.84 4 0.0 4(0.0) = 0.80 5 0.4 5(0.4) = 0.70 μ = ΣxP(x) =.94 Variance and Standard Deviation Variance of a discrete probability distribution σ = Σ(x μ) P(x) Standard deviation of a discrete probability distribution σ = σ = Σ( x μ) Px ( ) Example: Finding the Variance and Standard Deviation The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the variance and standard deviation. ( μ =.94) x P(x) 0.6 0. 3 0.8 4 0.0 5 0.4 7
Finding the Variance and Standard Deviation Recall μ =.94 x P(x) x μ (x μ) (x μ) P(x) 0.6.94 =.94 (.94) 3.764 3.764(0.6) 0.60 0..94 = 0.94 ( 0.94) 0.884 0.884(0.) 0.94 3 0.8 3.94 = 0.06 (0.06) 0.004 0.004(0.8) 0.00 4 0.0 4.94 =.06 (.06).4.4(0.0) 0.5 5 0.4 5.94 =.06 (.06) 4.44 4.44(0.4) 0.594 Variance: σ = Σ(x μ) P(x) =.66 Expected Value Expected value of a discrete random variable Equal to the mean of the random variable. E(x) = μ = ΣxP(x) Standard Deviation: σ = σ =.66.3 Example: Finding an Expected Value Finding an Expected Value At a raffle, 500 tickets are sold at $ each for four prizes of $500, $50, $50, and $75. You buy one ticket. What is the expected value of your gain? To find the gain for each prize, subtract the price of the ticket from the prize: Your gain for the $500 prize is $500 $ = $498 Your gain for the $50 prize is $50 $ = $48 Your gain for the $50 prize is $50 $ = $48 Your gain for the $75 prize is $75 $ = $73 If you do not win a prize, your gain is $0 $ = $ 8
Finding an Expected Value Probability distribution for the possible gains (outcomes) Gain, x $498 $48 $48 $73 $ P(x) 500 500 500 500 496 500 Ex ( ) =ΣxPx ( ) 496 = $498 + $48 + $48 + $73 + ( $) 500 500 500 500 500 = $.35 You can expect to lose an average of $.35 for each ticket you buy. Section 4. Summary Distinguished between discrete random variables and continuous random variables Constructed a discrete probability distribution and its graph Determined if a distribution is a probability distribution Found the mean, variance, and standard deviation of a discrete probability distribution Found the expected value of a discrete probability distribution Section 4. Objectives Binomial Experiments Determine if a probability experiment is a binomial experiment Find binomial probabilities using the binomial probability formula Graph a binomial distribution Find the mean, variance, and standard deviation of a binomial probability distribution. The experiment is repeated for a fixed number of trials, where each trial is independent of other trials.. There are only two possible outcomes of interest for each trial. The outcomes can be classified as a success (S) or as a failure (F). 3. The probability of a success P(S) is the same for each trial. 4. The random variable x counts the number of successful trials. 9
Notation for Binomial Experiments Example: Binomial Experiments Symbol n p = P(S) q = P(F) x Description The number of times a trial is repeated The probability of success in a single trial The probability of failure in a single trial (q = p) The random variable represents a count of the number of successes in n trials: x = 0,,, 3,, n. Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p, and q, and list the possible values of the random variable x.. A certain surgical procedure has an 85% chance of success. A doctor performs the procedure on eight patients. The random variable represents the number of successful surgeries. Binomial Experiments Binomial Experiments Binomial Experiment. Each surgery represents a trial. There are eight surgeries, and each one is independent of the others.. There are only two possible outcomes of interest for each surgery: a success (S) or a failure (F). 3. The probability of a success, P(S), is 0.85 for each surgery. 4. The random variable x counts the number of successful surgeries. Binomial Experiment n = 8 (number of trials) p = 0.85 (probability of success) q = p = 0.85 = 0.5 (probability of failure) x = 0,,, 3, 4, 5, 6, 7, 8 (number of successful surgeries) 0
Example: Binomial Experiments Binomial Experiments Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p, and q, and list the possible values of the random variable x.. A jar contains five red marbles, nine blue marbles, and six green marbles. You randomly select three marbles from the jar, without replacement. The random variable represents the number of red marbles. Not a Binomial Experiment The probability of selecting a red marble on the first trial is 5/0. Because the marble is not replaced, the probability of success (red) for subsequent trials is no longer 5/0. The trials are not independent and the probability of a success is not the same for each trial. Binomial Probability Formula Example: Finding Binomial Probabilities Binomial Probability Formula The probability of exactly x successes in n trials is n! P( x) = ncxp q = p q ( n x)! x! n = number of trials p = probability of success q = p probability of failure x = number of successes in n trials x n x x n x Microfracture knee surgery has a 75% chance of success on patients with degenerative knees. The surgery is performed on three patients. Find the probability of the surgery being successful on exactly two patients.
Finding Binomial Probabilities Method : Draw a tree diagram and use the Multiplication Rule P( successful surgeries) = 3 9 64 0.4 Finding Binomial Probabilities Method : Binomial Probability Formula 3 n = 3, p =, q = p =, x = 4 4 3 P( successful surgeries) = 3 C 4 4 3! 3 = (3 )!! 4 3 4 = 3 9 6 4 = 7 64 0.4 Binomial Probability Binomial Probability List the possible values of x with the corresponding probability of each. Example: Binomial probability distribution for Microfracture knee surgery: n = 3, p = x 0 3 P(x) 0.06 0.4 0.4 0.4 Use the binomial probability formula to find probabilities. 3 4 Example: Constructing a Binomial In a survey, U.S. adults were asked to give reasons why they liked texting on their cellular phones. Seven adults who participated in the survey are randomly selected and asked whether they like texting because it is quicker than calling. Create a binomial probability distribution for the number of adults who respond yes.
Constructing a Binomial Constructing a Binomial 56% of adults like texting because it is quicker than calling. n = 7, p = 0.56, q = 0.44, x = 0,,, 3, 4, 5, 6, 7 P(0) = 7 C 0 (0.56) 0 (0.44) 7 = (0.56) 0 (0.44) 7 0.003 P() = 7 C (0.56) (0.44) 6 = 7(0.56) (0.44) 6 0.084 P() = 7 C (0.56) (0.44) 5 = (0.56) (0.44) 5 0.086 P(3) = 7 C 3 (0.56) 3 (0.44) 4 = 35(0.56) 3 (0.44) 4 0.304 P(4) = 7 C 4 (0.56) 4 (0.44) 3 = 35(0.56) 4 (0.44) 3 0.93 P(5) = 7 C 5 (0.56) 5 (0.44) = (0.56) 5 (0.44) 0.39 P(6) = 7 C 6 (0.56) 6 (0.44) = 7(0.56) 6 (0.44) 0.0950 P(7) = 7 C 7 (0.56) 7 (0.44) 0 = (0.56) 7 (0.44) 0 0.073 x P(x) 0 0.003 0.084 0.086 3 0.304 4 0.93 5 0.39 6 0.0950 7 0.073 All of the probabilities are between 0 and and the sum of the probabilities is. Example: Finding Binomial Probabilities Finding Binomial Probabilities A survey indicates that 4% of women in the U.S. consider reading their favorite leisure-time activity. You randomly select four U.S. women and ask them if reading is their favorite leisure-time activity. Find the probability that at least two of them respond yes. n = 4, p = 0.4, q = 0.59 At least two means two or more. Find the sum of P(), P(3), and P(4). P() = 4 C (0.4) (0.59) = 6(0.4) (0.59) 0.35094 P(3) = 4 C 3 (0.4) 3 (0.59) = 4(0.4) 3 (0.59) 0.6654 P(4) = 4 C 4 (0.4) 4 (0.59) 0 = (0.4) 4 (0.59) 0 0.0858 P(x ) = P() + P(3) + P(4) 0.35094 + 0.6654 + 0.0858 0.54 3
Example: Finding Binomial Probabilities Using a Table About ten percent of workers (6 years and over) in the United States commute to their jobs by carpooling. You randomly select eight workers. What is the probability that exactly four of them carpool to work? Use a table to find the probability. (Source: American Community Survey) Finding Binomial Probabilities Using a Table A portion of Table is shown Binomial with n = 8, p = 0.0, x = 4 The probability that exactly four of the eight workers carpool to work is 0.005. Example: Graphing a Binomial Graphing a Binomial Sixty percent of households in the U.S. own a video game console. You randomly select six households and ask each if they own a video game console. Construct a probability distribution for the random variable x. Then graph the distribution. (Source: Deloitte, LLP) x 0 3 4 5 6 P(x) 0.004 0.037 0.38 0.76 0.3 0.87 0.047 Histogram: n = 6, p = 0.6, q = 0.4 Find the probability for each value of x 4
Mean, Variance, and Standard Deviation Example: Finding the Mean, Variance, and Standard Deviation Mean: μ = np Variance: σ = npq Standard Deviation: σ = npq In Pittsburgh, Pennsylvania, about 56% of the days in a year are cloudy. Find the mean, variance, and standard deviation for the number of cloudy days during the month of June. Interpret the results and determine any unusual values. (Source: National Climatic Data Center) n = 30, p = 0.56, q = 0.44 Mean: μ = np = 30 0.56 = 6.8 Variance: σ = npq = 30 0.56 0.44 7.4 Standard Deviation: σ = npq = 30 0.56 0.44.7 Finding the Mean, Variance, and Standard Deviation μ = 6.8 σ 7.4 σ.7 On average, there are 6.8 cloudy days during the month of June. The standard deviation is about.7 days. Values that are more than two standard deviations from the mean are considered unusual. 6.8 (.7) =.4, a June with cloudy days or fewer would be unusual. 6.8 + (.7) =., a June with 3 cloudy days or more would also be unusual. Section 4. Summary Determined if a probability experiment is a binomial experiment Found binomial probabilities using the binomial probability formula Found binomial probabilities using a binomial table Graphed a binomial distribution Found the mean, variance, and standard deviation of a binomial probability distribution 5