Outline Lecture 9 Probability Distributions 6-1 Introduction 6- Probability Distributions 6-3 Mean, Variance, and Expectation 6-4 The Binomial Distribution Outline 7- Properties of the Normal Distribution 7-3 The Standard Normal Distribution 7-4 Applications of the Normal Distribution 7-5 The Central Limit Theorem 6- Probability Distributions A variable is defined as a characteristic or attribute that can assume different values. A variable whose values are determined by chance is called a random variable. 6- Probability Distributions If a variable can assume only a specific number of values, such as the outcomes for the roll of a die or the outcomes for the toss of a coin, then the variable is called a discrete variable. Discrete variables have values that can be counted. 6- Probability Distributions If a variable can assume all values in the interval between two given values then the variable is called a continuous variable. - temperature between 68 0 to 78 0. Continuous random variables are obtained from data that can be measured rather than counted.
6- Probability Distributions - Tossing Two Coins H 6- Probability Distributions - Tossing Two Coins H T First Toss T Second Toss H T From the tree diagram, the sample space will be represented by HH, HT, TH, TT. If X is the random variable for the number of heads, then X assumes the value 0, 1, or. 6- Probability Distributions - Tossing Two Coins Sample Space TT TH HT HH Number of Heads 0 1 6- Probability Distributions - Tossing Two Coins OUTCOME PROBABILITY X P(X) 0 1/4 1 /4 1/4 6- Probability Distributions 6- Probability Distributions -- Graphical Representation A probability distribution consists of the values a random variable can assume and the corresponding probabilities of the values. The probabilities are determined theoretically or by observation. PROBABILITY 1 0.5.5 Experiment: Toss Two Coins 0 1 3 NUMBEROF HEADS
Two requirements The sum of the probabilities of all the events in the sample space must equal 1 The probability of each event in the sample space must be between 0 and 1. 6-33 Mean, Variance, and Expectation for Discrete Variable The mean of the random variableof a probability distribution is µ = X 1 P( X1) + X P( X ) +... + X n P( X n) = X P( X ) where X, X,..., X are the outcomes and 1 P( X ), P( X ),..., P( X ) are the corresponding 1 probabilities. n n 6-33 Mean for Discrete Variable - Find the mean of the number of spots that appear when a die is tossed. The probability distribution is given below. X 1 3 4 5 6 P(X) 1/6 1/6 1/6 1/6 1/6 1/6 6-33 Mean for Discrete Variable - µ = X P( X ) = 1 ( 1/ 6) + ( 1/ 6) + 3 ( 1/ 6) + 4 ( 1/ 6) + 5 ( 1/ 6) + 6 ( 1/ 6) = 1/ 6 = 35. That is, when a die is tossed many times, the theoretical mean will be 3.5. 6-33 Mean for Discrete Variable - In a family with two children, find the mean number of children who will be girls. The probability distribution is given below. X 0 1 P(X) 1/4 1/ 1/4 6-3 Mean for Discrete Variable 3 Mean for Discrete Variable - µ = X P( X) = 0 ( 1/ 4) + 1 ( 1/ ) + ( 1/ 4) = 1. That is, the average number of girls in a two-child family is 1.
Variance of a Probability Distribution The mean describes the measure of the long-run or theoretical average, but it does not tell anything about the spread of the distribution. 6-3 Formula for the Variance of a Probability Distribution The variance of a probability distribution is found by multiplying the square of each outcome by its corresponding probability, summing these products, and subtracting the square of the mean. 6-33 Formula for the Variance of a Probability Distribution The formula for the variance of a probability distribution is σ = [ X P X ] σ = ( ) µ. The standard deviation of a probability distribution is σ. 6-33 Variance of a Probability The probability that 0, 1,, 3, or 4 people will be placed on hold when they call a radio talk show with four phone lines is shown in the distribution below. Find the variance and standard deviation for the data. 6-33 Variance of a Probability 6-33 Variance of a Probability X 0 1 3 4 P (X ) 0.1 8 0.3 4 0. 3 0. 1 0.0 4 X P(X) X P(X) X P(X) 0 0.18 0 0 1 0.34 0.34 0.34 0.3 0.46 0.9 3 0.1 0.63 1.89 4 0.04 0.16 0.64 µ = 1.59 ΣX P(X) =3.79 σ = 3.79 1.59 = 1.6
6-33 Variance of a Probability Now, µ = (0)(0.18) + (1)(0.34) + ()(0.3) + (3)(0.1) + (4)(0.04) = 1.59. Σ X P(X) = (0 )(0.18) + (1 )(0.34) + ( )(0.3) + (3 )(0.1) + (4 )(0.04) = 3.79 1.59 =.53 (rounded to two decimal places). σ = 3.79.53 = 1.6 σ = 1.6 = 1.1 6-33 Expectation The expected value of a discrete random variable of a probability distribution is the theoretical average of the variable. The formula is µ = E( X ) = X P( X ) The symbol E( X ) is used for the expected value. 6-33 Expectation - A ski resort loses $70,000 per season when it does not snow very much and makes $50,000 when it snows a lot. The probability of it snowing at least 75 inches (i.e., a good season) is 40%. Find the expected profit. 6-33 Expectation - Profit, X 50,000 70,000 P(X) 0.40 0.60 The expected profit = ($50,000)(0.40) + ( $70,000)(0.60) = $58,000. 6-44 The Binomial Distribution 6-44 The Binomial Distribution A binomial experiment is a probability experiment that satisfies the following four requirements: Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. Each outcome can be considered as either a success or a failure. There must be a fixed number of trials. The outcomes of each trial must be independent of each other. The probability of success must remain the same for each trial.
6-44 The Binomial Distribution 6-44 The Binomial Distribution The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called a binomial distribution. Notation for the Binomial Distribution: P(S) ) = p, probability of a success P(F) ) = 1 p = q, probability of a failure n = number of trials X = number of successes. 6-44 Binomial Probability Formula In a binomial experiment, the probability of exactly X successes in n trials is n! P( X ) = ( n X )! X! p q X n X 6-44 Binomial Probability - If a student randomly guesses at five multiple-choice questions, find the probability that the student gets exactly three correct. Each question has five possible choices. Solution: n = 5, X = 3, and p = 1/5. Then, P(3) = [5!/((5 3)!3! )](1/5) 3 (4/5) 0.05. 6-44 Binomial Probability - A survey from Teenage Research Unlimited (Northbrook, Illinois.) found that 30% of teenage consumers received their spending money from part-time jobs. If five teenagers are selected at random, find the probability that at least three of them will have part-time jobs. 6-44 Binomial Probability - Solution: n = 5, X = 3, 4, and 5, and p = 0.3. Then, P(X 3) = P(3) + P(4) + P(5) = 0.133 + 0.084 + 0.004 = 0.1631. NOTE: You can use Table B in the textbook to find the Binomial probabilities as well.
6-44 Binomial Probability - A report from the Secretary of Health and Human Services stated that 70% of singlevehicle traffic fatalities that occur on weekend nights involve an intoxicated driver. If a sample of 15 single-vehicle traffic fatalities that occurred on a weekend night is selected, find the probability that exactly 1 involve a driver who is intoxicated. 6-44 Binomial Probability - Solution: n = 15, X = 1, and p = 0.7. From Table B, P(X =1) = 0.170 6-44 Mean, Variance, Standard Deviation for the Binomial 7- The Normal Distribution A coin is tossed four times. Find the mean, variance, and standard deviation of the number of heads that will be obtained. Solution: n = 4, p = 1/, and q = 1/. µ = n p = (4)(1/) =. σ = n p q = (4)(1/)(1/) = 1. σ = 1 = 1. Many continuous variables have distributions that are bell-shaped and are called approximately normally distributed variables. The theoretical curve, called the normal distribution curve, can be used to study many variables that are not normally distributed but are approximately normal. 7- Mathematical Equation for the Normal Distribution The mathematical equation for the normal distribution: ( x µ ) σ e y = σ π where e. 718 π 314. µ = population mean σ = population standard deviation 7- Properties of the Normal Distribution The shape and position of the normal distribution curve depend on two parameters, the mean and the standard deviation. Each normally distributed variable has its own normal distribution curve, which depends on the values of the variable s mean and standard deviation.
7- Properties of the Theoretical Normal Distribution The normal distribution curve is bell-shaped. The mean, median, and mode are equal and located at the center of the distribution. The normal distribution curve is unimodal (single mode). 7- Properties of the Theoretical Normal Distribution The curve is symmetrical about the mean. The curve is continuous. The curve never touches the x-axis. The total area under the normal distribution curve is equal to 1. 7- Properties of the Theoretical Normal Distribution 7- Areas Under the Normal Curve The area under the normal curve that lies within one standard deviation of the mean is approximately 0.68 (68%). two standard deviations of the mean is approximately 0.95 (95%). three standard deviations of the mean is approximately 0.997 (99.7%). µ 3σ 68% 95% 99.7% µ σ µ 1σ µ µ +1σ µ +σ µ +3σ 7-33 The Standard Normal Distribution The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. All normally distributed variables can be transformed into the standard normally distributed variable by using the formula for the standard score: (see next slide) 7-33 The Standard Normal Distribution value mean z = standard deviation or X µ z = σ
Normal Curve - Find the area under the standard normal curve between z = 0 and z =.34 P(0 z.34). Use your table at the end of the text to find the area. The next slide shows the shaded area. Normal Curve - 7-3 Area Under the Standard Normal Curve - 0.4904 0.34 Find the area under the standard normal curve between z = 0 and z = 1.75 P( 1.75 z 0). Use the symmetric property of the normal distribution and your table at the end of the text to find the area. The next slide shows the shaded area. Normal Curve - Normal Curve - 0.4599 0.4599 1.75 0 1.75 Find the area to the right of z = 1.11 P(z > 1.11). Use your table at the end of the text to find the area. The next slide shows the shaded area.
Normal Curve - 0.3665 0.1335 0.3665 0 1.11 Normal Curve - Find the area to the left of z = 1.93 P(z < 1.93) 1.93). Use the symmetric property of the normal distribution and your table at the end of the text to find the area. The next slide shows the area. Normal Curve - Normal Curve - 0.068 0.473 0.068 0.473 Find the area between z = and z =.47 P( z.47). Use the symmetric property of the normal distribution and your table at the end of the text to find the area. The next slide shows the area. 1.93 0 1.93 Normal Curve - Normal Curve - 0.493 0.477 0.47 0.493 0.477 0.0160 Find the area between z = 1.68 and z = 1.37 P( 1.37 z 1.68). Use the symmetric property of the normal distribution and your table at the end of the text to find the area. The next slide shows the area.
Normal Curve - Normal Curve - 0.4147 0.4535 1.37 0 1.68 0.4535 +0.4147 0.868 Find the area to the left of z = 1.99 P(z < 1.99). Use your table at the end of the text to find the area. The next slide shows the area. Normal Curve - Normal Curve - 0.4767 +0.4767 0.9767 Find the area to the right of z = 1.16 P(z > 1.16) 1.16). Use your table at the end of the text to find the area. The next slide shows the area. 0 1.99 Normal Curve - RECALL: The Standard Normal Distribution 0.377 1.16 0 + 0.3770 0.8770 value mean z = standard deviation or X µ z = σ
Each month, an American household generates an average of 8 pounds of newspaper for garbage or recycling. Assume the standard deviation is pounds. Assume the amount generated is normally distributed. If a household is selected at random, find the probability of its generating: More than 30. pounds per month. First find the z-value for 30.. z =[X µ]/σ = [30. 8]/ = 1.1. Thus, P(z > 1.1) = 0.5 0.3643 = 0.1357. That is, the probability that a randomly selected household will generate more than 30. lbs. of newspapers is 0.1357 or 13.57%. 0.3643 0.1357 Between 7 and 31 pounds per month. First find the z-value for 7 and 31. z 1 = [X µ]/σ = [7 8]/ = 0.5; z = [31 8]/ = 1.5 Thus, P( 0.5 z 1.5) = 0.1915 + 0.433 = 0.647. 0 1.1 0.1915 0.433 0.5 0 1.5 0.1915 + 0.433 0.647 The American Automobile Association reports that the average time it takes to respond to an emergency call is 5 minutes. Assume the variable is approximately normally distributed and the standard deviation is 4.5 minutes. If 80 calls are randomly selected, approximately how many will be responded to in less than 15 minutes?
First find the z-value for 15 is z = [X µ]/σ = [15 5]/4.5 =.. Thus, P(z <.) = 0.4868 = 0.013. The number of calls that will be made in less than 15 minutes = (80)(0.013) = 1.056 1. 0.013. 0. 0.4868 0.013 An exclusive college desires to accept only the top 10% of all graduating seniors based on the results of a national placement test. This test has a mean of 500 and a standard deviation of 100. Find the cutoff score for the exam. Assume the variable is normally distributed. 7-4 Applications of the Normal Work backward to solve this problem. Subtract 0.1 (10%) from 0.5 to get the area under the normal curve for accepted students. Find the z value that corresponds to an area of 0.4000 by looking up 0.4000 in the area portion of Table E. Use the closest value, 0.3997. 7-4 Applications of the Normal Substitute in the formula and solve for X. z = X µ σ The z-value for the cutoff score (X) is z = [X µ]/σ = [X 500]/100 = 1.8. (See next slide). Thus, X = (1.8)(100) + 500 = 68. The score of 68 should be used as a cutoff score. 0.4 0 X = 1.8 0.1
NOTE: To solve for X, use the following formula: X = z σ + µ. : For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure. (Continued on the next slide). (Continued)-- If the mean systolic blood pressure is 10 and the standard deviation is 8, find the upper and lower readings that would qualify people to participate in the study. (continued) Note that two values are needed, one above the mean and one below the mean. The closest z values are 0.84 and 0.84 respectively. X = (z)(σ) + µ = (0.84)(8) + 10 = 16.7. The other X = ( 0.84)(8) + 10 = 113.8. See next slide. i.e. the middle 60% of BP readings is between 113.8 and 16.7. 0.3 0.3 0. 0. 0.84 0 0.84 7-55 Distribution of Sample Means Distribution of Sample means: A sampling distribution of sample means is a distribution obtained by using the means computed from random samples of a specific size taken from a population. 7-55 Distribution of Sample Means Sampling error is the difference between the sample measure and the corresponding population measure due to the fact that the sample is not a perfect representation of the population.
7-55 Properties of the Distribution of Sample Means The mean of the sample means will be the same as the population mean. The standard deviation of the sample means will be smaller than the standard deviation of the population, and it will be equal to the population standard deviation divided by the square root of the sample size. 7-55 Properties of the Distribution of Sample Means - Suppose a professor gave an 8-point quiz to a small class of four students. The results of the quiz were, 6, 4, and 8. Assume the four students constitute the population. The mean of the population is µ = ( + 6 + 4 + 8)/4 = 5. 5 7-55 Properties of the Distribution of Sample Means - 7-55 Graph of the Original Distribution The standard deviation of the population is σ = { ( 5) + ( 6 5) + ( 4 5) + ( 8 5) /4} =.36. The graph of the distribution of the scores is uniform and is shown on the next slide. Next we will consider all samples of size taken with replacement. 7-55 Properties of the Distribution of Sample Means - Sample Mean Sample Mean, 6, 4, 4 3 6, 4 5, 6 4 6, 6 6, 8 5 6, 8 7 4, 3 8, 5 4, 4 4 8, 4 6 4, 6 5 8, 6 7 4, 8 6 8, 8 8 7-55 Frequency Distribution of the Sample Means - X-bar 3 4 5 6 7 8 (mean) f 1 3 4 3 1
7-55 Graph of the Sample Means Frequency 4 3 1 0 DISTRIBUTION OF SAMPLE MEANS (APPROXIMATELY NORMAL) 3 4 5 6 SAMPLE MEANS 7 8 7-55 Mean and Standard Deviation of the Sample Means Mean of Sample Means µ = + 3 +... + 8 80 = = 5 X 16 16 which is the same as the population mean. Thus µ = µ. X 7-55 Mean and Standard Deviation of the Sample Means The standard deviation of the sample means is ( 5) + ( 3 5) +... + ( 8 5) σ X = 16 = 1581.. σ This is the same as. 7-55 The Standard Error of the Mean The standard deviation of the sample means is called the standard error of the mean. Hence σ σ =. X n 7-55 The Central Limit Theorem As the sample size n increases, the shape of the distribution of the sample means taken from a population with mean µ and standard deviation of σ will approach a normal distribution. As previously shown, this distribution will have a mean µ and standard deviation σ / n. 7-55 The Central Limit Theorem The central limit theorem can be used to answer questions about sample means in the same manner that the normal distribution can be used to answer questions about individual values. The only difference is that a new formula must be used for the z - values. It is X µ z =. σ/ n
7-55 The Central Limit Theorem - A.C. Neilsen reported that children between the ages of and 5 watch an average of 5 hours of TV per week. Assume the variable is normally distributed and the standard deviation is 3 hours. If 0 children between the ages of and 5 are randomly selected, find the probability that the mean of the number of hours they watch TV is greater than 6.3 hours. 7-55 The Central Limit Theorem - The standard deviation of the sample means is σ/ n = 3/ 0 = 0.671. The z-value is z = (6.3-5)/0.671= 1.94. Thus P(z > 1.94) = 0.5 0.4738 = 0.06. That is, the probability of obtaining a sample mean greater than 6.3 is 0.06 =.6%. 7-55 The Central Limit Theorem - 0.4738 0.06 7-55 The Central Limit Theorem - The average age of a vehicle registered in the United States is 8 years, or 96 months. Assume the standard deviation is 16 months. If a random sample of 36 cars is selected, find the probability that the mean of their age is between 90 and 100 months. 0 1.94 7-55 The Central Limit Theorem - The standard deviation of the sample means is σ/ n = 16/ 36 =.6667. The two z-values are z 1 = (90 96)/.6667 =.5 and z = (100 96)/.6667 = 1.50. Thus P(.5 z 1.50) = 0.4878 + 0.433 = 0.91 or 9.1%. 7-55 The Central Limit Theorem -.5 0.4878 0.433 0 1.50