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Overview This chapter will deal with the construction of probability distributions Probability Distributions will describe what will probably happen instead of what actually did happen. 1 Random Variables 2 Definitions Random Variable a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure Probability Distribution a graph, table, or formula that gives the probability for each value of the random variable 3 Definitions Discrete random variable has either a finite number of values or countable number of values, where ‘countable’ refers to the fact that there might be infinitely many values, but they result from a counting process. Continuous random variable has infinitely many values, and those values can be associated with measurements on a continuous scale with no gaps or interruptions. 4 Requirements for Probability Distribution P(x) = 1 where x assumes all possible values 0 P(x) 1 for every value of x 5 Mean, Variance and Standard Deviation of a Probability Distribution Formula 6-1 µ = [x • P(x)] Formula 6-2 2 = [(x - µ) • P(x)] 2 6 The probability that a car insurance company pays $200,000 is .01, $100,000 is .02 and $50,000 is .04. .93 1. Draw the probability density function. 2. Determine the minimum value the insurance company should charge to break even. .04 .02 0 $50000 $100000 .01 $200000 µA= ($200000)(.01)+($100000)(.02)+($50000)(.04)+(0)(.93) = $6000 7 Page 243 #5 Amount Payment 92 68 51 17 522 None Percent $10,000.00 $25,000.00 $60,000.00 $90,000.00 12.27% 9.07% 6.80% 2.27% 69.60% .70 1. Draw the probability density function. 2. Determine the minimum value the insurance company should charge to break even. .12 .09 0 10 25 .07 60 .02 90 µA= ($10000)(.12)+($25000)(.09)+($60000)(.07)+ (.022) ($90000)+ (0)(.93) = $9450 8 Definition Expected Value The average value of outcomes E = [x • P(x)] 9 What is the expected value for a $1.00 ticket that pays off a $500.00 prize if 1000 tickets are sold? E = [x • P(x)] Event Win Lose 10 E = [x • P(x)] Event x Win $499 Lose - $1 11 E = [x • P(x)] Event x P(x) Win $499 0.001 Lose - $1 0.999 12 E = [x • P(x)] Event x P(x) x • P(x) Win $499 0.001 0.499 Lose - $1 0.999 - 0.999 13 E = [x • P(x)] Event x P(x) x • P(x) Win $499 0.001 0.499 Lose - $1 0.999 - 0.999 E = -$.50 14 Definitions Binomial Probability Distribution 1. The experiment must have a fixed number of trials. 2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.) 3. Each trial must have all outcomes classified into two categories. 4. The probabilities must remain constant for each trial. 15 Notation for Binomial Probability Distributions n = fixed number of trials r = specific number of successes in n trials p = probability of success in one of n trials q = probability of failure in one of n trials (q = 1 - p ) P(r) = probability of getting exactly r success among n trials Be sure that r and p both refer to the same category being called a success. 16 Binomial Probability Formula 17 Binomial Probability Formula P(x) = n! • (n - r )! r! pr • n-r q 18 Method 1 Binomial Probability Formula P(r) = n! • (n - r )! r! P(r) = nCr • pr • pr • n-r q qn-r for calculators with nCr key. 19 Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time. This is a binomial experiment where: n=5 r=3 p = 0.90 q = 0.10 20 Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time. This is a binomial experiment where: n=5 r=3 p = 0.90 q = 0.10 Using the binomial probability formula to solve: P(3) = 5C3 • 0.93 • 0.12 =.0729 21 Example: Using n = 5 and p = 0.90, find the following: a) The probability of exactly 3 successes b) The probability of at least 3 successes a) P(3) = 0.073 b) P(at least 3) = P(3 or 4 or 5) = P(3) or P(4) or P(5) = 0.073 + 0.328 + 0.590 = 0.991 22 For Binomial Distributions: • Formula 4-6 µ =n•p • Formula 4-7 = n • p • q 2 or = n•p•q 23 Example: Find the mean and standard deviation for the number of girls in groups of 14 births. • n = 14 • p = 0.5 • q = 0.5 • Using the binomial distribution formulas: 24 Example: • • • • • Find the mean and standard deviation for the number of girls in groups of 14 births. n = 14 p = 0.5 q = 0.5 Using the binomial distribution formulas: µ = (14)(0.5) = 7 girls (14)(.5)(.5) = 1.9 girls 25 Reminder • Maximum usual values = µ + 2 • Minimum usual values = µ - 2 26 Example: Determine whether 68 girls among 100 babies could easily occur by chance. • • • • • For this binomial distribution, µ = 50 girls = 5 girls µ + 2 = 50 + 2(5) = 60 µ - 2 = 50 - 2(5) = 40 The usual number girls among 100 births would be from 40 to 60. So 68 girls in 100 births is an unusual result. 27 Poisson Random Variable Poisson process is the random occurrence of a series of events over time (or space) in such a way that: 1. The number of successes can be any whole number. 2. The probability of success is equally spread out. λ e λ P(x r) r! r the mean number of successes r the desired number of successes 28 Example: On an average day, 6 people visit the barbershop a. What is the probability that exactly 4 will visit the barbershop? b. What is the probability that 2 or less will visit the barbershop? c. What is the probability that more than 3 will visit the barbershop? e-6 64 a. P(4) .1338 4! e-6 60 e-6 61 e-6 62 b. P(X 2) .0024 .0149 .0446 .0619 0! 1! 2! e-6 60 e-6 61 e-6 62 e-6 63 .849 c. P(X 3) 1 1! 2! 3! 0! 29 Example: A 100-foot of pipe has 20 leaks. A 25-foot section is chosen at random a. What is the probability it has exactly 3 leaks? b. What is the probability it has at most 2 leaks? c. What is the probability it has more than 3 leaks? =5 e-5 53 a. P(3) .1404 3! e-5 50 e-5 51 e-5 52 b. P(X 2) .0067 .0337 .0842 .1246 0! 1! 2! e-5 50 e-5 51 e-5 52 e-5 53 .735 c. P(X 3) 1 1! 2! 3! 0! 30 Overview Continuous random variable Normal distribution 31 Overview Continuous random variable Normal distribution Curve is bell shaped and symmetric µ Score 32 Overview Continuous random variable Normal distribution Curve is bell shaped and symmetric µ Score y= e 1 2 x-µ 2 ( ) 2p 33 Definitions Uniform Distribution a probability distribution in which the continuous random variable values are spread evenly over the range of possibilities; the graph results in a rectangular shape. 34 Definitions Density Curve (or probability density function) the graph of a continuous probability distribution 35 Definitions Density Curve (or probability density function) the graph of a continuous probability distribution 1. The total area under the curve must equal 1. 2. Every point on the curve must have a vertical height that is 0 or greater. 36 Because the total area under the density curve is equal to 1, there is a correspondence between area and probability. 37 Times in First or Last Half Hours 38 A continuous random variable,T, has a uniform (rectangular) probability distribution over the interval [2,10]. a. Graph the probability distribution. b. Find P(x>6) c. Find P(5<x<8) a. A = b*h, 1=b*8, h = .125 b. P(x>6) A = b*h = 4*.125 = .5 c. P(5<x<8) A = b*h = 3*.125 = .375 2 10 39 A continuous random variable,T, has a uniform (triangular) probability distribution over the interval [0,6]. a. Graph the probability distribution. b. Find P(x > 4) c. Find P(1 < x < 4) (0, 1/3) (1, 5/18) (2, 2/9) (3, 1/6) (4, 1/9) (5, 1/18) 0 6 a. A = ½ b*h, 1= ½* 6*h, h = 1/3 b. P(x > 4) A = ½ b*h = ½ *2*(1/9) = 1/9 c. P(1 < x < 4) A = ½ b1*h1 - ½ b2*h2 = ½* 5*(5/18) – ½ *2*(1/9) = 25/36 – 1/9 = 21/36 = 7/12 40 Heights of Adult Men and Women Women: µ = 63.6 = 2.5 Figure 5-4 Men: µ = 69.0 = 2.8 63.6 69.0 Height (inches) 41 Definition Standard Normal Deviation a normal probability distribution that has a mean of 0 and a standard deviation of 1 42 Table A-2 Formulas and Tables card Appendix A2 43 Table A-2 Standard Normal Distribution =1 µ=0 0 x z 44 Definition Standard Normal Deviation a normal probability distribution that has a mean of 0 and a standard deviation of 1 Area found in Table A-2 Area = 0.3413 0.4429 -3 -2 -1 0 1 2 3 0 z = 1.58 Score (z ) Figure 5-5 Figure 5-6 45 Table A-2 Standard Normal (z) Distribution z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 .0000 .0398 .0793 .1179 .1554 .1915 .2257 .2580 .2881 .3159 .3413 .3643 .3849 .4032 .4192 .4332 .4452 .4554 .4641 .4713 .4772 .4821 .4861 .4893 .4918 .4938 .4953 .4965 .4974 .4981 .4987 .0040 .0438 .0832 .1217 .1591 .1950 .2291 .2611 .2910 .3186 .3438 .3665 .3869 .4049 .4207 .4345 .4463 .4564 .4649 .4719 .4778 .4826 .4864 .4896 .4920 .4940 .4955 .4966 .4975 .4982 .4987 .0080 .0478 .0871 .1255 .1628 .1985 .2324 .2642 .2939 .3212 .3461 .3686 .3888 .4066 .4222 .4357 .4474 .4573 .4656 .4726 .4783 .4830 .4868 .4898 .4922 .4941 .4956 .4967 .4976 .4982 .4987 .0120 .0517 .0910 .1293 .1664 .2019 .2357 .2673 .2967 .3238 .3485 .3708 .3907 .4082 .4236 .4370 .4484 .4582 .4664 .4732 .4788 .4834 .4871 .4901 .4925 .4943 .4957 .4968 .4977 .4983 .4988 .0160 .0557 .0948 .1331 .1700 .2054 .2389 .2704 .2995 .3264 .3508 .3729 .3925 .4099 .4251 .4382 .4495 .4591 .4671 .4738 .4793 .4838 .4875 .4904 .4927 .4945 .4959 .4969 .4977 .4984 .4988 .0199 .0596 .0987 .1368 .1736 .2088 .2422 .2734 .3023 .3289 .3531 .3749 .3944 .4115 .4265 .4394 .4505 .4599 .4678 .4744 .4798 .4842 .4878 .4906 .4929 .4946 .4960 .4970 .4978 .4984 .4989 .0239 .0636 .1026 .1406 .1772 .2123 .2454 .2764 .3051 .3315 .3554 .3770 .3962 .4131 .4279 .4406 .4515 .4608 .4686 .4750 .4803 .4846 .4881 .4909 .4931 .4948 .4961 .4971 .4979 .4985 .4989 .0279 .0675 .1064 .1443 .1808 .2157 .2486 .2794 .3078 .3340 .3577 .3790 .3980 .4147 .4292 .4418 .4525 .4616 .4693 .4756 .4808 .4850 .4884 .4911 .4932 .4949 .4962 .4972 .4979 .4985 .4989 .0319 .0714 .1103 .1480 .1844 .2190 .2517 .2823 .3106 .3365 .3599 .3810 .3997 .4162 .4306 .4429 .4535 .4625 .4699 .4761 .4812 .4854 .4887 .4913 .4934 .4951 .4963 .4973 .4980 .4986 .4990 .0359 .0753 .1141 .1517 .1879 .2224 .2549 .2852 .3133 .3389 .3621 .3830 .4015 .4177 .4319 .4441 .4545 .4633 .4706 .4767 .4817 .4857 .4890 .4916 .4936 .4952 .4964 .4974 .4981 .4986 .4990 * * 46 To find: z Score the distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of Table A-2 Area the region under the curve; refer to the values in the body of Table A-2 47 Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. 48 Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. P ( 0 < x < 1.58 ) = 0 1.58 49 Table A-2 Standard Normal (z) Distribution z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 .0000 .0398 .0793 .1179 .1554 .1915 .2257 .2580 .2881 .3159 .3413 .3643 .3849 .4032 .4192 .4332 .4452 .4554 .4641 .4713 .4772 .4821 .4861 .4893 .4918 .4938 .4953 .4965 .4974 .4981 .4987 .0040 .0438 .0832 .1217 .1591 .1950 .2291 .2611 .2910 .3186 .3438 .3665 .3869 .4049 .4207 .4345 .4463 .4564 .4649 .4719 .4778 .4826 .4864 .4896 .4920 .4940 .4955 .4966 .4975 .4982 .4987 .0080 .0478 .0871 .1255 .1628 .1985 .2324 .2642 .2939 .3212 .3461 .3686 .3888 .4066 .4222 .4357 .4474 .4573 .4656 .4726 .4783 .4830 .4868 .4898 .4922 .4941 .4956 .4967 .4976 .4982 .4987 .0120 .0517 .0910 .1293 .1664 .2019 .2357 .2673 .2967 .3238 .3485 .3708 .3907 .4082 .4236 .4370 .4484 .4582 .4664 .4732 .4788 .4834 .4871 .4901 .4925 .4943 .4957 .4968 .4977 .4983 .4988 .0160 .0557 .0948 .1331 .1700 .2054 .2389 .2704 .2995 .3264 .3508 .3729 .3925 .4099 .4251 .4382 .4495 .4591 .4671 .4738 .4793 .4838 .4875 .4904 .4927 .4945 .4959 .4969 .4977 .4984 .4988 .0199 .0596 .0987 .1368 .1736 .2088 .2422 .2734 .3023 .3289 .3531 .3749 .3944 .4115 .4265 .4394 .4505 .4599 .4678 .4744 .4798 .4842 .4878 .4906 .4929 .4946 .4960 .4970 .4978 .4984 .4989 .0239 .0636 .1026 .1406 .1772 .2123 .2454 .2764 .3051 .3315 .3554 .3770 .3962 .4131 .4279 .4406 .4515 .4608 .4686 .4750 .4803 .4846 .4881 .4909 .4931 .4948 .4961 .4971 .4979 .4985 .4989 .0279 .0675 .1064 .1443 .1808 .2157 .2486 .2794 .3078 .3340 .3577 .3790 .3980 .4147 .4292 .4418 .4525 .4616 .4693 .4756 .4808 .4850 .4884 .4911 .4932 .4949 .4962 .4972 .4979 .4985 .4989 .0319 .0714 .1103 .1480 .1844 .2190 .2517 .2823 .3106 .3365 .3599 .3810 .3997 .4162 .4306 .4429 .4535 .4625 .4699 .4761 .4812 .4854 .4887 .4913 .4934 .4951 .4963 .4973 .4980 .4986 .4990 .0359 .0753 .1141 .1517 .1879 .2224 .2549 .2852 .3133 .3389 .3621 .3830 .4015 .4177 .4319 .4441 .4545 .4633 .4706 .4767 .4817 .4857 .4890 .4916 .4936 .4952 .4964 .4974 .4981 .4986 .4990 * * 50 Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Area = 0.4429 P ( 0 < x < 1.58 ) = 0.4429 0 1.58 51 Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Area = 0.4429 P ( 0 < x < 1.58 ) = 0.4429 0 1.58 The probability that the chosen thermometer will measure freezing water between 0 and 1.58 degrees is 0.4429. 52 Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Area = 0.4429 P ( 0 < x < 1.58 ) = 0.4429 0 1.58 There is 44.29% of the thermometers with readings between 0 and 1.58 degrees. 53 Using Symmetry to Find the Area to the Left of the Mean Because of symmetry, these areas are equal. Figure 5-7 (a) (b) 0.4925 0.4925 0 z = - 2.43 0 Equal distance away from 0 z = 2.43 NOTE: Although a z score can be negative, the area under the curve (or the corresponding probability) can never be negative. 54 Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads freezing water between -2.43 degrees and 0 degrees. Area = 0.4925 P ( -2.43 < x < 0 ) = 0.4925 -2.43 0 The probability that the chosen thermometer will measure freezing water between -2.43 and 0 degrees is 0.4925. 55 The Empirical Rule Standard Normal Distribution: µ = 0 and = 1 56 The Empirical Rule Standard Normal Distribution: µ = 0 and = 1 68% within 1 standard deviation 34% x-s 34% x x+s 57 The Empirical Rule Standard Normal Distribution: µ = 0 and = 1 95% within 2 standard deviations 68% within 1 standard deviation 34% 34% 13.5% x - 2s 13.5% x-s x x+s x + 2s 58 The Empirical Rule Standard Normal Distribution: µ = 0 and = 1 99.7% of data are within 3 standard deviations of the mean 95% within 2 standard deviations 68% within 1 standard deviation 34% 34% 2.4% 2.4% 0.1% 0.1% 13.5% x - 3s x - 2s 13.5% x-s x x+s x + 2s x + 3s 59 Probability of Half of a Distribution 0.5 0 60 Finding the Area to the Right of z = 1.27 Value found in Table A-2 0.3980 0 This area is 0.5 - 0.3980 = 0.1020 z = 1.27 61 Finding the Area Between z = 1.20 and z = 2.30 0.4893 (from Table A-2 with z = 2.30) Area A is 0.4893 - 0.3849 = 0.1044 0.3849 A 0 z = 1.20 z = 2.30 62 Notation P(a < z < b) denotes the probability that the z score is between a and b P(z > a) denotes the probability that the z score is greater than a P (z < a) denotes the probability that the z score is less than a 63 Figure 5-10 Interpreting Area Correctly 64 Interpreting Area Correctly ‘greater than ‘at least x’ x’ ‘more than Subtract from 0.5 Add to 0.5 x’ ‘not less than x’ 0.5 x x 65 Interpreting Area Correctly ‘greater than ‘at least x’ Add to 0.5 x’ ‘more than Subtract from 0.5 x’ ‘not less than x’ 0.5 x Add to 0.5 x ‘less than ‘at most x’ x’ ‘no more than x’ ‘not greater than Subtract from 0.5 x’ 0.5 x x 66 Interpreting Area Correctly ‘greater than ‘at least x’ Add to 0.5 x’ ‘more than Subtract from 0.5 x’ ‘not less than x’ 0.5 x Add to 0.5 x ‘less than ‘at most x’ x’ ‘no more than x’ ‘not greater than Subtract from 0.5 x’ 0.5 x x Add C ‘between x1 and Use A=C-B x2’ A x1 x2 B x1 x2 67 Finding a z - score when given a probability Using Table A-2 1. Draw a bell-shaped curve, draw the centerline, and identify the region under the curve that corresponds to the given probability. If that region is not bounded by the centerline, work with a known region that is bounded by the centerline. 2. Using the probability representing the area bounded by the centerline, locate the closest probability in the body of Table A-2 and identify the corresponding z score. 3. If the z score is positioned to the left of the centerline, make it a negative. 68 Finding z Scores when Given Probabilities 95% 5% 5% or 0.05 0.45 0.50 z 0 ( z score will be positive ) Finding the 95th Percentile 69 Finding z Scores when Given Probabilities 95% 5% 5% or 0.05 0.45 0.50 0 1.645 (z score will be positive) Finding the 95th Percentile 70 Finding z Scores when Given Probabilities 90% 10% Bottom 10% 0.10 0.40 z 0 (z score will be negative) Finding the 10th Percentile 71 Finding z Scores when Given Probabilities 90% 10% Bottom 10% 0.10 0.40 -1.28 0 (z score will be negative) Finding the 10th Percentile 72 Other Normal Distributions 0 1 If or (or both), we will convert values to standard scores using Formula 6-7, then procedures for working with all normal distributions are the same as those for the standard normal distribution. z= x-µ 73 Converting to Standard Normal Distribution P (a) x 74 Converting to Standard Normal Distribution x- z= P P (a) x (b) 0 z 75 Probability of Weight between 143 pounds and 201 pounds z= x = 143 σ = 29 143 201 201 - 143 29 = 2.00 Weight z 0 2.00 76 Probability of Weight between 143 pounds and 201 pounds Value found in Table A-2 x = 143 σ = 29 143 201 Weight z 0 2.00 77 Probability of Weight between 143 pounds and 201 pounds 0.4772 x = 143 σ = 29 143 201 Weight z 0 2.00 78 Probability of Weight between 143 pounds and 201 pounds There is a 0.4772 probability of randomly selecting a woman with a weight between 143 and 201 lbs. x = 143 σ = 29 143 201 Weight z 0 2.00 79 Probability of Weight between 143 pounds and 201 pounds OR - 47.72% of women have weights between 143 lb and 201 lb. x = 143 σ = 29 143 201 Weight z 0 2.00 80 Cautions to keep in mind 1. Don’t confuse z scores and areas. Z scores are distances along the horizontal scale, but areas are regions under the normal curve. Table A-2 lists z scores in the left column and across the top row, but areas are found in the body of the table. 2. Choose the correct (right/left) side of the graph. 3. A z score must be negative whenever it is located to the left of the centerline of 0. 81 Finding z Scores when Given Probabilities 95% 5% 5% or 0.05 0.45 0.50 0 1.645 (z score will be positive) Finding the 95th Percentile 82 Finding z Scores when Given Probabilities 90% 10% Bottom 10% 0.10 0.40 -1.28 0 (z score will be negative) Finding the 10th Percentile 83 Procedure for Finding Values Using Table A-2 and Formula 6-7 1. Sketch a normal distribution curve, enter the given probability or percentage in the appropriate region of the graph, and identify the value(s) being sought. x 2. Use Table A-2 to find the z score corresponding to the region bounded by x and the centerline of 0. Cautions: Refer to the BODY of Table A-2 to find the closest area, then identify the corresponding z score. Make the z score negative if it is located to the left of the centerline. 3. Using Formula 5-2, enter the values for µ, , and the z score found in step 2, then solve for x. x = µ + (z • ) (Another form of Formula 6-7) 4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and the context of the problem. 84 Finding P10 for Weights of Women 10% 90% 40% x=? 143 50% Weight 85 Finding P10 for Weights of Women 0.10 0.40 0.50 x=? 143 -1.28 0 Weight 86 Finding P10 for Weights of Women x = 143 + (-1.28 • 29) = 105.88 0.10 0.40 0.50 x=? 143 -1.28 0 Weight 87 Finding P10 for Weights of Women The weight of 106 lb (rounded) separates the lowest 10% from the highest 90%. 0.10 0.40 x = 106 -1.28 0.50 143 Weight 0 88 Forgot to make z score negative??? x = 143 + (1.28 • 29) = 180 0.10 0.40 0.50 x=? 143 1.28 0 Weight 89 Forgot to make z score negative??? x = 143 + (1.28 • 29) = 180 0.10 0.40 x = 180 1.28 0.50 143 Weight 0 90 Forgot to make z score negative??? UNREASONABLE ANSWER! 0.10 0.40 x = 180 1.28 0.50 143 Weight 0 91 REMEMBER! Make the z score negative if the value is located to the left (below) the mean. Otherwise, the z score will be positive. 92 A pizza company advertises that it puts .5 lb of cheese on its medium pizzas. In reality, the amount of cheese is normally distributed with a mean value of .5 lbs with a standard deviation of .025 lbs. a. What is the probability that the amount of cheese on a medium pizza is between .525 and .550 lbs? b. What is the probability that a medium pizza has less than .47 lbs of cheese? c. If the pizza has more cheese than 95% of the other pizzas, how many pounds of cheese does it have? a. z xμ σ z .525 .5 1 .025 z .550 .5 2 .025 .4772-.3413=.1359 93 A pizza company advertises that it puts .5 lb of cheese on its medium pizzas. In reality, the amount of cheese is normally distributed with a mean value of .5 lbs with a standard deviation of .025 lbs. a. What is the probability that the amount of cheese on a medium pizza is between .525 and .550 lbs? b. What is the probability that a medium pizza has less than .47 lbs of cheese? c. If the pizza has more cheese than 95% of the other pizzas, how many pounds of cheese does it have? b. z xμ σ z .47 .5 -1.2 .025 .5-.3849=.1150 94 A pizza company advertises that it puts .5 lb of cheese on its medium pizzas. In reality, the amount of cheese is normally distributed with a mean value of .5 lbs with a standard deviation of .025 lbs. a. What is the probability that the amount of cheese on a medium pizza is between .525 and .550 lbs? b. What is the probability that a medium pizza has less than .47 lbs of cheese? c. If the pizza has more cheese than 95% of the other pizzas, how many pounds of cheese does it have? c. z xμ σ z = 1.645 x = μ + zσ x = .5 + (1.645)(.025) x =.5411 95 A car’s gas mileage averages 30 mi/gal with a standard deviation of 1.2 mi/gal a. What is the probability that the mileage of a randomly selected car is between 31.2 mi/gal and 33 mi/gal? b. What is the probability that a car selected at random exceeds 32 mi/gal? c. What is the miles per gallon if the car is in the bottom 20%? a. xμ 31.2 30 z z 1 σ 1.2 z 33 30 2.5 1.2 . 4938-.3413=.1525 96 A car’s gas mileage averages 30 mi/gal with a standard deviation of 1.2 mi/gal a. What is the probability that the mileage of a randomly selected car is between 31.2 mi/gal and 33 mi/gal? b. What is the probability that a car selected at random exceeds 32 mi/gal? c. What is the miles per gallon if the car is in the bottom 20%? b. z xμ σ z 32 30 1.67 1.2 . 5 - .4525 = .0475 97 A car’s gas mileage averages 30 mi/gal with a standard deviation of 1.2 mi/gal a. What is the probability that the mileage of a randomly selected car is between 31.2 mi/gal and 33 mi/gal? b. What is the probability that a car selected at random exceeds 32 mi/gal? c. What is the miles per gallon if the car is in the bottom 20%? c. z xμ σ x = μ + zσ x = 30 – (.84)(1.2) = 28.99 98