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are two available models, A and B. The dealer normally sells at least twice as many model A cars as model B cars. If the dealer makes a profit of $1300 on model A cars and $1700 on model B cars, how many of each car should the dealer have in the lot? 23. A 30-acre orchard is to contain two types of trees, peach and almond. The profit per year is $16.80 per peach tree and $21.60 per almond tree. An acre can sustain 1080 peach trees or 900 almond trees. If the grower has available labor to plant a maximum of 30,000 trees, how many of each type of tree should be planted? 24. An investor has $12,000 to invest into two different funds. Fund A, which is a high-risk fund, yields an average return of 14%. Fund B, which is a low-risk fund, yields an average return of 6%. To reduce the risk, the investor wants the amount in fund B to be at least twice the amount in fund A. How much should be invested in each fund to maximize the return? What is the maximum return? C H A P T E R 12 R E V I E W Important Concepts Section 12.1 System of equations...................... 779 Solution of a system...................... 780 Number of solutions of a system............ 782 Substitution method...................... 782 Elimination method...................... 783 Inconsistent system...................... 784 Section 12.1.A z-axis................................. 790 Coordinate plane........................ 790 Octants...................... |
.......... 790 Section 12.2 Section 12.3 Section 12.4 Section 12.5 Matrix................................ 795 Augmented matrix....................... 795 Equivalent systems....................... 795 Elementary row operations................ 796 Reduced row-echelon form................ 797 Gauss-Jordan elimination.................. 797 m n matrix........................... 804 Equality of matrices...................... 804 Addition and subtraction of matrices........ 804 Scalar multiplication of matrices............ 805 Matrix multiplication..................... 806 Directed networks....................... 809 Adjacency matrix........................ 809 Matrix equation........................ 814 Identity matrices........................ 815 Invertible matrix......................... 815 Inverse of a matrix....................... 816 Matrix solutions of a square system......... 817 Curve fitting... |
......................... 818 Algebraic solutions of systems of nonlinear equations...................... 821 Graphical solution of systems of nonlinear equations...................... 822 Section 12.5.A Solutions of systems of linear inequalities.... 826 Linear programming..................... 829 834 Review Exercises In Exercises 1–10, solve the system of linear equations by any method. Chapter Review 835 Section 12.1 1. 5x 3y 4 2x y 3 3. 3x 5y 10 4x 3y 6 5. 7. 9. 3x y z 13 x y 2z 9 3x y 2z 9 4x 3y 3z 2 5x 3y 2z 10 2x 2y 3z 14 x 2y 3z 1 5y 10z 0 8x 6y 4z 8 2. 3x y 6 2x 3y 7 4. 6. 8. 1 4 1 10 2y 3z 1 4x 4y 4z 2 10x 8y 5z 4 x y 4z 0 2x y 3z 2 3x y 2z 4 10. 4x y 2z 4 x y 1 z 1 2 2x y z 8 11. The sum of one number and three times a second number is 20. The sum of the second number and two times the first number is 55. Find the two numbers. 12. You are given $144 in $1, $5, and $10 bills. There are 35 bills. There are two more $10 bills than $5 bills. How many bills of each type do you have? 13. Let L be the line with equation 4x 2y 6 and M the line with equation 10x 5y 15. a. L and M do not intersect. Which of the following statements is true? b. L and M intersect at a single point. d. All of the above are true. c. L and M are the same line. e. None of the above are true. 14. Which of the following statements about the given system of equations are false? x 4 |
y z 2 6x 4y 14z 24 2x y 4z 7 z 0 z 1 x 2, x 1, x 1, y 3, y 1, y 3, is a solution. is a solution. a. b. c. d. The system has an infinite number of solutions. e. is not a solution. is a solution. z 1 x 2, y 5, z 3 15. Tickets to a lecture cost $1 for students, $1.50 for faculty, and $2 for others. Total attendance at the lecture was 460, and the total income from tickets was $570. Three times as many students as faculty attended. How many faculty members attended the lecture? 16. An alloy containing 40% gold and an alloy containing 70% gold are to be mixed to produce 50 pounds of an alloy containing 60% gold. How much of each alloy is needed? 836 Chapter 12 Systems and Matrices Section 12.2 For Exercises 17–20, write the system of linear equations represented by the augmented matrix. 17. 2 2 a 6 3 16 7b 19 10 2 1 3 ¢ 18. 2 3 a 1 2 4 1b 20 ¢ For Exercises 21–26, write the system represented by each matrix, find the solutions, if any, and classify each system as consistent or inconsistent. 21. 4 1 a 2 5 14 9b 23 ¢ 25. 2 1 4 3 1 1 a 4 7b 22. 9 12 a 6 8 3 4b 24 ¢ 26 ¢ Section 12.3 In Exercises 27–30, perform the indicated matrix multiplication or state that the product is not defined 1b 1 0 2 4b B 2 4 a E ° 3 1b 4b 3 3 1 ¢ 27. AB 28. CD 29. AE 30. DF In Exercises 31–34, find the inverse of the matrix, if it exists. 31. 33. 3 4 a 3 1 2 ° 7 9b 2 1 2 6 2 5 ¢ 32. 2 1 a 6 3b 34 ¢ Section 12.4 In Exercises 35–38, use matrix inverses to solve the system. 35. x 2y 3z 4 2x y 4z 3 3x 4y z 2 36. 2x y 2z 2u 0 x 3y 2z u 0 x 4y 2z 3u 0 x 4y 2z 3u 0 37. x 4y 2 |
z 6w 2 3x 4y 2z w 0 5x 4y 2z 5w 4 4x 4y 2z 3w 1 39. Find the equation of the parabola passing through the points, (2, 17), (8, 305). 3, 52 1 2 Chapter Review 837 38. 2x y 2z u 2 x 3y 4z 2u 2v 2 2x 3y 5z 4u v 1 x 3y 2z 4u 4v 4 2x 3y 6z 4u 5v 0 40. The table shows the number of hours spent per person per year on home video games. Find a quadratic equation that contains x 6 this data, with to 1996. corresponding Year 1996 2000 2004 Hours 25 76 161 [Source: Statistical Abstract of the U.S.: 2001] Section 12.5 In Exercises 41–46, solve the system. 41. x2 y 0 y 2x 3 43. x2 y2 16 x y 2 45. x3 y3 26 x2 y 6 Section 12.5.A 47. Minimize and maximize F 2x 2y 4 x y 36 2x y 10 x 0 y 0 subject to ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 42. x2 y2 25 x2 y 19 44. 6x2 4xy 3y2 36 x2 xy y2 9 46. x2 3xy 2y2 y x 0 5x2 10xy 5y2 8 30x 10y x, y 1 2 48. Minimize and maximize 8x 7y x, y 2 subject to F 1 4x 3y 24 3x 4y 8 x 0 y 0 ⎧ ⎪ ⎨ ⎪ ⎩ 49. Animal feed is to be made from corn and soybeans. One pound of corn has 30 units of fat and 20 units of protein, and one pound of soybeans has 20 units of fat and 40 units of protein. What is the minimum total weight of feed to supply a daily requirement of 2800 units of fat and 2200 units of protein? 50. A home supply store sells two models of dehumidifiers, standard and 3 deluxe. The standard model comes in a 10-ft box and weighs 10 lb, and 3 the deluxe model comes in a 9-ft box and weighs 12 lb. The store’s delivery van has |
248 ft of space and can hold a maximum of 440 lb. If the store makes a profit of $20 on the standard model and $30 on the deluxe model, how many boxes of each model can the van carry to maximize the profit for each load? 3 C H A P T E R 12 Partial Fractions In calculus it is sometimes necessary to write a complicated rational expression as the sum of simpler ones. Two forms of rational expressions whose sum is a rational expression will be introduced in this section: denominators with nonrepeated factors and those with repeated factors. Example 1 Denominators with Nonrepeated Factors Find the constants A and B such that 7x 6 2 x 6 x˛ A x 2 B x 3. Solution The denominators on the right side of the equation are linear factors of the denominator on the left side. Multiply both sides of the equation by the common denominator, x 3 x 2 7x and collect like terms. x 2 2 2 Ax 3A Bx 2B Ax Bx 3A 2B x 3A 2B A B 1 2 Because the polynomials on the left and right sides of the last equation are equal, their coefficients must be equal term by term. 2 1 1 A B 7 3A 2B 6 Coefficient of x Constant term The two equations above form a system of equations with unknowns A A 4 and B. Solving the system yields Therefore, B 3. and 7x ■ In Example 1, 4 x 2 3 x 3 is called the partial fraction decomposi- tion, or simply the partial fractions, of 7x 6 2 x 6. x If the denominator of a rational expression can be expressed as a product of nonrepeated linear factors, each term of the decom- position has the form A x a. Figure 12.C-1 Nonrepeated Linear Factor Denominators 838 5 Solution 5 5 5 Figure 12.C-2 Figure 12.C-3 Nonrepeated Quadratic Factor Denominators When a factor of the denominator is repeated, every power less than or equal to the multiplicity of the factor must be considered. Example 2 Repeated Linear Factors Find the partial fraction decomposition of 2x x 2 15x 10 2 4 3 3x. The denominator can be factored into ure 12.C-2. Because 2 must be considered as possible denominators of the decomposition. The process of finding the numerators is the same as that shown |
in Example 1. x 2 x 1 2, 2 x 2 is a repeated factor, both as shown in Fig2 x 2 x 2 and 2 2 1 1 1 1 2 1 2 Multiply both sides of the equation by the common denominator, x 1 2, and collect like terms on the right side. C x 2 1 2 2 15x 10 A x 1 x 2 2 B x 2 2x 4x 4 A x x 1 1 2 Ax 2 Bx 2B Cx C 2 4Ax 4A Bx 4A 2B C 1 The polynomials on the left and right sides are equal, so coefficients must be equal term by term. 4A 4A B C 15 4A 2B C 10 Coefficient of x 2 Coefficient of x Constant term This is a system of equations with unknowns A, B, and C. The augmented matrix of the system and an equivalent reduced row echelon form matrix are shown in Figure 12.C-3. A 3, Therefore, C 4, and B 1, 2 15x 10 2 x 2 2x ■ In theory, any polynomial with real coefficients can be written as a product of real linear factors and real quadratic factors. (See Section 4.2.) If the denominator contains a nonrepeated quadratic factor, then the decomposition will contain a term of the form where ax 2 bx c is irreducible over the set of Ax B ax2 bx c real numbers., Like repeated linear factors, if an irreducible quadratic factor is repeated, every power less than or equal to the multiplicity of the factor must be considered. The numerator of all quadratic factors has the form Ax B. 839 Example 3 Nonrepeated Quadratic Factor Find the partial fraction decomposition of x 2 2 10x 8 3 6x. x Solution 10 The denominator has the nonrepeated linear factor as shown in x Figure 12.C-4, and the nonrepeated quadratic factor found by using synthetic division. Thus, the partial fraction decomposition has the x 4, 2 2x 2 –10 10 form x 3 6x x 2 2 10x 8 A x 4 Bx C 2 2x 2 x. –10 Figure 12.C-4 1 2 1 x Multiplying both sides of the equation by the common denominator, x 4, and collecting like terms yields x 4 2 Cx 4 |
Bx 4C 2A 4C 2 2x 2 2 2 2x 2 x 2 A x 1 Ax 2 2Ax 2A Bx 2A 4B C Bx C A B 1 x Because there is no sponding coefficient must be 0. Therefore, 2 term in the original rational expression, the corre 4B C 1 2 A 4C 2 Coefficient of x 2 Coefficient of x Constant term Figure 12.C-5 Figure 12.C-5. The solution of the system is A 3 5, B 3 5, and C 1 5, as shown in x 2 x3 6x2 10x x2 2x 2 1 5 3 x 4 a 3x 1 x2 2x 2b ■ All the rational expressions in the previous examples have been proper fractions, which means the degree of the numerator was less than the degree of the denominator. If the rational expression is improper, then divide the numerator by the denominator and decompose the remainder that is a proper fraction. Example 4 Decomposing an Improper Rational Expression Find the partial fraction decomposition of 3 x 3 2 x 2 x x. Solution The rational expression is an improper fraction because the degree of the numerator is greater than the degree of the denominator. Therefore, divide the numerator by the denominator (shown in the margin) and write the remainder as a fraction of the divisor. x 1 2 x 2x 2 x 3 3 0x x 2 2x 2x 1 840 Figure 12.C-6 Decomposition into Partial Fractions 3 x 3 x 2 x 2 x 2x 1 2 x 2 Now decompose x discussed in Example 1. x 1 2x 1 x 2 x 2 into A x 2 B x 1 by using the procedure 2x 1 A 1 B x 1 x 2 2 Ax A Bx 2B A B x A 2B 2B 1 A 1 and B 1. As shown in Figure 12.C-6 ■ When the denominator contains a power of a linear or a quadratic factor, every integral power of that factor must be taken into consideration when finding partial fractions, as outlined below. 1. Divide numerator by denominator if the fraction is improper, and find partial fractions of the remainder. 2. Factor the denominator into factors of the form (px q)m (ax2 bx c)n, and over the set of real numbers. where ax2 bx c is irreducible 3 |
. For each linear factor of the form (px q)m, the partial fraction must include the following sum: A1 px q A2 (px q)2 p Am (px q)m (ax2 bx c)n, the 4. For each quadratic factor of the form partial fraction must include the following sum: B2x C2 (ax2 bx c)2 B1x C1 ax2 bx c Bnx Cn (ax2 bx c)n p Exercises In Exercises 1–7, find the partial fraction decomposition of each expression. 5. 5x x 2 1 3 1 x 2 2 10x 8 3 6x 6. x 1. 3. x 2 3x 2 x 2x 1 2 3x 18 3 4x x 2. 4. 1 2 1 x 2x 7. 2 x 3 3 4x x 2 2x 3 2 x 21 x 3 x 2x 2 8x 4 841 C H A P T E R 13 Statistics and Probability What are the odds? Suppose a dart player can hit the bulls-eye about 25% of the time. How likely is the event shown above? The number of bulls-eyes in a given number of tries can be described as a binomial experiment, and the probabilities can be easily calculated. See Exercises 5–8 in Section 13.4.A. 842 Chapter Outline Basic Statistics 13.1 13.2 Measures of Center and Spread 13.3 Basic Probability 13.4 Determining Probabilities 13.4.A Excursion: Binomial Experiments 13.5 Normal Distributions Chapter Review can do calculus Area Under a Curve Interdependence of Sections 13.1 > 13.2 13.3 > 13.4 13.5 > > S tatistics and probability are essential tools for understanding the modern world. Both involve studying a group of individuals or objects, known as a population, and a subset of the population, known as a sample. In statistics, information from a sample is used to draw conclusions about the population. In probability, information from the population is used to draw conclusions about a sample. Population statistics probability Sample 13.1 Basic Statistics Objectives • Identify data types • Create displays of qualitative and quantitative data • Describe the shape of a distribution data qualitative quantitative discrete continuous Statistics is used to make sense of information, or data, by using techniques to organize, summarize, and draw conclusions |
from the data. Most statistical data is gathered by taking a random sample of the population. In a random sample, all members of the population and all groups of members of a given size have an equal chance of being in the sample. Data can be divided into two types: qualitative and quantitative. Quantitative data is numerical, such as “the number of hours spent studying each night” or “the distance from home to school.” Qualitative data can be divided into categories, such as “liberal,” “moderate,” “conservative,” or “blue eyes,” “brown eyes.” Quantitative data can be further classified as either discrete or continuous. If the difference between two values can be arbitrarily small, the data is continuous. If there is a minimum increment between two different values, the data is discrete. 843 844 Chapter 13 Statistics and Probability Example 1 Types of Data In each example, identify the data as either qualitative or quantitative. If quantitative, then identify it as discrete or continuous. a. the height of each player on a basketball team b. the style of shoes worn by each student in a classroom c. the number of people in each household in the United States Solution a. The data is quantitative, because each value can be written as a number, such as 68.32 inches. It is continuous, because there is no minimum difference between two values; two players could have heights 68.32 inches and 68.33 inches, or 68.323 inches and 68.324 inches. b. The data is qualitative, since it can be grouped into categories, such as tennis shoes, sandals, and high heels. c. The data is quantitative, because each value is a number. It is discrete, because there is a minimum difference of 1 between two different values; two households could have 4 and 5 members, but not 4 and 4.1 members. ■ NOTE Continuous data is sometimes treated as discrete, and vice versa. For example, heights are usually rounded to the nearest inch, so there is a minimum difference of 1 inch between measurements. In the discrete case, for amounts of money, the minimum increment of $0.01 is so small that the data can often be treated as continuous. Data Displays One of the most important uses of statistics is to organize data and display it visually. Most displays show the data values or categories and some measure of how often each value or category occurs. The number of times a |
value occurs is known as the frequency of that value. If the frequency is divided by the total number of responses, the result is the relative frequency of that value, which can be expressed as a fraction, a decimal, or a percent. A frequency table displays the categories with frequencies, relative frequencies, or both. Example 2 Frequency Table A group of 30 people were asked their favorite flavor of ice cream. Of these, 6 chose vanilla, 12 chose chocolate, 4 chose butter pecan, and 8 chose mint chocolate chip. Create a table with frequencies and relative frequencies for each flavor. Section 13.1 Basic Statistics 845 Solution Flavor Frequency Relative frequency Vanilla Chocolate Butter pecan Mint chocolate chip 6 12 4 8 6 30 1 5 12 30 2 5 0.2 20% 0.4 40% 4 30 2 15 8 30 4 15 0.13 13% 0.27 27% ■ Displaying Qualitative Data Two common ways of displaying qualitative data are bar graphs and pie charts. A bar graph displays the categories on a horizontal axis and the frequencies or relative frequencies on a vertical axis, or vice versa. The height or length of each bar shows the frequency of the value. All bars should have the same width. Example 3 Bar Graph Use the data in the frequency table from Example 2 to make a bar graph. Solution y c n e u q e r F 14 12 10 8 6 4 2 0 Vanilla Chocolate Butter pecan Mint chocolate chip Flavor ■ A pie chart displays the categories and their relative frequencies. The “pie” is divided into sectors whose central angle measure equals the fraction of represented by the relative frequency of each category. 360° The central angle measure of the sector that represents a category with relative frequency r is r 360°. 846 Chapter 13 Statistics and Probability Example 4 Pie Chart Create a pie chart using the data in the frequency table from Example 2. Label each sector with the category and its relative frequency. Solution The central angle measures of the categories of the sectors are: Vanilla: 0.2 360° 72° Chocolate: 0.4 360° 144° Butter pecan: 0.13 360° 47° Mint chocolate chip: 0.27 360° 97° Mint chocolate chip 0.27 Vanilla 0.20 72° 97° 47° 144° Butter pecan 0.13 Chocolate 0.40 ■ Displaying Quantitative Data Numerical data can also be displayed in a variety of ways to indicate each value and its frequency. The shape of a smooth curve over the display indicates characteristics of |
the data values. The most common distribution shapes are shown below. Uniform: All the data values have approximately the same frequency. Symmetric: The right and left sides of the distribution have frequencies that are mirror images of each other. Skewed right: The right side of the distribution has much lower frequencies than the left. Skewed left: The left side of the distribution has much lower frequencies than the right. NOTE Before statistical analysis, numerical data is usually arranged in order from the lowest value to the highest value. An arrangement of numerical data is called a distribution. Common Distribution Shapes NOTE If an outlier is caused by an error in measurement or other type of error, it is usually removed from the data set before further analysis. In general, an outlier should not be removed without justification. Section 13.1 Basic Statistics 847 An outlier is a data value that is far removed from the rest of the data, which usually indicates that the value needs investigation. Outliers may be caused by errors or by unusual members of the population. Example 5 The Shape of Data From the four given shapes, choose the best distribution for the data. a. the last digit of each number in the phone book b. the salaries of the employees of a corporation c. the age of retirement for all people in the U.S. d. the heights of all adult women in the U.S. Solution a. The last four digits of a phone number are assigned randomly, so all digits have about the same frequency. The distribution is uniform. b. In a typical corporation, most employees earn relatively low salaries while a few executives make high salaries. The distribution is skewed right. c. Few people make enough money to retire young, and most people retire in their 60’s or later. The distribution is skewed left. d. The average height of an adult woman is at the middle of the distribution, which is symmetric with respect to this value. ■ Two common displays of quantitative data are the stem plot and the histogram. A stem plot is commonly used to display small data sets. The data below shows 31 test scores for a class exam: 32, 67, 89, 90, 87, 72, 75, 88, 95, 83, 97, 72, 85, 93, 79, 63 70, 87, 74, 86, 98, 100, 97, 85, 77, 88, 92, 94, 81, 76, 64 The stem plot for the class test data is shown below: 3 4 |
5 6 7 8 9 10 Figure 13.1-1 6 0 3 The entry sents the scores 63, 64, and 67. There are 31 scores represented. represents the score 32. Similarly, the row 3 0 2 4 7 repre- 848 Chapter 13 Statistics and Probability Note that the score of 32 is far below the remaining data and so it could be an outlier in this distribution. It may be the score of a student who didn’t study for the test. However, it is most likely not an error in measurement, so the value cannot be removed from the data set. The distribution is skewed left (Figure 13.1-1). Creating a Stem Plot To create a stem plot: 1. Choose the leading digit or digits to be the stems. Arrange the stems vertically from lowest to highest value from top to bottom. 2. The last digit is the leaf. Record a leaf for each data value on the same horizontal line as its corresponding stem. Arrange the leaves from lowest to highest value from left to right. 3. Provide a key that indicates the total number of data elements and an interpretation of one stem and leaf indicating appropriate units. Example 6 Stem Plot A company uses a 3-minute recorded phone message to advertise its product. A random sample of 40 calls is used to determine how much of the message was heard before the listener hung up. Create a stem plot of the data below and discuss its shape. 2.4, 0.2, 3.0, 2.8, 1.5, 1.9, 0.7, 1.0, 2.5, 1.3, 0.8, 2.1, 3.0, 0.4, 1.2, 3.0, 1.1, 0.3, 0.7, 1.8, 0.3, 1.0, 2.1, 3.0, 2.9, 0.5, 1.4, 3.0, 2.8, 1.2, 0.5, 0.5, 1.5, 0.9, 1.8, 0.6, 0.6, 0.7, 0.8, 0.8 Solution Key: 0 8 represents a time of 2 2.8 minutes. 40 times are represented. The distribution is skewed right, as shown in Figure 13.1-2. The values on the right side have lower frequencies than the values on the left. Notice that the |
distribution is cut off at 3, because no phone calls last longer than 3 minutes, the length of the entire message. ■ A histogram, which can be thought of as a bar graph with no gap between adjacent bars, is often used with large sets of quantitative data. First, the Figure 13.1-2 Section 13.1 Basic Statistics 849 data is divided into a convenient number of intervals of equal width. The frequencies (or relative frequencies) of the data in the intervals are the heights of the rectangles. For example, the test scores given on page 847 can be represented by the histogram in Figure 13.1-3 12 10 8 6 4 2 0 0 10 20 30 40 50 60 70 80 90 100 Score Figure 13.1-3 Here, the test scores have been divided into 10-point intervals, 0 through 9, 10 through 19, and so on. The histogram indicates, for example, that there are no scores between 50 and 59, eight scores between 70 and 79, and one score of 100. Each bar on the graph has width 10, so the class interval is said to be 10. Creating a Histogram To create a histogram: Technology Tip Most graphing calculators will produce histograms. Check the Xmin, Xmax, and Xscl to make sure the class intervals are relevant to the data. If you do not know how to create a histogram with a graphing calculator, refer to the Technology Appendix. 1. Divide the range of the data into classes of equal width, so that each data value is in exactly one class. The width of these intervals is called the class interval. 2. Draw a horizontal axis and indicate the first value in each class interval. 3. Draw a vertical scale and label it with either frequencies or relative frequencies. 4. Draw rectangles with a width equal to the class interval and height equal to the frequency of the data within each interval. Example 7 Histogram Create a histogram of the following scores. 580, 490, 590, 390, 410, 370, 470, 540, 490, 660, 500, 670, 430, 670, 490, 720, 580, 680, 590, 480, 560, 480, 400, 440, 560, 540, 330, 490, 540, 540, 520, 650, 540, 600, 630, 580, 540, 500, 270, 600, 390, 540, 300, 350, 600, 540, 510, 410, 370, 390, 160, |
500, 740, 510, 540, 560, 510, 430, 440, 590, 560, 510, 600, 460, 450, 510, 420, 430, 560, 680, 610, 600, 600, 520, 480, 490, 320, 450, 500, 490 850 Chapter 13 Statistics and Probability Solution The smallest value is 160 and the largest value is 740, so the range of the data is 580 points. A convenient choice for the classes is 150 through 199, 200 through 249, 250 through 299, and so on—a class interval of 50 points. The frequency table below shows how many data values are in each class. Class 150– 199 200– 249 250– 299 300– 349 350– 399 400– 449 450– 499 500– 549 550– 599 600– 649 650– 699 700– 749 Frequency 1 0 1 3 6 9 13 20 11 8 6 2 The histogram is shown in Figure 13.1-4. The shape is approximately symmetric 25 20 15 10 5 0 150 200 250 300 350 400 450 500 550 600 650 700 Score Figure 13.1-4 ■ Figure 13.1-5 shows three histograms of the data in Example 7, created on a calculator. Notice that the choice of the scale on the x-axis determines the width of the class intervals. Which do you think is the best representation of the data? 12 25 150 0 Xscl 25 800 150 0 800 150 Xscl 50 Figure 13.1-5 50 0 800 Xscl 100 Remember that it is more important to be able to interpret a distribution than it is to simply produce the display. Technology can easily produce a histogram, but the purpose of the display is to help interpret the data. Section 13.1 Basic Statistics 851 Exercises 13.1 In Exercises 1–4, identify the population and the sample. Exercise Frequency Relative frequency 1. There are three schedule options for classes at a high school: 90-minute classes every other day for a year, 90-minute classes every day for a semester, or 45-minute classes every day for a year. Out of 1200 students, 50 students from each grade level are chosen at random and asked their preference. Aerobics Kickboxing Tai chi Stationary bike 20 8 8 14???? 2. The manager of a convenience store wishes to determine how many cartons of eggs are damaged in shipment and delivery. For ten shipments of 1000 cartons of eggs, she |
examines every cracked eggs. carton to see how many cartons contain 50th 3. A survey is taken to determine the number of pets in the typical American family. A computer is used to randomly select 5 states, then 10 counties in each state, then 50 families in each county. Each of these families is asked how many pets they own. 4. A biologist tranquilizes 400 wild elephants and measures the lengths of their tusks to determine their ages. 11. Complete the table to show the relative frequency for each category. 12. Create a bar graph for the data with the vertical axis showing the frequencies. 13. Create a bar graph with the vertical axis showing the relative frequencies. 14. Create a pie chart for the data. For Exercises 15–18, suppose 25 people are asked their favorite color. The results are: 6 red, 8 blue, 5 purple, 4 green, 1 yellow, and 1 orange. 15. Create a frequency table for the given data. Include relative frequencies. Use the descriptions in Exercises 1–4 to answer Exercises 5–10. 16. Create a bar graph for the data with the vertical axis showing the frequencies. 5. Determine whether the data in each description is qualitative or quantitative. If the data is quantitative, determine whether it is continuous or discrete. 17. Create a bar graph with the vertical axis showing the relative frequencies. 18. Create a pie chart for the data. 6. Describe two ways in which the data from Exercise 1 could be displayed. 7. How large is the sample in Exercise 2? 8. Describe two ways in which the data from Exercise 2 could be displayed. In Exercises 19–24, state whether the shape of the distribution is best described as uniform, symmetric, skewed right, or skewed left. 19. The scores of a national standardized test 20. The age at which students get a driver’s license 9. How large is the sample in Exercise 3? 21. 100 10. Which would be more appropriate to display the data from Exercise 4: a stem plot or a histogram? Explain your reasoning. The following frequency table gives the preferred type of exercise for 50 women at a local gym. 80 60 40 20 0 852 Chapter 13 Statistics and Probability 22 23. The position of the second hand of a clock at 100 randomly chosen times in a 12-hour period 24. 8 9 10 11 12 In Exercises 25–28, create a stem plot for the given data |
. 25. 23, 45, 38, 41, 24, 67, 42, 46, 51, 33, 43, 47, 54, 49, 47, 36, 27, 33, 41, 29 26. 1.8, 2.0, 1.4, 5.6, 1.1, 2.6, 0.8, 1.5, 1.4, 2.6, 0.7, 1.6, 0.4, 1.1, 0.5, 1.3 27. 98, 87, 100, 86, 92, 78, 56, 100, 90, 88, 93, 99, 76, 83, 86, 91, 72, 85, 79, 81, 82, 91, 86, 70, 84 During summer semester, a community college surveyed its students to determine travel time to campus. A random sample of 30 students gave the following times in minutes: 12 15 32 42 12 25 55 40 65 28 48 75 18 17 42 45 60 35 25 37 6 27 22 35 45 90 30 8 40 55 During fall semester, the survey was repeated. The new times in minutes are: 63 46 9 35 104 52 40 25 31 7 43 29 40 69 52 48 20 20 86 55 32 75 46 63 29 14 48 37 17 14 28. Create a stem plot of the summer semester data. 29. Does the data set of the summer semester times contain any outliers? Explain. 30. Create a stem plot of the fall semester data. 31. Compare the shape of the two data sets. Which type of distribution do you think best describes these data sets? What might explain the differences in these data sets? 32. Create a histogram with a class interval of 10 for the data below. 68, 84, 59, 72, 62, 76, 61, 63, 68, 56, 70, 79, 54, 65, 66, 71, 70, 58, 63, 68, 84, 63, 53, 68, 63, 76, 66, 66, 70, 72, 88, 68, 75, 63, 76, 58, 86, 65, 66, 73, 53, 76, 59, 81, 59, 65, 67, 73, 62, 75, 89, 58 33. Create a histogram for the following ACT scores. Be sure to choose an appropriate class interval. 14, 25, 15, 18, 17, 11, 15, 10, 25, 6, 11, 4, 12, 24, 19 |
, 14, 20, 13, 23, 19, 13, 20, 14, 24, 10, 18, 30, 22, 16, 26, 10, 23, 22, 19, 23, 21, 16, 18, 18, 20, 25, 14, 19, 7, 16, 18, 31, 14, 7, 10, 16, 13, 18, 10 34. Create a histogram with a class interval of 5 for the data in the stem plot below 35. Critical Thinking How does the shape of the histogram you created in Exercise 34 compare to the shape of the data in the stem plot? Which do you think is a better representation of the data, and why? Section 13.2 Measures of Center and Spread 853 13.2 Measures of Center and Spread Objectives • Calculate measures of center • Calculate measures of spread • Choose the most appropriate measure of center or spread • Create and interpret a box plot While the shape of a stem plot or a histogram gives a picture of a data set, numerical measures are more precise and can be calculated easily (using technology for large data sets). These measures help to further summarize and interpret data. Two quantities are commonly used to describe a data set: a measure of the “center” of the data and a measure of how spread out the data is. Measures of Center A sample may contain hundreds, or even thousands of data values. This information is often summarized by one value that represents the center, or central tendency, of the data. The three most common measures of center are mean, median, and mode. Mean The mean is more commonly known as the average. The mean is calculated by adding all values and dividing by the total number of values. © Recall from Chapter 1, the symbol is used to indicate the sum of a set of values. mean: x x1 x2 p xn n ©xi n x, The mean is represented by x1, the data variable, while each data value is represented as so on. The sum is divided by n, the number of data elements. read as “x bar.” The x is used to represent and x3, x2, Example 1 Mean Number of Accidents A six-month study of a busy intersection reports the number of accidents per month as 3, 8, 5, 6, 6, 10. Find the mean number of accidents per month at the site. Solution x1 x 3, x2 ©xi n 8, x3 5, |
x4 3 8 5 6 6 10 6 6, x5 6, x6 38 6 10 6.3 The data shows an average of 6.3 accidents per month at the given intersection. ■ One problem with the mean as a measure of center is that it may be distorted by extreme values, as shown in the following example. 854 Chapter 13 Statistics and Probability Example 2 Mean Home Prices In the real-estate section of the Sunday paper, the following houses were listed: 2-bedroom fixer-upper: 2-bedroom ranch: 3-bedroom colonial: 3-bedroom contemporary: 4-bedroom contemporary: 8-bedroom mansion: $98,000 $136,700 $210,000 $289,900 $315,500 $2,456,500 Find the mean price, and discuss how well it represents the center of the data. Solution x ©xi n 98,000 136,700 210,000 289,900 315,500 2,456,500 6 3,506,600 6 584,433.33 In the data set, 5 out of the 6 values are below $350,000, but the mean is over $550,000, so the mean does not seem to be a very good representation of the center of the data set. Notice that the value $2,456,500 is more than twice the rest of the data combined. Thus, it has a very large effect on the mean, “pulling” it away from the other values. ■ Median As shown in Example 2, the mean is not always the best way to represent the center of a distribution. If the distribution is skewed or contains extreme values, the median, or middle value of the data set is often used. To determine the median, the data must be in order from smallest to largest (or largest to smallest). If the number of values is odd, then one number will be the middle number, as shown below. 3, 4, 7, 8, 9, 11, 15 There are 7 values, and the median, which is in the position, is 8. Notice that three values are less than the median and three values are greater than the median. 4th If the number of values is even, there are two middle numbers, as shown below. 17, 22, 24, 30, 35, 40 There are 6 values, and the median is the average of the middle numbers, 27. positions. So the median is which are in the and 3rd 4th 24 30 |
2 Notice that three values are less than the median and three values are greater than the median. Median Section 13.2 Measures of Center and Spread 855 p xn x3 x2, x1, If are ordered from smallest to largest, then the median is the middle entry when n is odd and the average of the two middle entries when n is even. for n odd, the value in the n 1 2 position median for n even, the average of the values in the n 2 and n 2 1 positions d The median is said to be a more resistant measure of center than the mean, since it is less affected by a skewed distribution or extreme values in a data set. Example 3 Median Home Prices Find the median of the data in Example 2, and discuss how well it represents the center of the data. Solution The data is already in order from smallest to largest, with n 6. 98,000 1st position 136,700 2nd position 210,000 3rd position 289,900 4th position 315,500 5th position 2,456,500 6th position The median is the average of the values in positions which are 210,000 and 289,900, so n 2 3 and n 2 1 4, median 210,000 289,900 2 499,900 2 249,950 A price of $249,950 is much more representative of the houses in this listing. The most expensive house does not have the same strong effect that it had in the calculation of the mean. ■ Mode The mode is the data value with the highest frequency. It is most often used for qualitative data, for which the mean and median are undefined. The mode can be thought of as the “most typical” value in the data set. NOTE If every value in a data set occurs the same number of times, there is no mode. If two or more scores have equal frequencies that are higher than those of all other values, the data set is called bimodal (two modes), trimodal (three modes), or multimodal. 856 Chapter 13 Statistics and Probability Example 4 Mode of a Data Set Find the mode of the data represented by the bar graph below 14 12 10 8 6 4 2 0 Purple Orange Red Favorite color Green Blue Solution The height of each bar represents the frequency, so the mode is the category with the tallest bar, which is red. ■ Mean, Median, and Mode of a Distribution Recall the shapes of symmetric, skewed left, and skewed right distributions. y |
median mode y y mode mode median median x x x mean symmetric mean skewed left Figure 13.2-1 mean skewed right The mean is the balance point of a distribution. Notice that on a skewed distribution, the mean moves toward the tail to balance out the “weight” of the outlying data. The median divides the area under the distribution into 2 equal areas. The mode is the highest point on the distribution. • If a distribution is symmetric, then the mean and median are equal. • If a distribution is skewed left, then the mean is to the left of the median. • If a distribution is skewed right, then the mean is to the right of the median. Technology Tip If necessary, see the Technology Appendix to learn how to find the mean and median of a data set. Section 13.2 Measures of Center and Spread 857 Calculator Exploration Use the statistics functions of your calculator to find the mean and median of the data represented by the following stem plot. 6|7 represents a score of 67 points. 43 scores are represented. 3 4 5 6 7 8 9 10 Key Measures of Spread Finding the shape and center of a data set still gives an incomplete picture of the data. The following stem plots show three data sets with a symmetric distribution and center 105. NOTE If the mean and median of a distribution are the same value, as in the data sets represented by the stem plots at right, their value is often referred to as the center. 6 7 8 9 10 11 12 13 14 10 11 12 13 14 10 11 12 13 14 1 5 1 3 5 7 9 5 9 The data has a different spread in each stem plot. The spread of the data, or variability, is an important characteristic of a data set. The second plot has the most variability because the data is very spread out, while the third has the least because the date is clustered very near the center. The three most common measures of spread are the standard deviation, the range, and the interquartile range. Standard Deviation The standard deviation of a data set is the most common measure of variability. It is best used if the data is symmetric about a mean. Standard deviation measures the average distance of a data element from the mean. x. is the difference, from the mean x xi The deviation of a data value xi Consider the following data set: The mean of the data is 2, 5, 7, 8, 10 2 5 7 8 10 5 6.4. The points are shown with |
their deviations on a number line in Figure 13.2-2. 858 Chapter 13 Statistics and Probability NOTE The distance from xi x data value to the mean is the absolute value of the deviation. deviation = 0.6 deviation = −1.4 deviation = 1.6 deviation = −4.4 deviation = 3.6 0 1 2 3 4 5 6 6.4 7 8 9 10 Figure 13.2-2 The average of the deviations is 4.4 1.4 1 2 0.6 1.6 3.6 5 0, because the positive and negative values cancel each other out. To avoid this, each deviation is squared, then the average is found. This quantity is called the variance. The square root of the variance is the standard deviation, denoted by the Greek letter s (sigma). For the data set {2, 5, 7, 8, 10}, first square each deviation. 4.4 2, 2 1 1.4 2 51 2, 0.62, 1.62, 3.62 5 6 19.36, 1.96, 0.36, 2.56, 12.96 6 Average the squared deviations to find the variance, s2 19.36 1.96 0.36 2.56 12.96 5 s2 : 7.44 Take the square root of the variance to find the standard deviation: s 27.44 2.73 Population versus Sample If data is taken from a sample instead of the instead of n when averentire population, it is common to divide by aging the squared deviations. The result is called the sample standard deviation and is denoted by s. For large data sets, the sample standard deviation is very close to the population standard deviation. n 1 Standard Deviation To find the standard deviation of a data set with n values, 1. Subtract each value from the mean to find the deviation. 2. Square each deviation, and find the mean of the squared n 1 deviations. If the data is from a sample, divide by instead of n. The result is called the variance. 3. Take the square root of the variance. These steps are summarized in the following formulas: Population standard deviation s B π(xi x)2 n Sample standard deviation π(xi x )2 s B n 1 Section 13.2 Measures of Center and Spread 859 Example 5 Standard Deviation Find the population standard deviation of the data in the first stem plot on page 857 using the formula |
. Then use a calculator to find the population standard deviations of the data in the other two plots. Solution For the first stem plot, squared deviations. x 105 and n 9. The following table shows the xi xi 85 91 x 20 14 xi 1 x 2 2 400 196 99 6 36 101 4 16 105 109 111 119 125 0 0 4 16 6 36 14 20 196 400 s 400 196 36 16 0 16 36 196 400 9 B 1296 9 B 12 An informal interpretation is that the average distance from the data values to the mean is 12 units. The population standard deviations of the data in the second and third stem plots are shown in Figure 13.2-3. Figure 13.2-3 Notice that the second stem plot, which is the most spread out, has the largest standard deviation, and the third stem plot, which is the most clustered together, has the smallest standard deviation. ■ Range The range is the difference between the maximum and minimum data values. The main advantage of the range is that it is easy to compute. Example 6 The Range Find the range of the data in each stem plot on page 857. 860 Chapter 13 Statistics and Probability Solution The range of the data in the first stem plot is The range of the data in the second stem plot is The range of the data in the third stem plot is 125 85 40. 145 65 80. 119 91 28. ■ Interquartile Range Like the mean, the standard deviation and range are strongly affected by extreme values in the data. The interquartile range is a measure of variability that is resistant to extreme values, yet gives a good indication of the spread of the data. Recall that the median is the middle value of the data set. Thus, the median divides the data into two halves, the lower half and the upper half. The quartiles further divide the data into fourths. The quartile, is the is the median of the upper median of the lower half. The half. (The median may be considered to be the quartile.) quartile, Q1, Q3, 2nd 3rd 1st Figure 13.2-4 shows the quartiles for n even or n odd. lower half upper half n even: 3, 5, 7, 9, 11, 12, 13, 16, 17, 19, 23, 27, 29, 31 Q1 lower half Q3 upper half n odd: 2, 3, 6, 8, 9, 14, 15, 16, 20, 21 |
, 23, 26, 28, 30, 33 Q1 Q3 Figure 13.2-4 The interquartile range is the difference between the quartiles, IQR Q3 Q1 which represents the spread of the middle 50% of the data. A value that is less than considered an outlier, as shown in Figure 13.2-5. IQR Q1 1 2 or greater than 1.5 Q3 1.5 IQR is 2 1 1.5 IQR IQR median 1.5 IQR outlier Q1 Q3 Figure 13.2-5 outlier Section 13.2 Measures of Center and Spread 861 Example 7 Interquartile Range Find the interquartile range of the data in the first stem plot on page 857. Solution The quartiles of the data in the first stem plot are shown below. 85 91 99 101 105 109 111 119 125 = Q1 91 + 99 2 = 95 = Q3 111 + 119 2 = 115 The interquartile range is 115 95 20. ■ Calculator Exploration Use a graphing calculator to find the interquartile range of the data in each of the other two stem plots on page 857. Five-Number Summary and Box Plots The five-number summary of a data set is the following list: minimum, Q1, median, Q3, maximum These values are used to construct a display called a box plot, as follows: 1. Construct a number line and locate each value of the five-number sum- mary. minimum Q1 median Q3 maximum 2. Construct a rectangle whose length equals the interquartile range, with a vertical line to indicate the median. minimum Q1 median Q3 maximum 862 Chapter 13 Statistics and Probability Technology Tip See the technology appendix, if necessary, for instruction on constructing a box plot using a graphing calculator. 3. Construct horizontal whiskers to the minimum and maximum values. minimum Q1 median Q3 maximum If a data element is an outlier, it may be marked with a or other mark. The whiskers then extend to the farthest value on each side that is not an outlier. Example 8 Constructing a Box Plot Construct a box plot for the data in the first stem plot on page 857. Solution The five-number summary of the data is 85, 95, 105, 115, 125. The box plot is shown below. 80 85 90 95 100 105 110 115 120 125 ■ Calculator Exploration Use a graphing calculator to construct a box |
plot for the data in each of the other two stem plots on page 857. Exercises 13.2 In Exercises 1–4, find the mean of each data set. 6. Find the median of the data set in Exercise 2. 1. 23, 25, 38, 42, 54, 57, 65 7. Find the median of the data set in Exercise 3. 2. 3, 5, 6, 2, 10, 9, 7, 5, 11, 6, 4, 2, 5, 4 8. Find the median of the data set in Exercise 4. 3. 3.6, 7.2, 5.9, 2.8, 21.6, 4.4 4. 78, 93, 87, 82, 90 5. Find the median of the data set in Exercise 1. 9. Find the mean, median, and mode of the following data set: 13, 13, 12, 6, 14, 9, 11, 19, 13, 9, 7, 16, 11, 12, 15, 12, 11, 12, 14, 9, 11, 13, 17, 13, 13 5910ac13_842-903 9/21/05 2:32 PM Page 863 Section 13.2 Measures of Center and Spread 863 In Exercises 10–13, find the mode of the data set represented by each display. 10. 30 25 20 15 10 5 0 A 11 12. C y D 8 6 4 2 24. 6, 8, 4, 11, 8, 8, 9, 6, 6, 8, 8, 12, 10, 10, 7 25. 50, 72, 86, 92, 86, 77, 57, 80, 93, 74, 53, 69, 65, 57, 73, 60, 66, 94, 81, 81 26. Find the range of the data set in Exercise 22. 27. Find the range of the data set in Exercise 23. 28. Find the range of the data set in Exercise 24. 29. Find the range of the data set in Exercise 25. 30. Find the interquartile range of the data set in Exercise 22. 31. Find the interquartile range of the data set in Exercise 23. 32. Find the interquartile range of the data set in Exercise 24. 0 1 2 3 4 5 6 x 33. Find the interquartile range of the data set in Exercise 25. 13. Vegetables Fruit Gr |
ains Dairy Meat Fat For each distribution shape, indicate whether the mean is larger, the median is larger, or the mean and median are equal. 14. symmetric 15. skewed left 16. skewed right 17. uniform Find the population standard deviation of the following data sets without using a calculator. 18. 8, 9, 10, 11, 12 19. 6, 8, 10, 12, 14 34. Find the five-number summary of the data set in Exercise 22, and create a box plot for the data. 35. Find the five-number summary of the data set in Exercise 23, and create a box plot for the data. 36. Find the five-number summary of the data set in Exercise 24, and create a box plot for the data. 37. Find the five-number summary of the data set in Exercise 25, and create a box plot for the data. For Exercises 38–43, the wait times of 30 people in a doctor’s office are given below, rounded to the nearest five minutes: 40, 35, 65, 40, 40, 5, 50, 85, 30, 50, 60, 60, 10, 65, 15, 45, 20, 40, 45, 70, 70, 25, 40, 45, 70, 65, 45, 25, 15, 25 38. Construct a histogram of the data. Describe the shape of the data set. Based on the shape, discuss the relative positions of the mean and median. 39. Find the mean and median of the data set. 40. Which measure of central tendency is preferred 20. 10, 10, 10, 10, 10 21. 0, 5, 10, 15, 20 for this data set? Why? Use a calculator to find the population and sample standard deviations of the following data sets. 22. 3, 6, 3, 5, 7, 8, 2, 6, 3, 6, 8, 4, 8, 2, 6, 9 23. 24, 17, 18, 18, 19, 26, 19, 8, 25, 15, 17, 11, 27, 20 41. Find the sample standard deviation, range, and interquartile range of the data set. 42. Construct a box-plot of the data. 43. Explain why the sample standard deviation is or is not a good measure of dispersion for this data set. 864 Chapter 13 Statistics and Probability 44. During a baseball game, 9 players had 1 hit each, |
3 players had 2 hits each, and 6 players had no hits. Find the mean number of hits per player. Find the sample standard deviation and the population standard deviation of the data, and interpret your results. 45. A teacher has two sections of the same course. The average on an exam was 94 for one class with 20 students, while the average was 88 for the other class with 30 students. Find the combined average exam score. 46. The mean score of a class exam was 78, and the median score was 82. Sketch a possible distribution of the scores. 47. Over the last year, 350 lawsuits for punitive damages were settled with a mean settlement of $750,000 and a median settlement of $60,000. Sketch a possible distribution of the settlements. 48. A restaurant employs six chefs with the salaries in dollars shown below: 25,000, 27,000, 35,000, 105,000, 40,000, 45,000 Determine the mean and median salaries. 49. Which measure of center more accurately describes the “typical” salary at the restaurant in Exercise 48? 50. The speed of a computer is primarily determined by a chip in the CPU. A manufacturer tested 12 chips and reported the following speeds in megahertz units: 11.6, 11.9, 12.0, 12.0, 14.0, 15.2, 13.0, 14.3, 13.6, 13.8, 12.8, 12.9 51. Create two data sets of five numbers each that have the same mean but different standard deviations. 52. Create two data sets of five numbers each that have the same standard deviations but different means. 53. Critical Thinking How is the mean of a data set affected if a constant k is added to each value? 54. Critical Thinking How is the standard deviation of a data set affected if a constant k is added to each value? 55. Critical Thinking How is the mean of a data set affected if each value is multiplied by a constant k? 56. Critical Thinking How is the standard deviation of a data set affected if each value is multiplied by a constant k? 57. Critical Thinking What must be true about a data set in order for the standard deviation to equal 0? 13.3 Basic Probability Objectives Definitions • Define probability and use properties of probability • Find the expected value of a random variable • Use probability density functions to estimate probabilities In the study of probability, an experiment is any process that generates |
one or more observable outcomes. The set of all possible outcomes is called the sample space of the experiment. Some examples of experiments and their sample spaces are shown in the following table. Section 13.3 Basic Probability 865 3 Figure 13.3-1 Experiment tossing a coin Sample space heads and tails, written as {H, T} rolling a number cube (Figure 13.3-1) {1, 2, 3, 4, 5, 6} choosing a name from the phone book all the names in the phone book counting the number of fish in a lake the set of non-negative integers An event is any outcome or set of outcomes in the sample space. For example, in the experiment of rolling a number cube, the set {1, 3, 5} is an event, which can be described as “rolling a 1, 3, or 5,” or simply “rolling an odd number.” The probability of an event is a number from 0 to 1 (or 0% to 100%) inclusive that indicates how likely the event is to occur. • A probability of 0 (or 0%) indicates that the event cannot occur. • A probability of 1 (or 100%) indicates that the event must occur. never happens 0 as likely as not 1 8 0.125 1 4 0.25 3 8 0.375 1 2 0.5 5 8 0.625 3 4 0.75 7 8 0.875 always happens 1 • The sum of the probabilities of all outcomes in the sample space is 1. • The probability of an event is the sum of the probabilities of the outcomes in the event. Probability Distributions The probability of an event E can be described by a function P, where the domain of the function is the sample space and the range of the function is the closed interval [0, 1]. denotes the probability of the outcome X, and X 2 denotes the probability of the event E. P P E 1 1 2 The rule of the function P can be described by a table, called a probability distribution. Example 1 Probability Distribution Suppose that 100 marbles are placed in a bag; 50 red, 30 blue, 10 yellow, and 10 green. An experiment consists of drawing one marble out of the bag and observing its color. a. What is the sample space of the experiment? 866 Chapter 13 Statistics and Probability NOTE Probabilities expressed as percents are often called chances. According to this probability distribution, there is a 50% chance of drawing a |
red marble, a 30% chance of drawing a blue marble, a 10% chance of drawing a yellow marble, and a 10% chance of drawing a green marble. b. Write out a reasonable probability distribution for this experiment, and verify that the sum of the probabilities of the outcomes is 1. c. What is the probability that a blue or green marble will be drawn? Solution a. The sample space is all possible outcomes: red, blue, yellow, green 6 5 b. A reasonable probability distribution is shown below, which is based on the relative frequency of marbles of each color. Color of marble Red Blue Yellow Green Probability 50 100 0.5 30 100 0.3 10 100 0.1 10 100 0.1 The sum of the probabilities of the outcomes is 0.5 0.3 0.1 0.1 1 c. The event “a blue or green marble will be drawn” can be written as the set of outcomes {blue, green}. The probability of the event is the sum of the probabilities for blue and green. blue, green P 15 P 1 62 blue P 1 2 green 2 0.3 0.1 0.4 ■ Mutually Exclusive Events Two events are mutually exclusive if they have no outcomes in common. Two mutually exclusive events cannot both occur in the same trial of an experiment. If two events E and F are mutually exclusive, then the probability of the event (E or F) is the sum of the individual probabilities The complement of an event is the set of all outcomes that are not contained in the event. The complement of event E can be thought of as “the event that E does not occur.” An event and its complement are always mutually exclusive, and together they contain all the outcomes in the sample space. Thus, the probability of an event and the probability of its complement must add to 1, which leads to the following fact. Probability of a Complement If an event E has probability p, then the complement of the event has probability 1 p. Section 13.3 Basic Probability 867 Example 2 Mutually Exclusive Events An experiment consists of spinning the spinner in Figure 13.3-2. The following table shows the probability distribution for the experiment. Outcome Probability A 0.4 S 0.3 C 0.2 E 0.1 a. Which of the following pairs of events E and F are mutually exclusive? E E A, C, E 6 5 a vowel 1 2 F F E 1 a vowel 2 |
F 5 6 6 C, S 5 in the first five letters 1 of the alphabet C 2 b. What is the complement of the event {A, S}? c. What is the probability of the event “the spinner does not land on A?” Solution A E C S Figure 13.3-2 a. The events E {A, C, E} and F {C, S} are not mutually exclusive because they have a common outcome, C. F E (a vowel) and The events alphabet) are not mutually exclusive because they have two common outcomes, A and E. E {C} are mutually exclusive (in the first five letters of the (a vowel) and The events because they have no common outcome. F b. The complement of the event {A, S} is the set of outcomes that are not in the event, {C, E}. c. The complement of the event “the spinner does not land on A” is {A}, which has a probability of 0.4. Thus, the probability of the event is 1 0.4 0.6. ■ Independent Events Two events are independent if the occurrence or non-occurrence of one event has no effect on the probability of the other event. For example, if an experiment is repeated several times under exactly the same conditions, the outcomes of the individual trials are independent. If two events E and F are independent, then the probability of the event is the product of the individual probabilities, P and 868 Chapter 13 Statistics and Probability NOTE The terms mutually exclusive and independent are often confused. Some important differences are detailed below. Mutually exclusive Independent The term often refers to two possible results for a single trial of a given experiment. The term often refers to the results from two or more trials of an experiment or from different experiments. The word “or” is often used to describe a pair of mutually exclusive events. The word “and” is often used to describe a pair of independent events. For mutually exclusive events E and F, E or For independent events E and F, E and If two events are mutually exclusive, they cannot be independent, because the occurrence of one would cause the other to have a probability of 0. Example 3 Independent Events The probability of winning a certain game is 0.1. Suppose the game is played on two different occasions. What is the probability of a. winning both times? b. losing both times? c. winning once and losing |
once? Solution a. The results of the two different trials are independent, so the probability of winning both times can be found by multiplying the probability of winning each time. winning both games P 1 2 0.1 0.1 0.01 b. Since losing is the complement of winning, the probability of losing 1 0.1 0.9. is by multiplying the probability of losing each time. The probability of losing both times can be found losing both games P 1 2 0.9 0.9 0.81 c. The complement of the event (winning once and losing once) is the set of the two events in parts a and b. The events in parts a and b are mutually exclusive, because it is impossible to win both times and lose both times, so their probabilities may be added. winning once and losing once P 1 1 1 2 0.01 0.81 0.18 2 ■ Section 13.3 Basic Probability 869 Random Variables In many cases, the characteristics of an experiment that are being studied are numerical, such as the total on a roll of two number cubes. In other cases, the outcomes of an experiment may be assigned numbers, such as heads 1, tails 0. A random variable is a function that assigns a number to each outcome in the sample space of an experiment. Example 4 Random Variable An experiment consists of rolling two number cubes. A random variable assigns to each outcome the total of the faces shown. a. Write out the sample space for the experiment. b. Find the range of the random variable. c. List the outcomes to which the value 7 is assigned. Solution a. The sample space may be written as a set of ordered pairs. (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1) (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4) (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (1, 6) (2, 6) (3, 6) (4 |
, 6) (5, 6) (6, 6) b. The smallest possible value is 2, which is assigned to the outcome (1, 1). The largest possible value is 12, which is assigned to the outcome (6, 6). The range is the set of integers from 2 to 12. c. The value 7 is assigned to the outcomes (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). ■ Expected Value of a Random Variable The expected value, or mean, of a random variable is the average value of the outcomes. In the experiment of rolling two number cubes, suppose the experiment was repeated 10 times, resulting in the following values for the random variable: 8, 5, 8, 6, 11, 11, 3, 9, 9, 7 The average value is 8 5 8 6 11 11 3 9 9 7 10 7.7. If the experiment is repeated a large number of times, the average approaches the expected value. A simulation was used to run a large number of trials, and the results are shown in Figure 13.3-3. The averages seem to be approaching 7, which is a reasonable estimate of the expected value. Average sum of two number cubes 7.3 6.98 7.26 7.005 6.978 Number of trials 100 200 300 400 500 Figure 13.3-3 870 Chapter 13 Statistics and Probability To calculate the expected value of a random variable from a probability distribution, multiply each value by its probability, and add the results. Example 5 Expected Value A probability distribution for the random variable in the experiment in Example 4 is given below. Find the expected value of the random variable. Sum of faces Probability 2 1 36 3 1 18 4 1 12 5 1 9 6 5 36 7 1 6 8 5 36 9 1 9 10 1 12 11 1 18 12 1 36 Solution Multiply each value by its probability, and add. 1 36b 2 a 3 1 18b a 4 1 12b a 5 1 9b a 6 5 36b a 7 1 6b a 8 5 36b a 9 1 9b a 10 1 12b a 11 1 18b a 12 1 36b a 7 ■ The expected value is not always in the range of the random variable, as shown in the following example. Example 6 Expected Value of a Lottery Ticket The probability distribution for a $1 instant-win lottery ticket is |
given below. Find the expected value and interpret the result. Solution Win $0 $3 $5 Probability 0.882746 0.06 0.04 $10 0.01 $20 $40 $100 $400 $2500 0.005 0.002 0.0002 0.00005 0.000004 0.882746 0 1 100 0.0002 2 3 0.06 1 400 1 2 0.04 5 2 0.00005 1 1 2 10 1 2 2500 1 20 0.01 2 0.000004 0.005 2 1 0.71 2 40 0.002 2 1 The average amount won is $0.71, though it is not possible to win exactly 71 cents on one ticket. However, since the ticket costs $1, there is an average net loss of $1 $0.71 $0.29 per play. ■ Section 13.3 Basic Probability 871 Probability Density Functions In Example 1, colored marbles were drawn from a bag and the probability of each color being drawn was determined by its relative frequency. red: 0.5 blue: 0.3 yellow: 0.1 green: 0.1 This probability distribution is displayed in a bar graph in Figure 13.3-4, in which each bar is 1 unit wide. Thus the area of each rectangular bar represents the probability of the corresponding color. The sum of the areas of the bars is 1. If rectangles in the bar graph have width 1 unit, then the area of each rectangle represents the probability of the corresponding category ll o Y n e r e G A function with the property that the area under the graph corresponds to a probability distribution is called a probability density function. Figure 13.3-4 Example 7 Discrete Probability Density Functions Draw a probability density function for the distribution in Example 5. Solution The probability density function is a piecewise-defined function, shown in Figure 13.3-5, where the height of each piece is the probability of the value on the left endpoint of the interval. The area of the shaded rectangle represents the probability that the sum is 9. y 6 36 4 36 2 36 x 0 1 2 43 5 6 7 8 9 10 11 12 13 Figure 13.3-5 ■ y When a random variable has infinitely many values within a certain interval, the probability distribution can be represented by a continuous density function, as in the following example. Example 8 Continuous Probability Density Function The probability density function in Figure 13.3 |
-6 can be used to estimate the probability that a customer calling a company’s customer service line will have to wait for a given amount of time. The area of each square on the grid is and the total area under the curve is 1. Estimate the probability that a customer will have to wait between 2 and 3 minutes. 0.5 0.05 0.025, x 0 1 2 43 5 6 7 (minutes) Figure 13.3-6 0.6 0.5 0.4 0.3 0.2 0..9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 872 Chapter 13 Statistics and Probability Solution is the area under The probability the curve between 2 and 3, which is shaded 1 2 in Figure 13.3-7. The area is approximately 3 squares on the grid, or 3 1 2 1 0.025 2 0.0875 Thus, the probability that a customer will have to wait between 2 and 3 minutes is about 0.0875. y 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 2 43 5 6 7 Figure 13.3-7 x ■ Exercises 13.3 Use the following probability distribution for Exercises 1–4. Suppose the experiment is repeated three times. Assume the trials are independent. Outcome Probability A 0.5 B 0.3 C? D? 1. List the sample space for the probability distribution. 2. Suppose that outcomes C and D have the same probability. Complete the probability distribution. 3. What is the probability of the outcome (A or B)? 4. What is the probability of the outcome (not A)? Exercises 5–8 refer to the spinner at the right. The probability of landing on black is 1 2, and the proba- bility of landing on red is 1 3. 5. Create a probability distribution for the experiment of spinning the spinner. 6. What is the probability it will land on black all three times? 7. What is the probability it will land on white all three times? 8. What is the probability of the outcome (black, white, red)? of the outcome (black, red, white)? What is the probability that it will land once on each color? A doctor has assigned the following chances to a medical procedure: full recovery condition improves no |
change condition worsens 55% 24% 17% 4% Suppose the procedure is performed on 5 patients. Assume that the procedure is independent for each patient. 9. What is the probability that all five patients will recover completely? 10. What is the probability that none of the patients will get worse? A bag contains red and blue marbles, such that the 21. What is the probability that at least one person is Section 13.3 Basic Probability 873 probability of drawing a blue marble is An exper-. 3 8 iment consists of drawing a marble, replacing it, and drawing another marble. The two draws are independent. A random variable assigns the number of blue marbles to each outcome. 11. What is the range of the random variable? absent? 22. Find the expected value of the random variable, and interpret the result. An experiment consists of planting four seeds. A random variable assigns the number of seeds that sprout to each outcome. 12. What is the probability that the random variable has an output of 2? 23. Complete the following probability distribution for the experiment. 13. What is the probability that the random variable has an outcome of 0? 14. What is the probability that the random variable has an outcome of 3? Sprouted Probability 0? 1 2 3 4 0.154 0.345 0.345 0.130 15. Create a probability distribution for the random 24. Find the expected value of the random variable, variable. and interpret the result. 16. Calculate the expected value of the random 25. Draw a probability density function for the variable. In Exercises 17–20, find the expected value of the random variable with the given probability distribution. 17. 18. 19. 20. Outcome 0 1 5 10 1000 Probability 0.43 0.32 0.24 0.10 0.01 Outcome 0 1 2 3 Probability 0.25 0.25 0.25 0.25 Outcome 15 16 17 18 19 20 Probability 0.1 0.3 0.2 0.2 0.1 0.1 Outcome Probability 2 1 4 3 1 2 4 1 4 An office employs 5 people. A random variable is assigned to the number of people absent on a given day. The probability distribution is given below. Absent 0 1 2 3 Probability 0.59 0.33 0.07 0.01 4 0 5 0 random variable. Shade an area of the graph that corresponds to the probability that 3 or more seeds will |
sprout. 26. Use the probability distribution to determine the probability that each seed will sprout, assuming that they are independent. Hint: if the probability that one seed will sprout is p, what is the probability that all four seeds will sprout? A random variable with a uniform distribution has a probability density function that is constant over the range of the variable, and 0 everywhere else. Use the graph of the probability density function shown below for Exercises 27–30 27. What is the range of the random variable? 28. What is the height h of the probability density function? 29. What is the probability that the random variable is between 2 and 4? 30. What is the probability that the random variable is greater than 4? 874 Chapter 13 Statistics and Probability y 0.5 0.4 0.3 0.2 0. 10 11 12 13 14 15 16 17 18 19 20 (inchesminutes) 9 10 The probability density function at left above models the number of inches of rainfall per year for a certain location. The area of each square on the grid is 0.01. 31. Estimate the probability that the rainfall for a certain year is between 14 and 16 inches. 32. Estimate the probability that the rainfall for a certain year is greater than 17 inches. 33. The median of a probability density function is the point that divides the area under the curve into two equal areas. Estimate the median rainfall, based on the given probability density function. In commuting to work, Joshua takes one bus, then transfers to another bus. The probability density function at right above models the total length of time that he has to wait for both buses. 34. What is the height of the probability density function at t 5 minutes? 35. What is the probability that Joshua has to wait for less than 5 minutes? 36. What is the probability that Joshua has to wait for between 3 and 6 minutes? 13.4 Determining Probabilities Objectives • Estimate probability using experimental methods • Estimate probability using theoretical methods The exact probability of a real event can never be known. Probabilities are estimated in two ways: experimentally and theoretically. Experimental Estimates of Probability Suppose an outcome of an experiment has a probability of 0.3. If the experiment were repeated many times, that outcome would occur in approximately 30% of the trials. In 100 trials, for example, it might occur 30 times, or maybe 28 times or 34 times. Statistical analysis shows, however, that it is |
unlikely that it would occur fewer than 20 times or more than 40 times. Figure 13.4-1 Section 13.4 Determining Probabilities 875 The basis for experimental estimates of probability may be summarized as follows: As the number of trials of an experiment increases, the relative frequency of an outcome approaches the probability of the outcome. Thus, if an experiment is repeated n times, the experimental estimate of the probability of an event is P E 2 1 number of trials with an outcome in E n Example 1 Experimental Estimate of Probability An experiment consists of throwing a dart at the target in Figure 13.4-1. Suppose the experiment is repeated 200 times, with the following results: red yellow blue 43 86 71 Write a probability distribution for the experiment. Solution The probabilities may be estimated using the experimental formula. Outcome Probability red 43 200 0.2 yellow 86 200 0.4 blue 71 200 0.4 ■ Probability Simulations In order to estimate probability using the experimental approach, a large number of trials is needed. Because this approach is often time-consuming, computer simulations that duplicate the conditions of a single trial are often used. Most graphing calculators have random number generators that can be used to simulate simple probability experiments. To simulate an experiment with a large number of trials, it is easiest to use a program that can keep track of the frequency of each outcome. Suppose an experiment consists of tossing three coins and counting the number of heads, and that the probability of heads for each coin is 0.5. The possible outcomes of the experiment are 0 heads, 1 head, 2 heads, or 3 heads. One trial can be simulated by a command that randomly generates three values, which can be either 0 or 1, and adds them (see the Technology Tip on page 876). The three random integers represent the three coins, where 1 represents heads and 0 represents tails. 876 Chapter 13 Statistics and Probability Technology Tip To randomly generate three values that can be 0 or 1 and add them: TI Casio sum (randInt (0, 1, 3)) Sum {Int 2Ran#, Int 2Ran#, Int 2Ran#} The following is a sample program for running a simulation with n trials of the experiment. The program displays a list of the probabilities for 0 heads, 1 head, 2 heads, and 3 heads. 1 22 S T 0, 1, 3 randInt Prompt N 0 S A: 0 S B: 0 S C: 0 S D For |
(K, 1, N, 1) sum 1 If T 0 A 1 S A If T 1 B 1 S B If T 2 C 1 S C If T 3 D 1 S D End N is the number of trials. T is the outcome of a single trial. A is the number of trials with 0 heads. B is the number of trials with 1 head. C is the number of trials with 2 heads. D is the number of trials with 3 heads. {A/N, B/N, C/N, D/N} Example 2 Probability Simulation Use the program to create a probability distribution for the experiment of tossing 3 coins and counting the number of heads. Assume that P(heads) P(tails) 0.5 for each coin. Solution The results will vary each time the program is run. Using the results from Figure 13.4-2, one approximate distribution is shown below. Outcome 0 heads 1 head 2 heads 3 heads Probability 0.14 0.38 0.36 0.12 Figure 13.4-2 ■ NOTE The assumption that all outcomes are equally likely was used in the probability simulation in Example 2. If P(heads) 1 2, then heads P(tails) and tails are equally likely. Also, for the given calculator commands, the outcomes 0 and 1 are equally likely each time. Probability for Equally Likely Outcomes Section 13.4 Determining Probabilities 877 Calculator Exploration Run the probability simulation in Example 2, using 100 trials. Are your results similar to those in the example? Compare your results to those of your classmates. Theoretical Estimates of Probability In the theoretical approach, certain assumptions are made about the outcomes of the experiment. Then, the properties of probability are used to determine the probability of each outcome. The most common assumption is that all outcomes are equally likely, that is, they have the same probability of occurring. For example, in tossing a coin it is usually assumed that heads and tails are equally likely. Since the probabilities must add to 1, the probability of each outcome must equal 1 2. This idea is used to develop the following formula. Suppose an experiment has a sample space of n outcomes, all of which are equally likely. Then the probability of each outcome is 1 n, and the probability of an event E is given by P(E) number of outcomes in E n Example 3 Rolling a Number Cube An experiment consists of rolling a number cube. Suppose that all outcomes are equally likely. a. |
Write the probability distribution for the experiment. b. Find the probability of the event that an even number is rolled. Solution a. The sample space consists of the 6 outcomes {1, 2, 3, 4, 5, 6}. If the outcomes are equally likely, then the probability of each is probability distribution for the experiment is 1 6. The Outcome Probability 878 Chapter 13 Statistics and Probability b. The event that an even number is rolled consists of the 3 outcomes {2, 4, 6}. Thus, the probability that an even number is rolled is 3 6 1 2. ■ Of course, the outcomes of an experiment are not always equally likely. Based on the probability simulation of tossing 3 coins in Example 2, the outcomes 0, 1, 2, and 3 heads do not seem to be equally likely. However, it is possible to determine the probability theoretically by considering each coin separately. Example 4 Theoretical Probability Use properties of probability to write a theoretical probability distribution for the experiment in Example 2. Solution Each coin has no effect on the other coins, so the outcomes of the coins are independent. Thus, the probabilities can be multiplied. The only outcome with 0 heads is TTT. The probability of tails for each coin is 0.5, so the probability of the outcome TTT is T T 0.5 0.5 1 T 0.5 2 21 0.125 There are three outcomes with 1 head: HTT, THT, and TTH. Each outcome has a probability of 0.125. For example, H T 0.5 0.5 1 T 0.5 2 21 0.125 The probabilities of the three outcomes can be added, so the probability of 1 head is 0.125 0.125 0.125 0.375. There are three outcomes with 2 heads: HHT, HTH, and THH. The probability of 2 heads is also 0.125 0.125 0.125 0.375. There is one outcome with 3 heads, HHH, which also has a probability of 0.125. The probability distribution is shown below. Outcome 0 heads 1 head 2 heads 3 heads Probability 0.125 0.375 0.375 0.125 ■ How do the theoretical probabilities obtained in Example 4 compare to the experimental ones you found in Example 2? Section 13.4 Determining Probabilities 879 Counting Techniques The probability formula for equally likely outcomes uses the size of the sample space. In simple experiments, this may be easily |
determined, but some experiments require more sophisticated counting techniques. The basis of most counting techniques is the Fundamental Counting Principle, which is also known as the Multiplication Principle. n1 Consider a set of k experiments. Suppose the first experiment outcomes, and so on. Then outcomes, the second has has n2 the total number of outcomes is experiments. p nk for all k n2 n1 Example 5 Using the Fundamental Counting Principle A catalog offers chairs in a choice of 2 heights, regular and tall. There are 10 colors available for the finish, and 12 choices of fabric for the seats. The chair back has 4 different possible designs. How many different chairs can be ordered? Solution Each option can be considered as an experiment. The number of choices for each option is the number of outcomes. According to the Fundamental Counting Principle, the number of different chairs is 2 10 12 4 960 ■ Consider the following experiment: Each letter of the alphabet is written on a piece of paper, and three letters are chosen at random. There are two important questions in determining the nature of the experiment: 1. Is each letter replaced before the next letter is chosen? 2. Does the order of the letters matter in the result? If the answer to Question 1 is yes, the letters are said to be chosen with replacement. In this case, letters may be repeated in the result. Also, the number of letters to choose from is always the same. If the answer is no, then the letters are said to be chosen without replacement. In this case, there will be no repeated letters, and the number of letters to choose from decreases by 1 for each letter chosen. If the answer to Question 2 is yes, the result is said to be order important. In this case, the outcome CAT is considered to be different from the outcome ACT. If the answer is no, the result is said to be in any order. In this case, the six outcomes CAT, CTA, TAC, TCA, ACT, ATC are considered to be the same. Fundamental Counting Principle NOTE Many times, it is necessary to use the context of the experiment to determine whether it is with replacement, or if order is important. 880 Chapter 13 Statistics and Probability Figure 13.4-3 NOTE equal 1. 0! is defined to Three of the four possible cases are shown in the table below. The fourth case, with replacement and in any order, will not be discussed. Use the Fundamental Counting Principle to explain the number of |
outcomes in each case. With replacement Order important Without replacement Order important Without replacement Any order 26 26 26 17,576 26 25 24 15,600 26 25 24 3 2 1 2600 Example 6 3 Coin Toss Use the Fundamental Counting Principle to verify the probability distribution in Example 4. Solution There are two possible outcomes for each coin: heads and tails. Thus, the To find the 8 number of outcomes for tossing three coins is different outcomes, it is helpful to use a tree diagram, as shown in Figure 13.4-3. The outcomes are given below. 2 2 2 8. HHH HHT 2 heads 3 heads HTH 2 heads HTT 1 head THH 2 heads THT 1 head TTH 1 head TTT 0 heads Each of these outcomes is equally likely, so the probability distribution is Outcome 0 heads 1 head 2 heads 3 heads Probability 1 8 0.125 3 8 0.375 3 8 0.375 1 8 0.125 ■ Permutations and Combinations The two cases without replacement are called permutations (order important) and combinations (any order). In order to write a formula for permutations and combinations, n!—read “n factorial”—is used to describe the product of all the integers from 1 to n. n! n n 1 n 2 p 2 21 1 2 3 1 21 21 1 2 In the example of drawing 3 letters without replacement where order is important, the number of permutations can be written using factorials as 26 25 24 26 25 24 23 22 p 3 2 1 23 22 p 3 2 1 26! 23! In the case where order is not important, the number of combinations can be written as 26 25 24 3 2 1 26! 3 2 1 1 23! 2 26! 3! 23! Section 13.4 Determining Probabilities 881 Permutation and combination formulas are also written using factorials. Permutations and Combinations Technology Tip Permutations and combinations are located on the PRB submenu of the MATH menu on TI and on the PROB submenu of the OPTN menu on Casio. Permutations If r items are chosen in order without replacement from n possible items, the number of permutations is nPr n! (n r)! Combinations If r items are chosen in any order without replacement from n possible items, the number of combinations is nCr n! r!(n r)! If each item is equally likely to be chosen, the perm |
utations and combinations are all equally likely for a given value of r. Note: may also be written as nPr Pn,r or P(n, r), and may be nCr written as Cn,r, C(n, r), or n r b a. Example 7 Matching Problem Suppose you have four personalized letters and four addressed envelopes. If the letters are randomly placed in the envelopes, what is the probability that all four letters will go to the correct addresses? Solution For all four letters to go to the correct addresses, they must be chosen in the exact same order as the envelopes. The size of the sample space is the number of permutations, 4P4. Thus, the probability is 1 24 4P4 4! 4 4 1 0.04. 4! 0! 24! 2 ■ Example 8 Pick-6 Lottery In a “pick-6” lottery, 54 numbered balls are used. Out of these, 6 are randomly chosen. To win, at least 3 balls must be matched in any order. What is the probability of winning the jackpot (all 6 balls)? What is the probability of matching any 5 balls? any 4 balls? any 3 balls? 882 Chapter 13 Statistics and Probability Solution The size of the sample space is the number of combinations. The probability of winning the jackpot is 54C6 25,827,165 1 25,827,165 0.00000004 The event of matching 5, 4, or 3 numbers can be determined as follows: matching k numbers P 1 number of combinations that match k numbers 54C6 2 6Ck ways to match k numbers out of 6. The remaining 6 k There are numbers do not match any of the 6 winning numbers, so there are 48 numways to choose the remaining numbers. bers to choose from, giving 6Ck combinaBy the Fundamental Counting Principle, there are tions that match k numbers. 48C6k 48C6k Figure 13.4-4 matching k numbers P 1 6Ck 48C6k 2 48C1 54C6 6 48 6C5 6C4 48C2 6C3 48C3 54C6 54C6 54C6 25,827,165 15 1128 25,827,165 20 17296 25,827,165 0.00001 0.0007 0.01 ■ matching 5 numbers matching 4 numbers matching 3 numbers P P P 1 1 1 |
2 2 2 Exercises 13.4 For Exercises 1–4, an experiment consists of drawing a marble out of a bag, observing the color, and then placing it back in the bag. Suppose the experiment is repeated 75 times, with the following results: 4. Suppose it is known that there is a total of 300 marbles in the bag. Estimate the number of each color of marble. red blue green yellow 38 23 11 3 1. Write a probability distribution of the experiment using the experimental formula. 2. Based on your distribution from Exercise 1, what is the probability of drawing either a blue or green marble? 3. Based on your distribution from Exercise 1, what is the probability of drawing two yellow marbles in a row? A dreidel is a top with four sides, used in a Hanukah game. The sides are labeled with the Hebrew letters nun, gimel, hay, and shin. A dreidel is spun 100 times, with the following results: nun gimel hay shin 10 45 24 21 dreidel 5. Write a probability distribution for the dreidel. 6. A player wins tokens if the dreidel lands on either gimel or hay. What is the probability of winning tokens? 7. A player loses tokens if the dreidel lands on nun. What is the probability of losing four times in a row? Exercises 8–11 refer to the following experiment: two number cubes are rolled, and a random variable assigns the sum of the faces to each outcome. 8. The following commands generate two random integers from 1 to 6 and add them. TI-83/86: sum(randInt(1, 6, 2)) TI-89/92: sum({rand(6), rand(6)}) Sharp 9600: Casio 9850: HP-38: sum Sum 1 5 INT 6 RANDOM 6 RANDOM 1 ©LIST INT 15 2 1 1 1 2 1 2 62 1, 6random int 22 Int 6Ran# 1, Int 6Ran# 1 2 1 Run 5 trials of the experiment and list your results. 9. Run a simulation of the experiment with at least 50 trials, and create a probability distribution. 10. Use your probability distribution from Exercise 9 to find the probabilities for the following values of the random variable. a. greater than 9 b. less than 6 c. at least 4 11. Use your probability distribution from Exercise 9 to find the expected value of the random |
variable. 12. A bag contains 3 red marbles and 4 blue marbles. Suppose each marble is equally likely to be chosen. What is the probability of the event of drawing a red marble? 13. Suppose that a person’s birthday is equally likely to be any day of the year. What is the probability that a randomly chosen person has the same birthday as you? A teacher writes the name of each of her 25 students on a slip of paper and places the papers in a box. To call on a student, she draws a slip of paper from the box. Each paper is equally likely to be drawn, and the papers are replaced in the box after each draw. 14. What is the probability of calling on a particular student? 15. What is the probability of calling on the same student twice in a row? Section 13.4 Determining Probabilities 883 16. If there are 9 students in the last row, what is the probability of calling on a student in the last row? 17. If the class contains 11 boys and 14 girls, what is the probability of calling on a girl? What is the probability of calling on 3 girls in a row? 18. A clothing store offers a shirt in 5 colors, in long or short sleeves, with a choice of three different collars. How many ways can the shirt be designed? 19. A quiz has 5 true-false questions and 3 multiple choice questions with 4 options each. How many possible ways are there to answer the 8 questions? 20. A license plate has 3 digits from 0 to 9, followed by 3 letters. How many different license plates are possible? 6 21. A gallery has 25 paintings in its permanent collection, with display space for 10 at one time. How many different collections can be shown? 22. A committee of 8 people randomly chooses 3 people in order to be president, vice president, and treasurer. In how many ways can the officers be chosen? 23. A baseball team has 9 players. How many different batting orders are there? 24. A small library contains 700 novels. In how many ways can you check out 3 novels? 25. A manufacturer is testing 4 brands of soda in a blind taste test. The participants know the brands being tested but do not know which is which. What is the probability that a participant will identify all 4 brands correctly by guessing? 26. A researcher is studying the abilities of people who claim to be able to read minds. He chooses 6 numbers between 1 and 50 (inclusive |
), and asks each participant to guess the numbers in order. What is the probability of guessing all 6 correctly? A botanist is testing two kinds of seeds. She divides a plot of land into 16 equal areas numbered from 1 to 16. She then randomly chooses 8 of these areas to plant seed A, and she plants seed B in the remaining areas 10 11 12 13 14 15 16 884 Chapter 13 Statistics and Probability 27. How many ways are there to choose 8 plots out of 16? 28. What is the probability of the event shown at right 29. What is the probability that all of the areas with 4 2 rectangle on one side of seed A will form a the plot, and all of the areas with seed B will form a rectangle on the other side? 4 2 Critical Thinking Exercises 30–35 refer to a famous probability problem: suppose a certain number of people are in a room. What is the probability that two or more people in the room will have the same birthday? “everybody has a different birthday.” How many ways are there to name a different date for each of 3 people? of 20 people? of n people? 32. The probability that everybody has a different birthday can be written as Number of ways to name n different dates Number of ways to name n dates Use your results from Exercises 30 and 31 to write a formula in terms of n for the probability that everybody has a different birthday. 33. Use your results from Exercise 32 to write a formula in terms of n for the probability that two or more people have the same birthday. Find the n 35. for probability for n 20, and for n 3, 34. How many people must be in the room for the 30. Suppose each day of the year is equally likely to probability to be approximately 1 2 that two or be a person’s birthday. How many ways are there to name one date per person (not necessarily all different) for each of 3 people? of 20 people? of n people? 31. The complement of the event “two or more people will have the same birthday” is more have the same birthday? 35. How many people must be in the room for the probability to be 1 that two or more have the same birthday? Hint: do not use the formula. 13.4.A Excursion: Binomial Experiments Objectives • Calculate the probability of a binomial experiment Many experiments can be described in terms of just two outcomes, such as |
winning or losing, heads or tails, boy or girl. These experiments determine a group of problems called binomial or Bernoulli experiments, named after Jacob Bernoulli, a Swiss mathematician who studied these distributions extensively in the late 1600’s. Binomial Experiments Here is a typical binomial experiment: in a basketball contest, each contestant is allowed 3 free-throws. If a certain individual has a 70% chance of making each free-throw, what is the probability of making exactly 2 out of 3? Binomial Experiment NOTE The terms “success” and “failure” are often used in experiments to designate outcomes such as heads or tails, even if neither outcome is preferred. Section 13.4.A Excursion: Binomial Experiments 885 The essential elements in a binomial experiment are given below. A set of n trials is called a binomial experiment if the following are true. 1. The trials are independent. 2. Each trial has only 2 possible outcomes, which may be designated as success (S) and failure (F). 3. The probability of success p is the same for each trial. The probability of failure is q 1 p. In the example of the basketball contest, the outcome SFS indicates that the first free-throw is a success, the second a failure, and the third a success. The trials are independent, so the probability of the outcome SFS is the product of the probabilities for each trial. SFS.7 1 21 0.3 21 0.7 2 0.147 Example 1 Basketball Contest Refer to the basketball contest described on page 884. Suppose that the probability of making each free-throw is 0.7. What is the probability of making exactly 2 free-throws in 3 tries? Solution The outcomes in the event “2 free-throws in 3 tries” are SSF, SFS, FSS. The probability of the event is the sum of the probabilities of the three outcomes SSF P P SFS P P FSS 2 free-throws.7 0.7 0. SSF 2 S 21 2 1 F 21 2 1 S 21 2 1 P SFS 0.147 0.147 0.147 0.441 0.147 2 0.147 2 0.147 2 FSS 0.3 21 0.7 21 0.7 21 P 1 1 1 P 0.7 0.3 0.7 1 1 1 2 2 2 ■ In Example 1, |
note that the probability of SSF is the same as the probability of SFS or FSS. In general, the probability of any outcome with r successes and n–r failures in n trials is qq p q 2 n r times 21 1 ⎧⎨⎩ ⎧⎨⎩ r times prqnr pp p p 1 2 1 2 To develop a general formula for the probability of r successes in n trials, it is necessary to determine how many different outcomes have r suc- 886 Chapter 13 Statistics and Probability cesses. Consider the number of outcomes with 3 successes in 5 trials, as shown below. For clarity, the F’s are left as blanks. SSS SS S S SS SSS SS SS S S S SS S S S S SS SSS The number of outcomes is the same as the number of ways to choose 3 positions for the S out of 5 possible positions. The order of the S’s does not matter, because they are all the same. This is the number of combinations, 5C3 10. Probability of a Binomial Experiment In a binomial experiment, P(r successes in n trials) nCr prqnr where p is the probability of success, and probability of failure. q 1 p is the Example 2 Lottery Tickets A lottery consists of choosing a number from 000 to 999. All digits of the number must be matched in order, so the probability of winning is 0.001. 1 1000 the lottery 1000 times in a row. A ticket costs $1, and the prize is $500. Suppose you play a. Write a probability distribution for the number of wins. b. What is the probability that you will break even or better? Solution a. The number of wins could be anything from 0 to 1000. However, the probability of winning more than a few times is so small that it is essentially 0. Thus, the sample space will be considered as 0, 1, 2, 3, 4, and 5 or more wins. The probabilities are calculated using the binomial probability formula, with 1000C01 0 wins 2 1000C11 1 win 2 1000C21 2 wins 2 1000C31 3 wins 2 1000C41 4 wins 2 5 or more wins 2 0 and p 0.001. n 1000 1000 0.3677 2 999 0.3681 998 0.1840 997 0.0613 996 0.0153 0.999 1 0. |
999 2 1 2 0.999 0.999 0.999 0.001 2 1 0.001 2 0.001 0.001 0.001 1 0.3677 0.3681 0.1840 0.0613 0.0153 0.0036 Outcome 0 wins 1 win 2 wins 3 wins 4 wins 5 or more wins Probability 0.3677 0.3681 0.1840 0.0613 0.0153 0.0036 Section 13.4.A Excursion: Binomial Experiments 887 b. In order to break even or better, you must win 2 or more times. The probability is the sum of the probabilities of winning 2, 3, 4, or 5 or more times. break even or better P 1 0.1840 0.0613 0.0153 0.0036 0.2642 ■ 2 Example 3 Multiple Choice Exam Morgan is taking a 10-question multiple choice test but has not studied. Each question has 4 possible responses, only one of which is correct. Find the probability of getting the results below if he answers all questions randomly. a. exactly 6 questions correct b. 4 or fewer questions correct c. 8 or more questions correct Solution The probability of getting each question correct is p 1 4 0.25. Technology Tip The command binompdf( in the DISTR menu of TI finds the probability for r successes in n trials of a binomial experiment, given n, p and r. The command binomcdf( finds the cumulative probability for r or fewer successes in n trials. a. b. c. Figure 13.4.A-1 6 P 1 6 correct 10C61 2 4 or fewer correct 8 1 8 or more correct 10 2 0.25 3 0.75 0.25 1 2 10C01 2 10C31 0.25 2 1 10C81 0.25 2 0 0.0004 0.25 0.75 0.25 0.75 2 2 2 2 P 1 10C21 P 1 10C101 2 1 2 0 4 0.016 0.75 7 1 2 0.75 2 8 0.75 1 10 2 10C41 2 0.25 4 10C11 0.25 2 1 10C91 0.25 2 0.75 9 2 1 2 1 9 0.75 2 1 6 0.922 2 1 0.75 2 ■ Binomial Distributions Consider the following binomial experiment: a coin is tossed 4 times, |
and the probability of heads on each toss is A probability distribution for. 1 2 the number of heads is shown below. Number of heads Probability 0 1 16 16 A probability density function that represents this distribution is shown in Figure 13.4.A-2. Notice that the shape of the graph is symmetric. The expected value of this distribution is x 1 16b 0 a 1 1 4b a 2 3 8b a 3 1 4b a 4 1 16b a 2 1 2 3 4 5 which is the (approximate) center of the probability distribution. y 0.5 0.25 0 Figure 13.4.A-2 888 Chapter 13 Statistics and Probability For large values of n or when p is near 1 2, the shape of a binomial dis- tribution is approximately symmetric, with its center at the expected value. For small values of n if p is different from 1 2, the distribution will be skewed. However, as n increases, the distribution becomes more symmetric. The graphs in Figure 13.4.A-3 show the distributions of a binomial experiment with p 1 3 for n 3, 10, and 30. The shape of the distribution approaches a special curve, called the normal curve, which is developed in the next section. Characteristics of a Binomial Distribution n = 3 n = 10 n = 30 Figure 13.4.A-3 The distribution of a binomial experiment with n trials and probability of success p is approximately symmetric for large values of n. The center of the distribution is the expected value. The expected value of the binomial distribution is np. The standard deviation of the binomial distribution is 1npq. Example 4 Multiple Choice Exam Find the expected value and standard deviation of the number of questions correct on the multiple choice exam in Example 3. Solution In this case, n 10, p 0.25, and q 0.75. The expected value is and the standard deviation is np 10 0.25 1 2 2.5 1npq 110 0.25 1 21 0.75 2 1.4. 11 This means that if a large number of students guessed on the exam, the average number correct would be 2.5, with a standard deviation of 1.4. A graph of the distribution is shown in Figure 13.4.A-4. ■ Figure 13.4.A-4 0.3 0 0 Section 13.4.A Excursion: Binomial Experiments 889 Exerc |
ises 13.4.A For Exercises 1–4, a binomial experiment consists of planting 4 seeds. The probability of success (that a given seed will sprout) is The sample outcome SFSS means that the first seed sprouted, the second seed did not sprout, and the third and fourth seeds sprouted. p 0.65. 1. What is the probability of failure? 2. Write out the two successes. 4C2 6 outcomes that have exactly 3. Complete the probability distribution below. 8. What is the expected value of the number of bulls-eyes in 4 tries? A true-false exam has 100 questions, and for each question What is the probability of answering P(true) P(false) 0.5. 9. 50 questions correct? 10. 70 questions correct? 11. 30 questions correct? 12. 90 questions correct? Sprouted Probability 0? 1? 2? 3? 4? 13. Find the expected value and standard deviation of the number of correct answers. 4. Find the expected value of the number of seeds that will sprout, and interpret the result. For Exercises 5–8, suppose the probability of a certain dart player hitting a bulls-eye is 0.25. 5. Write a probability distribution for the number of bulls-eyes in 4 tries. 6. What is the probability of hitting at least 3 bulls- eyes in 4 tries? 7. What is the probability of hitting less than 2 bulls- eyes in 4 tries? An experiment consists of rolling a number cube 30 times. The outcome 6 is considered a success, and all other outcomes are considered failures. Assume all p P(success) 1 6 faces are equally likely, so that. 14. Find the expected value of the probability distribution. What is the probability of e successes? m 15. What is the probability of m 1 is the probability of m 1 successes? successes? What 16. Find the standard deviation of the probability distribution. What is the probability that the outcome is between m s m s and s? 13.5 Normal Distributions Objectives • Draw a normal distribution given its mean and standard deviation • Use normal distributions to find probabilities To statisticians, the most important probability density function is the normal curve (sometimes called the bell curve). Normal distributions are used to predict the outcomes of many events, such as the probability that a student scores a 750 on the SAT, or the probability that you will grow to be 67 inches tall. For statisticians, |
it is also a valuable tool for predicting if an outcome is statistically significant or just caused by chance. 890 Chapter 13 Statistics and Probability Properties of the Normal Curve A normal distribution is bell-shaped and symmetric about its mean. The x-axis is a horizontal asymptote, and the area under the curve and above the x-axis is 1. The maximum value occurs at the mean, and the curve has two points of inflection, at 1 standard deviation to the right and left of the mean. Because a normal distribution is symmetric, the mean, median, and mode have the same value. This value is also called the center. The Greek letter m (mu) is used to represent the mean of a normally distributed population. Also, the population standard deviation is used for the standard deviation. s Figure 13.5-1 shows three normal curves, with m and s labeled. y 1 0.8 0.6 0.4 0..5 x −4 −2 0 2 4 6 8 10 Figure 13.5-1 The normal curve with a mean of 0 and standard deviation of 1 is called the standard normal curve. The equation of the standard normal curve is y 1 12p x 2 e 2 y 0.4 0.2 −4 −3 −2 −1 0 1 2 3 4 Figure 13.5-2 x The standard normal curve can be thought of as a parent function for all normal curves. • A change in • A change in m s results in a horizontal translation of the curve. results in a horizontal stretch and vertical compression of the curve, or vice versa, so that the resulting area is still 1. Section 13.5 Normal Distributions 891 Graphing Exploration The normal curves below have the same value of and different values of Graph the curves in the same viewing window and describe the results. m. s y 1 12p x2 e 2 y 1 12p e 2 x 3 2 1 2 y 1 12p 2 x 4 2 1 2 e The normal curves below have the same value of and different values of Graph the curves below in the same viewing window and describe the results. s. m y 1 12p x2 2 e y 1 312p x2 18 e y 3 12p 9x2 2 e Equation of Normal Curve A random variable is said to have a normal distribution with mean given by the equation if its density function is and standard deviation S M y 1 S12P (x M)2 2S2 |
e Most of the time, the mean and standard deviation of an entire population cannot be measured. The population mean and standard deviation are often estimated by using a sample. Example 1 Using Sample Information A paper in Animal Behavior gives 11 sample distances, in cm, from which a bat can first detect a nearby insect. The bat does this by sending out high-pitched sounds and listening for the echoes. Assume the population is normally distributed. 62, 23, 27, 56, 52, 34, 42, 40, 68, 45, 83 a. Compute the mean and standard deviation of this sample. b. Draw a normal curve to represent the distribution. Figure 13.5-3 Solution a. The mean is 48.36 and the sample standard deviation is 18.08, as shown in Figure 13.5-3. b. The normal curve is shown in Figure 13.5-4. 892 Chapter 13 Statistics and Probability 0.04 0 0 Figure 13.5-4 102 ■ The Empirical Rule Consider the intervals formed by one, two, and three standard deviations on either side of the mean, as shown below. The empirical rule describes the areas under the normal curve over these intervals. µ σ− 3 µ σ− 2 µ σ+ 3 68% 95% 99.7% Empirical Rule In a normal distribution: NOTE Each percentage in the empirical rule can also be interpreted as the probability that a data value chosen at random will lie within one, two, or three standard deviations of the mean. • about 68% of the data values are within one standard deviation of the mean. • about 95% of the data values are within two standard deviations of the mean. • about 99.7% of the data values are within three standard deviations of the mean. Example 2 Running Shoes A pair of running shoes lasts an average of 450 miles, with a standard deviation of 50 miles. Use the empirical rule to find the probability that a new pair of shoes will have the following lifespans, in miles. a. between 400 and 500 miles b. more than 550 miles Section 13.5 Normal Distributions 893 Solution Figure 13.5-5 shows the normal distribution with the intervals of one, two, and three standard deviations on either side of the mean. 300 350 400 450 500 550 600 Figure 13.5-5 x miles a. The area under the curve between 400 and 500 miles is approximately 68% of the total area. Since the area under a density |
function corresponds to the probability, the probability that a pair of shoes will last between 400 and 500 miles is about 0.68. b. The area under the curve between 350 and 550 miles is 95% of the total area, which leaves 5% for the area less than 350 and greater than 550. Since the normal curve is symmetric, the area greater than 550 is exactly half of this, or 2.5%. Thus, the probability that a pair of shoes will last more than 550 miles is about 0.025. ■ The Standard Normal Curve In general, determining the area under a normal curve is very difficult. Because of this, it is common to standardize data to match the normal curve with a mean of 0 and standard deviation of 1, for which these areas are known. Comparing Data Sets Sarah and Megan are high school juniors. Sarah scored 660 on the SAT, and Megan scored 29 on the ACT. Who did better? Although the scores on both tests are normally distributed, the mean and standard deviation are very different. One way to compare the distributions is to adjust the scales of the axes, as shown in Figure 13.5-6. SAT µ σ = 500 = 100 ACT = 18 µ = 6 σ 200 300 400 500 600 700 800 0 6 12 18 24 30 36 Figure 13.5-6 x x 894 Chapter 13 Statistics and Probability A more precise way to compare two scores from different data sets is to use the standard deviation as a unit of measurement. Each score is represented by the number of standard deviations above or below the mean. Example 3 Comparing Scores Use the standard deviation as a unit of measurement to compare the SAT and ACT scores for Sarah and Megan. Solution Sarah’s score of 660 is 160 points above the mean, which is 500. The stan- dard deviation is 100, so Sarah’s score is 160 100 1.6 standard deviations above the mean. Megan’s score of 29 is 11 points above the mean, which is 18. The stan- dard deviation is 6, so Megan’s score is 11 6 1.83 standard deviations above the mean. Thus, Megan’s score is better than Sarah’s score. ■ The number of standard deviations that a data value is above the mean is called the z-value. In Example 3, Sarah’s z-value is 1.6 and Megan’s z-value is 1.83. For a normal distribution, the |
z-values correspond to values on the standard normal curve, as shown in Figure 13.5-7. z-values: 200 −3 300 −2 400 −1 500 0 600 1 700 2 Figure 13.5-7 z-Values The z-value of the value x in a data set with mean standard deviation is S x 800 3 M and z x M S The area under a normal curve between x a and x b is equal to the area under the standard normal curve between the z-value of a and the z-value of b. Once z-values are determined, corresponding areas under the normal curve are often found using a table. Because of the symmetry of the normal curve, it is only necessary to include positive z-values in the table. Section 13.5 Normal Distributions 895 The following table gives the area under the normal curve between 0 and the given z-values, as shown in Figure 13.5-8. y −3 −2 −1 0 z 1 2 3 Figure 13.5-8 x z 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Area 0.04 0.08 0.12 0.16 0.19 0.23 0.26 0.29 0.32 0.34 Example 4 Using z-Values to Determine Area A normally distributed data set has a mean of 25 and a standard deviation of 5. Find the probability that a data value chosen at random is between 23 and 28. Solution First, find the z-values for 23 and 28, given z 23 25 5 0.4 and s 5. 0.6 m 25 z 28 25 5 0.4 Divide the area into two parts: the area from to 0, and the area from 0 to 0.6, as shown in Figure 13.5-9. The areas are found in the table above. Because the normal curve is symmetric with respect to the area from 0.4 to 0 is the same as the area from 0 to 0.4, or 0.16. The area from 0 to 0.6 is 0.23. The total area is the sum of the two areas, 0.39. Thus, the probability that a randomly chosen data value is between 23 and 28 is 0.39. x 0, Technology Tip Normalcdf under the DISTR menu of TI will calculate the area under the normal curve in a given interval. The parameters are Normal |
cdf (lower bound, upper ).sm, bound, 0.16 y 0.23 x −0.4 0.6 Figure 13.5-9 ■ Example 5 Response Times The EMT response time for an emergency is the difference between the time the call is received and the time the ambulance arrives on the scene. Suppose the response times for a given city have a normal distribution 896 Chapter 13 Statistics and Probability m 6 s 1.2 with estimate the probability of the following response times. minutes and minutes. For a randomly received call, a. between 6 and 7 minutes c. less than 7 minutes b. less than 5 minutes Solution a. The z-values are z 6 6 1.2 0 and z 7 6 1.2 0.8. The probability of a response time between 6 and 7 minutes is the area under the standard normal curve from 0 to 0.8, which is approximately 0.29. b. The z-value is z 5 6 1.2 0.8. The area from 0 to 0.8 is approxi- mately 0.29, so the probability of a response time under 5 minutes is (see Figure 13.5-10). about 0.5 0.29 0.21 half of total area = 0.5 y 0.21 0.29 −0.8 Figure 13.5-10 x c. The area under the curve from 0 to 6 minutes is 0.5. From part a, the area under the curve from 6 to 7 minutes is 0.29. Thus, the probability of a response time under 7 minutes is 0.5 0.29 0.79. ■ Exercises 13.5 In Exercises 1–4, refer to the normal curve below. The area of each square of the grid is 0.01. 3. Estimate the area under the curve from 16 to 24. 4. Estimate the area under the curve from 12 to 28. Graph the normal curves for the following values of M and S. 8 12 16 20 24 28 32 1. Find m. 2. Estimate s. x 5. m 10, s 12 6. m 40, s 12 7. m 500, s 100 8. m 3, s 5 Suppose the heights of adult men are normally distributed. The heights of a sample of 30 men are shown below, in inches. 65, 83, 69, 67, 69, 67, 67, 72, 85, 68, 73, 65, |
67, 65, 72, 71, 67, 73, 68, 72, 61, 75, 66, 78, 65, 71, 68, 76, 67, 68 Section 13.5 Normal Distributions 897 9. Compute the mean and standard deviation of this 23. the probability that the wait time for a table is sample. between 25 and 38 minutes 10. Draw a normal curve that represents the 24. the probability that the wait time for a table is less distribution of adult male heights, based on the sample. than 22 minutes Suppose that the heights of adult women are normally inches. distributed with Use the properties of the normal curve and the Empirical Rule to find the probability that a randomly chosen woman is within the given range. inches and S 2.5 M 65 11. taller than 65 inches 12. shorter than 67.5 inches 13. between 62.5 inches and 67.5 inches 14. between 60 inches and 70 inches 15. between 57.5 inches and 67.5 inches 16. A student took two national standardized tests while applying for college. On the first test, m 32 and s 6. the second, on which test did he do better? m 475 and If he scored 630 on the first test and 45 on and on the second test, s 75, 17. Four students took a national standardized test for which the mean was 500 and the standard deviation was 100. Their scores were 560, 450, 640, and 530. Determine the z-value for each student. 18. If a student’s z-value was 1.75 on the test described in problem 17, what was the student’s score? 19. A sample of restaurants in a city showed that the average cost of a glass of iced tea is $1.25 with a 7¢. standard deviation of Three of the restaurants charge z-value for each restaurant. $1.00, and $1.35. Determine the 95¢, 20. If a new restaurant charges a price for iced tea that has a z-value of what is the tea’s actual cost? 1.25 (see Exercise 19), then At a certain restaurant, the wait time for a table is norM 30 S 10 mally distributed with minutes. Use the table on page 895 to estimate the following: minutes and 21. the probability that the wait time for a table is between 30 and 35 minutes 22. the probability that the wait time for a table is between 24 |
and 30 minutes Daytime high temperatures in New York in February are normally distributed with an average of and a standard deviation of 30.2 8.5. 25. Estimate the probability that the temperature on a given day in February is 39° or higher. 26. Estimate the probability that the temperature on a given day in February is 22° or lower. 27. Estimate the probability that the temperature on a given day in February is between 13° and 39°. 28. Estimate the probability that the temperature on a given day in February is between 25° and 30°. 29. Estimate the probability that the temperature on a given day in February is between 27° and 38°. The quartiles of a normal distribution are the values that divide the area under the curve into fourths4 −3 −2 Q1 Q3 2 3 4 Q2 (center) 1st and The the left and right of the mean, or quartiles are approximately 3rd 0.675S to Q1 M 0.675S and Q3 M 0.675S 30. Find Q1 s 4. and Q3 for a distribution with m 20 and 31. Suppose the scores on an exam are normally m 70 Q1 s 10. Find and distributed with Q3, and interpret the result. and 32. For the exam in Exercise 31, what exam score would place a student in the top 25% of the class? C H A P T E R 13 R E V I E W Important Concepts Section 13.1 Section 13.2 Section 13.3 Section 13.4 Data...................................843 Population.............................. 843 Sample................................ 843 Frequency table.......................... 844 Bar graph.............................. 845 Pie chart................ |
............... 845 Uniform, symmetric, and skewed distributions............................ 846 Stem plot............................... 847 Histogram.............................. 849 Mean.................................. 853 Median................................ 854 Mode.................................. 855 Standard deviation....................... 857 Range................................. 859 Interquartile range....................... 860 Five-number summary and box plot......... 861 Experiment............................. 864 Sample space............................ 864 Event.................................. 865 Probability.............................. 865 Probability distribution.................... 865 Mutually exclusive events... |
............... 866 Complement............................ 866 Independent events....................... 867 Random variable......................... 869 Expected value.......................... 869 Probability density function................ 871 Experimental estimate of probability......... 875 Probability simulation..................... 875 Equally likely outcomes................... 877 Fundamental counting principle............. 879 Permutations and combinations............. 880 Section 13.4.A Binomial experiment...................... 884 Binomial distribution..................... 887 898 Chapter Review 899 Section 13.5 Normal curve........................... 889 Normal distribution...................... 889 Standard normal curve.................... 890 Empirical rule........................... 892 z-value................................. 894 Important Facts and Formulas mean: x ©xi n x3, p xn x1, x2, are ordered from smallest |
to largest, then the median If is the middle entry when n is odd and the average of the two middle entries when n is even. for n odd, the value in the n 1 2 position median • for n even, the average of the values in the n 2 and n 2 1 positions Population standard deviation s B © xi 1 2 x n 2 Sample standard deviation s © 2 x xi 1 n 1 2 B Suppose an experiment has a sample space of n outcomes, all of 1 n, which are equally likely. Then the probability of each outcome is and the probability of an event E is given by number of outcomes in E n P E 2 1 Fundamental Counting Principle Consider a set of k experiments. Suppose the first experiment has outcomes, the second has n1 number of outcomes is n1 outcomes, and so on. Then the total p nk for all k experiments. n2 n2 Permutations If r items are chosen in order without replacement from n possible items, the number of permutations is nPr n! n r 1! 2 Combinations If r items are chosen in any order without replacement from n possible items, the number of combinations is nCr n! n r! 2 r! 1 900 Chapter Review nCrprqnr In a binomial experiment, where p is the probability of success, and is the probability of failure. The expected value of a binomial distribution is np, 1npq. and the standard deviation is r successes in n trials q 1 p P 2 1 A random variable is said to have a normal distribution with mean m if its density function is given by the equation and standard deviation s y 1 s12p x m 2s2 2 2 1 e Empirical Rule In a normal distribution: • about 68% of the data values are within one standard deviation of the mean. • about 95% of the data values are within two standard deviations of the mean. • about 99.7% of the data values are within three standard deviations of the mean. The z-value of the value x in a data set with mean m and standard deviation s is z x m s. Review Exercises Exercises 1–4 refer to the following description: A group of bird-watchers is trying to determine what types of birds are common to their area. The group observed 21 sparrows, 15 purple finches, 10 chickadees, 5 cardinals, and 2 blue jays. Section 13.1 1. Is the data |
qualitative or quantitative? 2. Create a frequency table for the data. 3. Create a bar graph for the data. 4. Create a pie chart for the data. Exercises 5–17 refer to the following description: a study is done to determine the average commuting time of employees at a company. A total of 34 employees are surveyed, with the following results (in minutes). 31.9, 34, 30.7, 39, 33.1, 35.2, 30.5, 32.7, 29.4, 33.4, 22.3, 31.9, 32.3, 29.4, 33, 18.2, 29.1, 32.5, 22.2, 36.1, 27.7, 36, 31.9, 26, 31.7, 23.2, 30.7, 24.4, 33, 28.4, 28.8, 23.3, 32.2, 22.8 5. Is the data qualitative or quantitative? 6. Is the data discrete or continuous? Chapter Review 901 7. Create a stem plot for the data. 8. Create a histogram for the data with class intervals of 5 minutes. 9. Use your histogram from Exercise 8 to describe the shape of the data. Section 13.2 10. Find the mean of the data. 11. Find the median of the data. 12. Find the mode of the data. 13. Find the sample standard deviation of the data. 14. Find the range of the data. 15. Find the first and third quartiles of the data. 16. Find the interquartile range of the data. 17. Create a box plot for the data. Section 13.3 Exercises 18–22 refer to the following probability distribution: Outcome Probability 1 0.1 2? 3 0.4 4? 5 0.1 18. List the sample space for the probability distribution. 19. Suppose the probabilities are the same for outcomes 2 and 4. Complete the probability distribution. 20. What is the probability that the outcome is an even number? 21. What is the probability that the outcome is greater than 2? 22. Suppose the experiment with the given probability distribution is repeated 3 times. Assuming the trials are independent, what is the probability of the outcome {1, 2, 3}? An experiment consists of spinning the spinner at left 3 times. A probability distribution for the outcomes is given below. W white, R red |
1 2 Outcome WWW WWR WRW RWW WRR RWR RRW RRR Probability 0.512 0.128 0.128 0.128 0.032 0.032 0.032 0.008 A random variable is assigned to the number of times the spinner lands on red. 23. What is the range of the random variable? 24. What is the probability that the value of the random variable is 2? 902 Chapter Review Section 13.4 25. Create a probability distribution for the random variable. 26. Find the expected value of the random variable, and interpret the result. 27. Graph the probability density function of the random variable, and shade the area of the graph that corresponds to the probability that the spinner lands on white 3 times. 28. Suppose the experiment is repeated 25 times, with the following results: WRW, RRW, WWW, WWW, WWW, WWW, WWR, RWW, WWW, WRW, WWW, WWR, WWW, RWW, WWW, RWR, WWW, WRW, WWW, RWW, WWW, WWW, WRW, WWW, WWR Write a probability distribution of the random variable, based on the experimental results. How do the probabilities compare to your results in Exercise 25? Exercises 29–33 refer to the following experiment: A sock drawer contains 6 identical black socks, 8 identical white socks, and 1 blue sock. A sock is chosen randomly from the drawer. Assume all socks are equally likely to be chosen. 29. What is the probability of choosing a black sock? a white sock? a blue sock? 30. Suppose a black sock is chosen. What is the probability that the next sock chosen will also be black? Hint: how many of each color are left in the drawer? 31. Suppose a white sock is chosen. What is the probability that the next sock chosen will also be white? Hint: how many of each color are left in the drawer? 32. Suppose a blue sock is chosen. What is the probability that the next sock chosen will also be blue? Hint: how many of each color are left in the drawer? 33. Use your results from Exercises 30 – 32 to determine the probability of choosing a pair if two socks are chosen randomly from the drawer. 34. A lottery ticket involves matching 5 numbers between 1 and 50 in any order. What is the probability of |
matching all 5 numbers? What is the probability of matching any 4 numbers? 35. Suppose there are 4 people on a subcommittee, and you do not know their last names. If you have a list of 10 last names of all of the people in the committee, what is the probability of correctly guessing the last names of the people in the subcommittee? Chapter Review 903 Section 13.4.A A binomial experiment consists of randomly choosing 7 tiles imprinted with letters of the alphabet. An outcome of a vowel is considered a success, and a consonant is considered a failure. The probability of success is p 0.44. 36. Complete the probability distribution below for the number of vowels. Number of vowels Probability? 37. Find the expected value and standard deviation of the number of vowels. S 8. Suppose the scores on an exam are normally distributed with Use properties of the normal curve, the empirical rule, and the table on page 895 to answer Exercises 38–46. M 75 and Section 13.5 38. Write the equation of the normal curve for the distribution of the scores. 39. Graph the normal curve for the distribution of the scores. 40. Estimate the probability that a randomly chosen score is greater than 75. 41. Estimate the probability that a randomly chosen score is between 67 and 83. 42. Estimate the probability that a randomly chosen score is greater than 59. 43. Estimate the probability that a randomly chosen score is less than 99. 44. Estimate the probability that a randomly chosen score is between 75 and 78. 45. Estimate the probability that a randomly chosen score is between 70 and 82. 46. Estimate the probability that a randomly chosen score is less than 74. C H A P T E R 13 Area Under a Curve y h x a b Figure 13.C-1 Many applications of calculus involve finding the area under a curve. In probability, for example, it is often necessary to find areas under a normal curve or other probability density function. In other areas, such as physics, the area under a curve can be used to determine total distance traveled or the total amount of force on an object. In this section, properties of probability are used to estimate the area under a curve. Example 1 Area Model for Probability Consider the graph in Figure 13.C-1. Suppose a point in the rectangle is chosen randomly. If A is the area of the shaded region, write a formula in terms of A, a, b |
, and h for the probability that the point is in the shaded region. Solution If the point is chosen randomly, then all points in the sample space are equally likely to be chosen. The sample space is all points in the rectangle, and the event above can be described as the set of all points in the shaded area, both of which are infinite. Since it is impossible to divide the number of outcomes in the event by the number of outcomes in the sample space, the areas of the regions are used instead. The area of the rectangle is area of shaded region area of rectangle A b a 2 h 1 ■ A probability simulation can also be used to find the probability in Example 1. Suppose points in the rectangle are chosen randomly, and 470 of them are in the shaded region. Recall that the experimental estimate of the probability of an event is n 1000 E P 1 2 number of trials with an outcome in E n 470 1000 0.47 By setting the two probability estimates equal to each other and solving, a formula can be found for the area A in terms of a, b, and h. h 1 904 A b a 0.47 2 A 0.47 h b a 2 1 Area Under a Curve In general, the area A under a curve between the x-values a and b with a b may be estimated as follows: 6 1. Draw a rectangle with a base of length b a that contains the desired area. Let h be the height of the rectangle. 2. Randomly choose n points in the rectangle, using a calculator or computer. Determine the number of points e that lie in the desired area. 3. The area is approximated by the formula below. A e n h 1 b a 2 Using Probability Simulations To generate a large number of points and determine the number of points that lie under a curve, a calculator or computer program is usually used. A sample program is given below. The equation of the graph must be entered in and the calculator window should contain the desired region. Note: the choice of window will not affect the answer. Y1, Prompt A Prompt B Prompt H Prompt N 0 S E For (K, 1, N) B A rand A S X ) ( H rand S Y Pt-On (X, Y) Y Y1 If E 1 S E End (X) Disp (E/N)H( B A ) A is the left bound of the rectangle. B is the right bound of the rectangle. H is the |
height of the rectangle. N is the number of points in the rectangle. E is the number of points in the shaded region. This command generates an x-coordinate, X. This command generates a y-coordinate, Y. This command displays the point (X, Y). The point is tested to determine whether it is under the curve. This command displays the estimate of the area. Example 2 Area of a Quarter-Circle Use a probability simulation to estimate the area under the curve with equa- tion y 21 x2 from 0 to 1. Since this region is one-fourth of a circle with radius 1, the area should be p 4 0.785. 905 Solution The result of a simulation with The area estimate is 0.782, which is very close to the predicted area. points is shown in Figure 13.C-2. n 500 1.2 1.175 1.175 0.25 Figure 13.C-2 ■ This method can also be used to find areas under a normal curve. Example 3 Area Under a Normal Curve a. Use a probability simulation to estimate the area under the normal curve with m 70 and s 10 between a 85 and b 95. b. Suppose the scores on an exam are normally distributed with m 70 Estimate the probability that a randomly chosen score is s 10. and between 85 and 95. Solution a. The normal curve has the equation y 0.03989e0.005(x70)2. Trace to find a good value of h. The closer the value of h is to the maximum value of the function over the given interval, the better the estimate will be. A good choice of h is 0.014. The result of a simulation with n 500 points is shown in Figure 13.C-3. The area estimate is 0.0616. 0.05 0 40 100 Figure 13.C-3 b. The probability that a randomly chosen score is between 85 and 95 is approximately the area under the normal curve, or about 0.06. ■ 906 Exercises In Exercises 1 – 6, estimate the area under the given curve between the given values of a and b. 8. The function y 1 1x is not defined at x 0, and 1. y x2 2. y sin x a 0, b 1 a 0, b p 3. y x2 5x 4 a 1, b 4 4. y e x 5. y 1 x 6. y 1x |
a 2, b 10 a 3, b 5 a 7, b 9 the graph has a vertical asymptote. However, calculus can be used to show that the area under the curve from 0 to 1 is finite. Use a probability a 0 simulation to estimate the area by letting b 1. h 3, 4, for each value of h to be sure you have a good estimate. and Compare the value of the area estimates for and 5. Run the simulation at least twice 9. Use a probability simulation to verify that the area under the standard normal curve is 1. Hint: according to the Empirical Rule, 99.7% of the area under a normal curve is within of the mean. 3s 7. Suppose the weights of apples from a certain tree are normally distributed with s 18 g. chosen apple weighs between 100 g and 150 g. Estimate the probability that a randomly and m 138 g 907 C H A P T E R 14 Limits and Continuity Terminal Velocity Parachutists have two forces acting on them, as does any free-falling body. One force is gravity, which causes the body to speed up as it falls, and the other is air resistance, which causes the body to slow down. As the body moves faster, the air resistance builds until it nearly balances the gravitational force. So the body speeds up very little after it has fallen some distance. Terminal velocity is achieved when the air resistance approaches the gravitational pull. See Exercise 43 in Section 14.5. 908 Chapter Outline 14.1 14.2 Limits of Functions Properties of Limits 14.2.A Excursion: One-Sided Limits 14.3 The Formal Definition of Limit (Optional) 14.1 > 14.2 14.4 Continuity 14.5 Limits Involving Infinity Chapter Review can do calculus Riemann Sums Interdependence of Sections > > > 14.3 14.4 14.5 The mathematics presented in previous chapters deals with static prob- lems, such as What is the size of the angle? What is the average speed of a car between t 0 and t 5 minutes? Calculus, on the other hand, deals with dynamic problems, such as At what rate is the angle increasing in size? How fast is the car going at time t 4.2 minutes? The key to dealing with such dynamic problems is the concept of limit, which is introduced in this chapter. 14.1 Limits of Functions Objectives • Use the informal definition of limit |
Many mathematical problems involve the behavior of a function at a particular value: What is the value of the function f when x c? x 1 2 rather than at The underlying idea of limit, however, is the behavior of the function near x c You have dealt with limits informally in previous chapters, but this section will discuss limits in more detail and give the notation used when talking about them. x c. Suppose you want to describe the behavior of 0.1x4 0.8x3 1.6x2 2x 8 x 4 f x 1 2 909 910 Chapter 14 Limits and Continuity Technology Tip Letting the value of x that produces the hole be the center of the displayed parts of the x-axis will produce a graph that shows the hole as shown in Figure 14.1-1. When holes are at integer values, a decimal window will normally show the holes. Technology Tip Once the function is entered into the calculator, use the calculator’s table feature to generate values of the function. Set the INPUT or INDPNT to USER or ASK, and enter the values of x. To see what happens to the values of when x is very close to 4. Notice that the function is not defined when x 4. when x is very close to 4, observe the graph of the function in the viewing window shown in Figure 14.1-1. (See the Technology Tip about how to produce a graph that shows the hole.) x f 1 2 3 0 0 8 Figure 14.1-1 To further explore the behavior of the function near following Graphing Exploration. x 4, perform the Graphing Exploration in the viewing window with x f Graph and 2 1 0 y 3. Use the trace feature to move along the graph, and examwhen x takes values close to 4. Your results ine the values of should be consistent with the following table of values. x f 2 1 3.5 x 4.5 x approaches 4 from the left > x approaches 4 from the right > x 3.9 3.99 3.999 f x 2 1 1.8479 1.9841 1.9984 4 * 4.001 4.01 4.1 2.0016 2.0161 2.168 The exploration and the table suggest that as x gets closer and closer to get closer and closer 4 from either side, the corresponding values of to 2. Furthermore, by taking x close enough to 4, the corresponding values can be made as close |
as you want to 2. x f 1 2 For instance, 3.99999 f 1 2 1.999984 and f 4.00001 1 2 2.000016 Notation The statement above is usually expressed by saying “ The limit of f x 2 1 as x approaches 4 is 2, ” which is written symbolically as Section 14.1 Limits of Functions 911 lim xS4 f x 1 2 2 or lim xS4 0.1x4 0.8x3 1.6x2 2x 8 x 4 2 Informal Definition of Limit The following definition of “limit” in the general case is similar to the situation previously described, but now f is any function, c and L are fixed real numbers, and the phrase “arbitrarily close” means “as close as you c 4 want.” In the previous discussion, and L 2. Informal Definition of Limit Let f be a function and let c be a real number such that defined for all values of x near, except possibly at itself. Suppose that x c is f(x) x c whenever x takes values closer and closer but not equal to c (on both sides of c), the corresponding values of f(x) get very close to, and possibly equal, to the same real number L and that the values of can be made arbitrarily close to L by taking values of x close enough to c, but not equal to c. f(x) Then it is said that The limit of the function f(x) as x approaches c is the number L, which is written f(x) L lim xSc NOTE The limit in Example 1 is very important in calculus. Example 1 Limit of a Function If f x 1 2 sin x f x, find lim xS0 x. 2 1 Solution f x is not defined when Although be used with the graph of the function to examine the values of x is very close to 0. The table feature can also be used. a calculator’s trace feature can when x f 2 1 2 1 x 0, Calculator Exploration f Create a table of values for x 0. than and larger than same value as x approaches 0 from both sides? 1 Are the function values approaching the with values of x both smaller x 2 sin x x 912 Chapter 14 Limits and Continuity 2 The exploration should suggest that lim xS0 f x 1 2 1 or equivalently, lim xS0 sin x x 1, 2 2 |
a fact that will be proved in calculus. ■ 2 Figure 14.1-2 Example 2 Limit of a Function Find lim xS2 f, where f is the function given by the following two-part rule. x 1 2 f x 1 2 0 if x is an integer 1 if x is not an integer e Solution A calculator is not much help here, but the function is easily graphed by hand. y 3 2 1 0 −1 −1 −2 −3 −3 −2 x 1 2 3 Figure 14.1-3 When x is a number very close but not equal to 2 (either greater than 2 or less than 2), the corresponding value of is 1, and this is true no matter how close x is to 2. Thus, x f 1 2 lim xS2 f x 1 2 1. Because same as the value of the function f at 2 2 1 f 0 x 2. by definition, the limit of f as x approaches 2 is not the ■ Limits and Function Values If the limit of a function f as x approaches c exists, this limit may not be equal to f(c). In fact, f(c) may not even be defined. Very often the limit of a function as x approaches a point is equal to the value of the function at that point. Section 14.1 Limits of Functions 913 2 Example 3 Limit of a Function and a Function Value If f x 2 1 sin2 px cos px, find f 0.5 1 2 and lim xS0.5 f x. 2 1 3 3 Solution 2 Figure 14.1-4 0.5 f 1 2 f˛ 1 2b a sin2 p 2 cos p 2 12 0 1. Using the trace feature on the graph of gests that the limit is a number near 1. f x 1 2, shown in Figure 14.1-4, sug- A much narrower viewing window is needed to determine the limit more precisely. Graphing Exploration 2 1 x f Graph 0.99 y 1.01. both sides of x 0.5, f 1 2 in a viewing window with and Use the trace feature to move along the graph on and confirm that as x gets closer and closer to x 0.5 0.49 x 0.51 gets closer and closer to 1. The Exploration suggests that lim xS0.5 f x 1 2 Thus, the value of the function at approaches 0.5. 1 f 1 x 0.5 0. |
5. 2 is the same as the limit as x ■ NOTE Whenever a calculator was used in preceding examples, it was said that the information provided by the calculator suggested that the limit of the function was a particular number. Although such calculator explorations provide strong evidence, they do not constitute a proof and in some instances can be very misleading. (See Exercise 36.) Nevertheless, a calculator can help you develop an intuitive understanding for limits, which is needed for a rigorous treatment of the concept. Nonexistence of Limits Not every function has a limit at every number. Limits can fail to exist for several reasons. 914 Chapter 14 Limits and Continuity Nonexistence of Limits The limit of a function f as x approaches c may fail to exist if: 1. f(x) becomes infinitely large or infinitely small as x approaches c from either side 2. f(x) approaches L as x approaches c from the right and f(x) as x approaches c from the left approaches M, with M L, 3. f(x) oscillates infinitely many times between two numbers as x approaches c from either side The examples that follow illustrate each of these possibilities. 10,000 Example 4 A Function that Approaches Infinity Discuss the existence of lim xS0 1 x2. Solution 0.1 0.1 Figure 14.1-5 shows the graph of f x 1 2 1 x2 near x 0. As x approaches 1,000 Figure 14.1-5 y 2 1 −1 −2 −2 −1 become larger 0 from the left or right, the corresponding values of and larger without bound — rather than approaching one particular number—which can be verified by using the trace feature. Therefore, x f 1 2 lim xS0 1 x2 does not exist. ■ Example 5 A Function that Approaches Two Values x Find 0 lim xS0 x 0x, if it exists. 1 2 Solution The function f 0 x 1 2 x 0x is not defined when x 0. Figure 14.1-6a According to the definition of absolute value, x x There are two possibilities. x 6 0. x when 0 0 0 x when x 7 0 and 0 If x 7 0, then f If x 6 0, then f 0 0 x 0x x 0x. 1 Consequently, the graph of f looks like Figure 14.1-6a. A table of values for is shown in Figure 14.1-6b. x f 1 2 Figure 14.1-6b |
If x approaches 0 from the right, that is, through positive values, then the Section 14.1 Limits of Functions 915 f x corresponding value of that is, from negative values, then the corresponding value of always ding values of the definition of limit. Therefore, the limit does not exist. is always 1. If x approaches 0 from the left, is Thus, as x approaches 0 from both sides of 0, the correspondo not approach the same real number, as required by 1 ■ Example 6 An Oscillating Function Find lim xS0 sin p x, if it exists. Solution To understand the behavior of f x is close to 0. sin p x x 1 2, consider what happens when As x takes values: Then p x takes values: from from from 1 2 1 4 1 6 to to to 1 4 1 6 1 8 2p to 4p 4p to 6p 6p to 8p p x completes one period of the sine function from Thus, the graph of sin x 1 x 1 2 4, to another from x 1 4 nomenon occurs for negative values of x. Consequently, the graph of f 1 oscillates infinitely often between and 1, with the waves becoming more and more compressed as x approaches 0, as shown in Figure 14.1-7. and so on. A similar phe- x 1 6 to, y 1 0 −1 −2 −1 x 2 1 Figure 14.1-7 As x approaches 0, the function takes every value between nitely many times. In particular, p x 1 does not exist. real number. Therefore, and 1 infidoes not approach one particular sin x f 2 lim xS0 1 ■ 916 Chapter 14 Limits and Continuity Exercises 14.1 In Exercises 1–10, complete the table and use the result to estimate the given limit. 1. lim xS3 x 3 x2 2x 3 2.9 2.99 2.999 3.001 3.01 3.1 x f x 2 1 2. lim xS3 x 3 x2 9 2.9 2.99 2.999 3.001 3.01 3.1 x f x 2 1 3. lim xS0 2x 2 22 x x f x 1 2 8. lim xS1 0.9 0.99 0.999 1.001 1.01 1..1 1.01 1.001 0.999 0.99 0.9 f x 1 |
2 9. lim xS0 cos x 1 x 0.1 0.01 0.001 0.001 0.01 0.1 x f x 1 2 0.1 0.01 0.001 0.001 0.01 0.1 10. lim xSp 4 tan. lim xS0 2x 5 25 x 0.1 0.01 0.001 0.001 0.01 0.1 x f x 2 1 5. lim xS7 22 x 3 x 7 x 7.1 7.01 7.001 6.999 6.99 6.9 f x 1 2 6. lim xS0 21 x 1 x x f x 2 1 7. lim xS1 0.1 0.01 0.001 0.001 0.01 0..78 0.785 0.7853 0.7854 0.7855 0.786 f x 1 2 In Exercises 11–26, use a calculator to find a reasonable estimate of the limit. If the limit does not exist, explain why. 11. lim xS1 x6 1 x4 1 12. lim xS2 x5 32 x3 8 13. lim xS3 x2 x 6 x2 2x 3 14. lim xS1 x2 1 x2 x 2 15. lim xS1 x3 1 x2 1 17. lim xS0 tan x x x3 19. lim xS0 x tan x x sin x 21. lim xS3 2x 23 x 3 23. lim xS0 x ln 0 x 0 16. lim xS2 x2 5x 6 x2 x 6 18. lim xS0 tan x x sin x 20. lim xS0 x sin 2x x sin 2x 22. lim xS25 2x 5 x 25 24. lim xS0 e2x 1 x 25. lim xS0 ex 1 sin x 26. lim xS0 a x sin 1 xb 30. In Exercises 27–32, use the graph of the function f to determine the following: lim xS3 f(x) lim xS0 f(x) lim xS2 f(x) −4 −3 −2 27. 28. 291 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 −4 −4 −3 − |
2 −4 −3 −2 −4 −3 − Section 14.1 Limits of Functions 917 y y y 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 − 31. 32. −4 −3 −2 −4 −3 −2 33. a. Graph the function f whose rule is if x 6 2 if 2 x 6 2 if x 2 if x 7 2 Use the graph in part a to evaluate the following limits. x lim xS2 d. lim xS2 lim xS1 c. Explain why value of c. lim xSc t 1 x 2 does not exist for every 36. Critical Thinking If f 1 cos x6 x12 x 1 2, then 1 2 1 2, x f as is shown in calculus. A calculator lim xS0 or computer, however, may indicate otherwise. f Graph 1 0.1 x 0.1, determine the values of What does this suggest that the limit is? and use the trace feature to in a viewing window with when x approaches 0. x x f 2 1 2 918 Chapter 14 Limits and Continuity 34. a. Graph the function g whose rule is g x 1 2 µ x2 x 2 2 3 x if x 6 1 if 1 x 6 1 if x 1 if x 7 1 Use the graph in part a to evaluate the following limits. x lim xS1 lim xS0 lim xS1 d. c. 35. Critical Thinking Consider the function t whose rule is x t 1 2 0 if x is rational 1 if x is irrational e 14.2 Properties of Limits Objectives • Find the limit of the constant function the identity function • Use the properties of limits • Find the limit of polynomial functions rational functions • Use the Limit Theorem Most of the functions that appear in calculus are combinations of simpler functions, such as sums, products, quotients, and compositions. This section presents rules for finding limits of combinations of functions without a table or a graph. There are two easy, but important, cases where the limit of a function may be found by evaluating the function. The first occurs with constant funcnote that as x lim tions, such as xS3 is always the number 5, so 5. The same 5. 2 approaches 3, the corresponding value of 5 5. that that is, x is the number |
To find Thus, lim xS3 5 lim xS3 lim xS3 2 thing is true for any constant function. 1 1 2 lim xS3 1 2 Limit of a Constant If d is a constant, then lim xSc d d. The same phenomenon occurs with the identity function, which is given If c is any real number, then the statement “x by the rule approaches c” is exactly the same as the statement “I(x) approaches c” because for every x. Thus, 2 1 x x. x x I I 1 2 Limit of the Identity Function For every real number c, lim xSc x c. Section 14.2 Properties of Limits 919 Properties of Limits There are a number of facts that greatly simplify the computation of limits. Rigorous proofs of these properties will not be given, knowing the central idea is more important now. For instance, suppose that as x approaches c, the values of a function f approach a number L and the values of a function g approach a number M. Then it is plausible that as x L M approaches c, the values of the function and that Following is a formal the values of the function statement in terms of limits. approach approach L M. f g f g Properties of Limits If f and g are functions and c, L, and M are numbers such that lim xSc f(x) L and lim xSc g(x) M, In Property 4, NOTE f gb1 x lim 2 xSc a M 0 and L 0, and may not exist. does not exist if L 0. M 0 If the limit may or then 1. lim xSc ( f g)(x) lim xSc L M [f(x) g(x)] 2. lim xSc ( f g)(x) lim xSc L M [f(x) g(x)] 3. lim xSc (f g)(x) lim xSc L M [f(x) g(x)] 4. lim xSc a f gb (x) lim xSc a f(x) g(x)b L M 2f (x) 2L 5. lim xSc (for M 0 ) (provided f(x) 0 for all x near c) x f L lim xSc These properties are often stated somewhat differently. For instance, M, because 2 lim xSc and similarly for the other properties. Property 1 can be written as lim xSc f g lim |
xSc lim xSc and 1 x g g 21 Limits of Polynomial Functions Properties 1–3, together with the facts about limits of constants and the identity function presented at the beginning of the section, now make it easy to find the limit of any polynomial function. Example 1 Limit of a Polynomial Function If f x 1 2 x2 2x 3, find lim xS4 f x 1 2 lim xS4 1 x2 2x 3. 2 920 Chapter 14 Limits and Continuity Solution lim xS4 1 2x lim 3 xS4 x lim xS4 lim xS4 x lim xS4 x 2 lim xS4 4 x2 lim xS4 2 lim xS4 x 3 3 2 1 2 x2 2x 3 lim xS4 lim xS4 4 21 1 16 27 2 x lim xS4 x lim xS4 4 2 8 2 1 3 Properties 1 and 2 3 Property 3 Limit of a constant Limit of x Note that the limit of value of the function at x f 2 x 4, x2 2x 3 namely, f 1 at 4 x 4 4 1 2 1 is the same as the 3 27. 2 2 ■ 4 2 1 2 Because any polynomial function consists of sums and products of constants and x, the same argument used in Example 1 works for any polynomial function and leads to the following conclusion. Limits of Polynomial Functions If f(x) is a polynomial function and c is any real number, then f(x) f(c). lim xSc In other words, the limit is the value of the polynomial function f at x c. Limits of Rational Functions The fact in the previous box and Property 4 of limits make it easy to compute limits of many rational functions. Example 2 Limit of a Rational Function If f x 1 2 x3 3x2 10 x2 6x 1, find lim xS2 f x. 2 1 3.1 Solution 4.7 4.7 3.1 Figure 14.2-1 0.86, The graph of f near as shown in Figure 14.2-1. You can determine the limit exactly by noting that suggests that the limit is a number near x 2 x f 1 2 is the quotient of the two functions x3 3x2 10 and h x g 1 2 x2 6x 1. x 1 2 As x approaches 2, the limit of each can be found by evaluation of |
the functions at Therefore, x 2. Section 14.2 Properties of Limits 921 lim xS2 f x 1 2 2 lim xS2 lim 1 xS2 lim xS2 x3 3x2 10 x2 6x 1 b Q x3 3x2 10 x2 6x 1 1 23 3 22 10 22 6 2 1 6 7 6 7 0.857 2 limit property 4 limits of polynomial functions Note that the limit of f x 1 2 as x approaches 2 is the number f 2. 2 1 ■ The procedure in Example 2 works for other rational functions as well. Limits of Rational Functions Let f(x) be a rational function and let c be a real number such that f(c) is defined. Then f(x) f(c). lim xSc In other words, the limit of a rational function as x approaches c is the value of the function at if the function is defined there. x c, When a rational function is not defined at a number, different techniques must be used to find its limit there—if it exists. Example 3 Limit of a Rational Function If f x 2 1 x2 2x 3 x 3, find lim xS3 f x. 2 1 Solution f x is not defined when x 3, Because the limit cannot be found by evaluating the function. Although it could be estimated graphically, it can be found by using algebraic methods. Begin by factoring the numerator. 2 1 x2 2x 3 x 3 1 x 1 x 3 21 x 3 2 Because the numerator and denominator have a common factor, the rational expression may be reduced. x 1 x 3 21 x 3 1 2 x 1, for all x 3. The definition of the limit as x approaches 3 involves only the behavior of the function near The preceding equation shows have exactly the same that both x 3. and not at g x and the function 922 Chapter 14 Limits and Continuity Figure 14.2-2 Limit Theorem NOTE Do not confuse the variable h with the function h. The meaning should be clear in context. values at all numbers except x approaches 3. x 3. So they must have the same limit as x2 2x 3 x 3 lim xS3 lim xS3 1 x 1 2 3 1 4, as illustrated in Figure 14.2-2. ■ The technique illustrated in Example 3 applies in many cases. When two functions have identical behavior, except possibly at they will have the same limit as |
x approaches c. x c, If f and g are functions that have limits as x approaches c and f(x) g(x) for all x c, then lim xSc f(x) lim xSc g(x). Recall that the difference quotient of a function f is given by The difference quotient can be evaluated for a specific value of x, say x c, to obtain a new form Limits of the difference quotient of a function f play an important role in calculus. When computing such limits, the variable is often the quantity h, as in the following example. Example 4 Limit of a Difference Quotient If f x 2 1 x2, find lim hS0. Solution Using algebra, write the difference quotient as follows 52 5 h 2 h 25 10h h2 1 10h h2 h h 25 2 Section 14.2 Properties of Limits 923 When the difference quotient is written this way, it is easy to see that it is a function of h, and the function is not defined when Find the limit of the difference quotient as h approaches 0. h 0 lim hS0 2 lim hS0 10h h2 h 10 h h 10 h h 1 lim hS0 lim hS0 2 1 10 0 10 2 Factor the numerator. Limit Theorem limit of a polynomial function ■ Exercises 14.2 In Exercises 1 – 8, use the following facts about the functions f, g, and h to find the required limit. h(x) 2 f(x) 5 lim xS4 g(x) 0 lim xS4 lim xS4 1. lim xS4 1 f g x 2 21 3. lim xS4 f g 1 x 2 2. lim xS4 1 g h x 2 21 4. lim xS4 g h 1 x 2 5. lim xS4 f g x 2 21 1 6. lim xS4 3 7. lim xS4 3h 2f gb 1 x 2 a 8. lim xS4 a 4h b 1 2 h x 2 4 1 f 2g 20. lim xS2 22. lim xS0 In Exercises 24–27, find 21. lim xS0 22 x 22 x 23. 0 lim xS3 x 3 x 3 0 f(2 h) f(2) h. lim hS0 x 2 24. 26. f f x x 1 1 2 |
2 x2 x2 x 1 25. 27. f f x x 1 1 2 2 x3 2x In Exercises 9–23, find the limit, if it exists. If the limit does not exist, explain why. 9. lim xS2 1 6x3 2x2 5x 3 2 10. lim xS11 x7 2x5 x4 3x 4 2 11. lim xS2 3x 1 2x 3 12. lim xS3 x2 x 1 x2 2x 13. lim xS1 2x3 6x2 2x 5 14. lim xS2 2x2 x 3 15. ° lim xS0 ¢ 16. ° lim xS0 17. lim xS1 a 1 x 1 2 x2 1b 18. lim xS2 c x2 x 2 2x x 2 d 19. lim xS0 x2 x 0 0 In Exercises 28–29, use a unit circle diagram to explain why the given statement is true. 28. sin t 1 lim tSp 2 29. cos t 0 lim tSp 2 Exercises 30–34 involve the greatest integer function, f(x) [x], which is defined to be the greatest integer that is less than or equal to a given number x. See Section 3.1. Use a calculator as an aid in analyzing these problems. 30. For x h 1 exists. 2 31. For x g 1 exists. 2 x 3 4 x 4 3, find lim xS2 h 1 x 2, if the limit x x 3 4, find lim xS2 g 1 x 2, if the limit 3 x 4 2 x x 3 4, find lim xS3 r 1 x 2, if the limit 32. For r x 1 exists. 924 Chapter 14 Limits and Continuity 33. For k x 1 exists. 2 x x 3 4 x 3 4, find lim xS1 k 1 x 2, if the limit 34. For f x 1 2 x 0 x 0, find lim hS0. 36. Critical Thinking Give an example of functions f x and g and a number c such that neither f lim xSc 1 2 nor lim xSc g 1 x 2 exists, but lim xSc f g x 2 21 1 does exist. 35. Critical Thinking Give an example of functions f x and g and a number c such that neither f nor lim xSc g 1 x 2 exists |
, but lim xSc 1 f g x 2 2 1 lim 2 xSc does exist. 1 14.2.A Excursion: One-Sided Limits Objectives • Find one-sided limits The function whose graph is shown in Figure 14.2.A-1 is defined for all values of x except x 4. y 4 2 −2 0 2 −2 −4 Figure 14.2.A-1 x 6 As x approaches 4 from the right, that is, takes values larger than but close f to 4, the graph shows that the corresponding values of get very close 1 as x approaches 4 from the right to 2. Consequently, “the limit of is 2.” x x f 2 1 2 xS4 f lim x 1 2 2 x 7 4 The small plus sign on 4 indicates that only the values of x with are considered. Similarly, as x approaches 4 from the left, the graph Conshows that the corresponding values of sequently, “the limit of get very close to as x approaches 4 from the left is 1. 1. x x f f 2 1 1 2 lim xS4 f x 1 2 1 The small minus sign indicates that only values of x with considered. x 6 4 are Section 14.2.A Excursion: One-Sided Limits 925 These “one-sided” limits are a bit different than the “two sided” limits discussed in Section 14.2. When x approaches 4 from the left and from the right, the corresponding values of do not approach a single number, so the limit as x approaches 4, as defined in Section 14.1, does not exist. x f 2 1 The same notation and terminology are used in the general case, where f is any function and c and L are real numbers. The definition of the “righthand limit,” xSc f lim x 2 1 L, ” in place of the phrase “on both sides of is obtained by inserting “ c” in the definition of limit in Section 14.2. The function f need not be defined when x 6 c. x 7 c Similarly, inserting “ definition produces the definition of the “left-hand limit.” ” in place of “on both sides of c” in the same x 6 c Again, the function f need not be defined when x 7 c. f lim xSc x 2 1 L Example 1 One-Sided |
Limit Find xS3 lim 1x 3 1 A B. Solution f 2x 3 1 The function x 3 and values of x to the right of 3. The graph of ure 14.2.A-2a, and a table of values is shown in 14.2.A-2b. is defined only when x x f 2 1 2 1 x 3, that is, for is shown in Fig- 2.2 2.9 0 4 Figure 14.2.A-2a Figure 14.2.A-2b The values of approach 1 as x approaches 3 from the right. There- 2 x f 1 1x 3 1 A fore, xS3 lim 1. B ■ Computing One-Sided Limits The computation of one-sided limits is greatly facilitated by the following fact. 926 Chapter 14 Limits and Continuity All the results about limits in Section 14.2, such as the properties of limits, limits of polynomial functions, and the Limit Theorem, x S c. ” remain valid if “ ” is replaced by either “ x S c x S c ” or “ Example 2 Using Properties of Limits Find xS319 x2. lim Solution 29 x2 1 2 f x The function because the quantity under the radical is negative for other values of x. Compute as x approaches 3 from the left by using the properties the limit of of limits: is defined only when x f 2 1 3 x 3, lim xS3 9 x2 29 x2 2 lim 2 1 xS3 9 lim 2 lim xS3 xS3 2 lim xS3 9 lim xS3 x2 x lim xS3 x B BA A property 5 property 2 property 3 29 32 0 limit of a constant limit of the identity function ■ Three Types of Limits Notice that there are three kinds of limits: left-hand limits, right-hand limits, and “two-sided” limits as defined in Section 14.1. Example 1 exhibits a function that has a right-hand limit at but no left-hand or two sided limit. Even when a function has both a left-hand and a right-hand limit at these limits may not be the same, as was shown in the introduction of this Excursion. x 3, x c, f x x c It is clear, however, that a function which has a two-sided limit L at x c. necessarily has L as both its left- |
hand and right-hand limit at If the can be made arbitrarily close to L by taking x close enough values of to c on both sides of c, then the same thing is true if you take only values of x on the left of c or on the right of c. Conversely, if a function has the same left-hand and right-hand limits at then this number must be a two-sided limit as well. In summary, x c, 1 2 Two-sided Limits Let f be a function and let c and L be real numbers. Then f(x) L lim xSc exactly when xSc f(x) L lim and lim xSc f(x) L. Section 14.2.A Excursion: One-Sided Limits 927 Example 3 Limits From the following graph, find a. d. lim xS2 f x 1 2 lim xS6 f x 1 2 b. e. lim xS2 f x 1 2 lim xS6 f x 1 2 c. lim xS2 2 − Figure 14.2.A-3 Solution The graph shows that a. d. lim xS2 f x 1 2 lim xS6 f x 2 1 3 5 b. e. lim xS2 f x 1 2 lim xS6 f x 1 2 3 4 c. lim xS2 f x 1 2 3 ■ Exercises 14.2.A In Exercises 1–6, use a calculator to find a reasonable estimate of the limit. 7. 1. lim xSp sin x 1 cos x 3. xS0 2x lim 1 ln x 2 5. xS0 lim sin 6x x 2. lim 1 xSp 2 sec x tan x 2 4. lim xS0 a x 1 xb 6. lim xS0 sin 3x 1 sin 4x In Exercises 7–10, use the graph of the function f to determine the required limits. a. c. xS2 f lim 1 x xS3 f lim 1 x 2 2 b. d. xS0 f lim xS3 f lim 1 −1 −2 −3 −4 −4 −3 −2 x 1 2 3 4 928 8. 9. 10. Chapter 14 Limits and Continuity y y y 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 |
−1 −2 −3 −4 −4 −3 −2 −4 −3 −2 −4 −3 −. a. b. xS2 f lim 1 x xS2 f lim 11. In Exercise 33a of Section 14.1, find xS2 f lim 1 x xS2 f lim 12. In Exercise 34a of Section 14.1, find xS1 g lim 1 x xS1 g lim xS1 g lim 1 x xS1 g lim d. d. b. c. a In Exercises 13–22, find the limit. 13. xS1 lim A 2x 1 3 B 15. xS4 lim x 4 x2 16 17. xS3 lim 3 x2 9 19. 20. 25 2x x B 2x 3 23x lim xS2.5A xS3 lim A 14. 16. xS3 23 x lim x 2 x 2 xS2 0 lim 0 18. xS2 lim x 1 x2 x 2 B 21. xS3 lim 0 a x 3 x 3 0 2x 3 1 b 22. xS4 lim 2 0 A x 0 4 x2 B Exercises 23 and 24 involve the greatest integer function, See Exercise 30 in Section 14.2. f(x) [x]. 23. Find xS2 lim 24. Find lim xS3 3 1 x 4 x xS2 and lim x. 4 3 x 3 4 2 and lim xS3 1 x x. 4 2 3 Section 14.3 The Formal Definition of Limit 929 14.3 The Formal Definition of Limit (Optional) Objectives • Use the formal definition of limit The informal definition of limit in Section 14.1 is quite adequate for understanding the basic properties of limits and for calculating the limits of many familiar functions. This definition, or one very much like it, was used for more than a century and played a crucial role in the development of calculus. Nevertheless, the informal definition is not entirely satisfactory; it is based on ideas that have been illustrated by examples but not precisely defined. Mathematical intuition, as exemplified in the informal definition of limit, is a valuable guide; but it is not infallible. On several occasions in the history of mathematics, what first seemed intuitively plausible has turned out to be false. In the long run, the only way to guarantee the accuracy of mathematical results is to base them on rigorously precise definitions |
and theorems. This section takes the first step in building this rigorous foundation by developing a formal definition of limit. In order to keep the discussion as concrete as possible, suppose f is a funcYou do not need to know the rule of f or tion such that 12. f x lim xS5 1 2 anything else about it to understand the following discussion. The informal definition of the statement in Section 14.1 has two components: –lim xS5 f x 1 2 12– that was given A. As x takes values very close to but not equal to 5, the f get very close—and possibly are x corresponding values of equal—to 12. B. The value of x f 1 2 1 2 want) to 12 by taking x sufficiently close (but not equal) to 5. can be made arbitrarily close (as close as you In a sense, Component A is unnecessary because it is included in Component B: If the values of can be made arbitrarily close to 12 by taking x close enough to 5, then presumably can be made to get very close 2 to 12 by taking values of x very close to 5. x x f f 1 1 2 Consequently, to obtain the formal definition of limit, begin with Component B of the informal definition. f(x) 12 means that lim xS5 as close as you want to 12 by taking x close enough to 5. the value of f(x) can be made [1] The definition above, referred to as Definition [1], will be modified step by step until the formal definition of limit is reached. Definition [1] says, in effect, that there is a two-step process involved in finding a limit, if it exists: 1. Know how close you want 2. Determine how close x must be to 5 to guarantee this. to be to 12, x f 1 2 Definition [1] can now be restated as Definition [2]. 930 Chapter 14 Limits and Continuity y 12.01 11.99 5 11.99 6 f x is in this interval when x 1 Figure 14.3-1 6 12.01 2 y ε ε 12 δ δ 5 Figure 14.3-2 x x f(x) 12 means that lim xS5 f(x) should be to 12, you know how close x must be to 5 to guarantee it. whenever you specify how close [2] For example, if you want to be within 0.01 of 12, that is |
, between 11.99 and 12.01, you can find how close x must be to 5 to guarantee that x f 1 2 2 1 Any value of x in the interval around 5 shown in Figure 14.3-1 will produce 11.99 6 f 6 12.01. x 11.99 6 f x 6 12.01. 1 2 But “arbitrarily close” implies much more. You must be able to find how close x is to 5 regardless of how close you want to be to 12. If you want to be within 0.002 of 12, or 0.0001 of 12, or within any distance of 12, you must be able to find how close x must be to 5 in each case to accomplish this. So, Definition [2] can be restated as follows(x) 12 means that no matter what positive number lim xS5 you specify in measuring how close you want f(x) to be to 12, you must be able to find how close x must be to 5 in order to guarantee that f(x) is that close to 12. [3] 2 1 e x d should be to 12 will be denoted by the Greek letter Hereafter, the small positive number you specify in measuring how close f (epsilon). When you know how close x should be to 5 to accomplish this, denote the (delta). Presumably the number which number by the Greek letter measures how close x must be to 5, will depend on the number which x measures how close you want to be to 12. Using this language, Def1 inition [3] becomes Definition [4]. f(x) 12 means that lim xS5 there is a positive number this property: If x is within of 5 but not equal to 5, then f(x) is within of 12, and possibly equal to 12. for every positive number that depends on with d, e, D E [4] D E E f 2 Although Definition [4] is essentially the formal definition, somewhat briefer notation is usually used. If you think of ment f x 1 2 means that and 12 as numbers on the number line, then the state- is within of 12 e f x 1 2 the distance from f x to 12 is less than e. 2 1 Because distance on the number line is measured by absolute value (see Sections 2.4 and 2.5.A), the last statement can be written as f x 1 2 0 12 0 6 |
e. Similarly, saying that x is within of 5 and not equal to 5 means that the distance from x to 5 is less than but greater than 0, that is 0 6 d d x 5 6 d. 0 0 e, NOTE d, Epsilon, and are Greek letters delta, that are often used to represent small amounts. Section 14.3 The Formal Definition of Limit 931 Using this notation, Definition [4] becomes the desired formal definition. f(x) 12 means that lim xS5 there is a number x 5 if 0 66 D 66 D, then 00 00 00 for each positive number E, 66 E. that depends on f(x) 12 such that 00 E, [5] Definition [5] is sufficiently rigorous because the imprecise terms “arbitrarily close” and “close enough” in the informal definition have been replaced by a precise statement using inequalities that can be verified in specific cases, as will be shown in the examples that follow. There is nothing special about 5 and 12 in the preceding discussion; the entire analysis applies equally well in the general case and leads to the formal definition of limit, which is just Definition [5] with c in place of 5, L in place of 12, and f for any function. Definition of Limit Let f be a function and let c be a real number such that f (x) is defined for all x, except possibly in some open interval containing c. The limit of f (x) as x approaches c is L, which is written x c, f(x) L, lim xSc provided that for each positive number number E that depends on with the property E, there is a positive D if 0 66 x c 00 00 66 D, then f(x) L 66 E. 00 00 This definition is often called the e-d definition of limits. Example 1 Proving a Limit Let f x 1 2 4x 8 and prove that lim xS5 f x 1 2 12. Solution c 5, L 12, Apply the definition of limit with e ure 14.3-3 illustrates the situation. Suppose that d Find a positive number with the property 6 d, then x 5 If 0 6 f f 4x 8. and Figis any positive number. x 2 1 x 1 12 0 2 6 e. 0 Let be the number d, and show that this will work. For now, do not d was found; just verify that the following argu- 0 |
0 e 4 d e 4 y 12 + ε 12 12 − ε x 5 5 + δ 5 − δ worry about how ment is valid. Figure 14.3-3 932 Chapter 14 Limits and Continuity If 0 x 5 0 6 d, then 4x 20 6 e 0 12 6 e 0 12 6 e 0 e 7 0, 6 d, then 12 lim xS5 1 2 0 1 4x 8 2 x f Multiply both sides by 4. 4 a 0 0 b ab 4 0 0 Rewrite 20 as 8 12. 4x 8 This verifies that for each x 5 If 0 6 This completes the proof that d there is a positive with the property 12 6 e. x f 0 1 2 0 ■ Proofs like the one in Example 1 often seem mysterious to beginners. d Although they can follow the argument after has been found, they do d. Example 2 gives a fuller picture of the processes not see how to find used in proving statements about limits. Example 2 proves that for any positive number d there exists such that x f ˛1 2 0 9 0 6 e whenever 0 6 0 6 d. 0 e, x 1 Example 2 Use the E-D Definition of Limit Prove that lim xS1 1 2x 7 9. 2 Solution and L 9. Let be a positive number e 2x 7, c 1, x f˛ 1 In this case, 2 d and find a with the property If 0 6 x 1 0 0 6 d, then 2x 7 9 0 2 0 1 6 e. In order to get some idea which might have this property, work backwards from the desired conclusion, namely, d The last statement is equivalent to 2x 7 9 0 2 0 1 6 e. 2x 2 6 e, 0 0 which in turn is equivalent to each of the following statements Section 14.3 The Formal Definition of Limit 933 When the conclusion is written this way, it suggests that the number e 2 would be a good choice for d. Everything up to here has been “scratch work.” Now give the actual proof, written forwards. Given a positive number e, let be the number d e 2 If 0 6. x 1 0 0 6 d, then 2x 2x 7 f ˛ Multiply both sides by 2. 2 a 0 0 b ab 2 0 0 0 0 0 0 Rewrite 2 as 7 9. 2x 7 f ˛1 x |
2 Figure 14.3-4 Therefore, d e 2 has the required property, and the proof is complete. ■ Proving Limit Properties Once the algebraic scratch work was done in Examples 1 and 2, the limit proofs were relatively easy. In most cases, however, a more involved argument is required. In fact, it can be quite difficult to prove directly from the definition, for example, that lim xS3 such complicated calculations can often be avoided by using the various limit properties given in Section 14.2. Of course, these properties must first be proved using the definition. Surprisingly, the proofs are comparatively easy. Fortunately,. x2 4x 1 x3 2x2 x 1 3 Example 3 Proving Limit Properties Let f and g be functions such that Prove that lim xSc x f ˛1 2 1 2 x f˛ 1 lim xSc g˛1 x 22 L and lim xSc L M. M. x g˛1 2 Solution Scratch Work: If erty e is any positive number, find a positive with the prop- d If 0 6 x c 0 0 6 d, then x f˛ 1 2 0 1 g˛1 x 22 1 L M 2 0 6 e. 934 Chapter 14 Limits and Continuity Note the following result of the triangle inequality. (See Section 2.4.) x f ˛1 2 0 1 g˛1 x 22 1 L M 2 0 x f˛ 1 0 1 x f1 1 g˛ If there is a with the property d If 0 6 x c 0 0 6 d, then x 2 f ˛1 0 g ˛1 x d then the smaller quantity 6 d. e x c when 0 6 x f ˛1 0 1 1 Such a can be found as follows. 2 0 22 2 0 0 L g ˛1 0 0 L M x M 6 e, 0 2 will also be less than Proof Let be any positive number. Because e inition of limit with e 2 in place of e: lim xSc f ˛1 x 2 L, apply the def- there is a positive number d1 with the property 6 d1, then 0 6 e 2 If 0 6 x c L f ˛1 x. 2 0 Similarly, because lim xSc 0 g˛1 x 2 0 M, there is a positive number with the property d2 6 d2, then 1 If 0 |
6 x c 0 0 d Then Now let be the smaller of the two numbers d d2. x c x c and it must be true that x c and 0 6 if 0 6 0 6 6 d, 6 d1 d1 0 0 0 0 6 d2. 0 d2, so that d d1 and Therefore, Consequently, if L 6 e 2 6 d˛ 1 2 0 6 g˛1 x 0 22 2 0 x f ˛1 2 0 1 0 1 and x g ˛˛1 1 x g˛1 2 M 2 M 0 2 0 0 then x f ˛1 0 1 x f ˛1 0 2 6 e e 2 2 e It has been shown that for any x c If 0 6 0 lim xSc f ˛1 1 6 d, then g˛1 L M 22 x 0 1. 0 x 2 Therefore, e 7 0, there is a g˛1 f ˛1 x 2 d 7 0 x 22 1 with the property: L M 6 e. 2 0 ■ The proofs of the other limit properties and theorems of Section 14.2 are introduced in calculus. Section 14.3 The Formal Definition of Limit 935 One-Sided Limits The formal definition of limit may easily be carried over to one-sided limits, as defined informally in Excursion 14.2.A, by using the following fact: x c 6 d exactly when d 6 x c 6 d, 0 which is equivalent to 0 x c 6 d exactly when c d 6 x 6 c d. 0 0 Thus, the numbers between c and tance of c, and the numbers between distance of c. d d c d lie to the right of c, within disand c lie to the left of c, within c d Consequently, the formal definition of right-hand limits can be obtained 6 d from the definition above by replacing the phrase “if ” c 6 x 6 c d. ” For a formal definition of the left-hand limits, with “ 0 6 replace the phrase “if c d 6 x 6 c. ” with “if Exercises 14.3 In Exercises 1–12, use the formal definition of limit to prove the given statement, as in Example 1. 1. 2. 3. 4. 5. 6. 7. 8. 9. lim xS1 lim xS5 lim x |
S0 lim xS2 lim xS7 lim xS1 lim xS2 lim xS4 10. lim xS1 3x 2 lim xS3 1 4x 6x 3 1 1 2 15 2 5 2 2x 19 1 4 4 p p x 6 2 2x 7 1 1 2 5 2 11. 12. lim xS2 lim xS3 1 1 2x 5 2x 4 2 2 1 10 In Exercises 13 and 14, use the formal definition of limit to prove the statement. 13. 14. lim xS0 lim xS0 x2 0 x3 0 In Exercises 15 and 16, let f and g be functions such that lim xSc f˛(x) L and lim xSc g˛(x) M. 15. Critical Thinking Prove that L M. g˛1 lim xSc f ˛1 22 x x 1 2 Hint: see Example 3. 16. Critical Thinking If k is a constant, prove that lim xSc k f x 2 1 kL. 936 Chapter 14 Limits and Continuity 14.4 Continuity Objectives • Determine if a function is continuous at a point • Determine if a function is continuous on an interval • Apply properties of continuous functions Calculus deals in large part with continuous functions, and the properties of continuous functions are essential for understanding many of the key theorems in calculus. This section presents the intuitive idea of continuity, its formal definition, and the various properties of continuity—which were used in the work with graphs in earlier chapters. Continuity Informally Let c be a real number in the domain of a function f. Informally, the function f is continuous at if you can draw the graph of f at and near c, f ˛1 the point without lifting your pencil from the paper. For example, 1 each of the four graphs in Figure 14.4-1 is the graph of a function that is continuous at x c. x c 22 c (c, f(c)) c y y x (c, f(c)) x c y y (c, f(c)) c (c, f(c)) c x x Figure 14.4-1 Thus, a function is continuous at is connected and unbroken. x c if its graph around the point 1 c, f ˛1 c 22 On the other hand, none of the functions whose graphs are shown in Figure 14.4 |
-2 is continuous at Try to draw one of these graphs at c and near without lifting your pencil from the paper. x c. c, f ˛1 1 22 (c, f(c)) y c a. y c. Section 14.4 Continuity 937 (c, f(c)) c y b. y x x x x c c Figure 14.4-2 d. Figure 14.4-2 shows that a function is discontinuous, that is, not continx c. uous, at if the graph has a break, gap, hole, or jump when x c Calculators and Discontinuity When a calculator uses “connected” mode to graph a function, it plots a number of points and then connects them with line segments to produce a curve. Thus, the calculator assumes that the function is continuous at any point it plots. For example, a calculator may not show the hole in graph d of Figure 14.4-2, or it may insert a vertical line segment where the graph jumps in graph b of Figure 14.4-2. Consequently, a calculator may present misleading information about the continuity of a function. Analytic Description of Continuity The goal is to find a mathematical description of continuity at a point that does not depend on having the graph given in advance. This is done by expressing in analytic terms the intuitive geometric idea of continuity given above. 938 Chapter 14 Limits and Continuity If the graph of f can be drawn at and near a point pencil from paper, then there are two possibilities: c, f ˛1 1 c 22 without lifting (c, f(c)) is an interior point of the graph. (c, f(c)) (c, f(c)) (c, f(c)) is an endpoint of the graph. Figure 14.4-3 Continuity at an Interior Point c c, f˛ 1 If the point 1 c, f ˛1 drawn around the point 1 the very least, the following two statments are true. is an interior point of the graph, and the graph can be without lifting pencil from paper, then at 22 22 c x c x t, 2 x x • • must be defined for must be defined for f ˛1 f ˛1 2 f ˛1 is not defined for some t near c, there will be a hole in the graph For if at the point which would require lifting the pencil when drawing the |
graph. See graphs c and d of Figure 14.4-2 for functions that are not The situation can be described more precisely by saying: defined at when t is any number near c 1 1 x c. t, f 22 t t, 2 f(x) is defined for all x in some open interval containing c. In other words, there are numbers a and b with such that f(x) is defined for all x with a 66 x 66 b. In particular, f(c) is defined. a 66 c 66 b [1] x c, Although condition [1] is necessary in order for f to be continuous at x c this condition by itself does not guarantee continuity. For instance, and the graphs a and b in Figure 14.4-2 show functions whose graphs are defined for all values of x near c, but are not continuous at x c. is defined, there are two conditions that can prevent a function from c f ˛1 If being continuous at 2 x c: • There is a jump at does not exist. x c, that is, the limit of x f ˛1 2 as x approaches c • There is a hole in the graph at c f ˛1 but it is not equal to x c, 2 x c,. that is, limit of x f˛ 1 2 exists at The conditions that prevent a function from being continuous at a point are shown in Figure 14.4-4. Notice the reason that the graph is not continuous at and at x c. x b, x a, at Section 14.4 Continuity 939 y lim f(x) does not exist. x→b f(a) is not defined. lim f(x) ≠ f(c) x→c x ba c Figure 14.4-4 The preceding analysis leads to the following formal definition of continuity for interior points. Definition of Continuity Let f be a function that is defined for all x in some open interval containing c. Then f is said to be continuous at under the following conditions: x c 1. f(c) is defined 2. lim xSc 3. lim xSc exists f˛(x) f˛(x) f˛(c) Example 1 Continuity at a Point Without graphing, show that the function 2x2 x 1 x 5 x f ˛1 2 is continuous at x 3. Solution To show continuity of f at x 3, 3 |
f˛ 1 3 f ˛1 2 2 show that 2x2 x 1 x 5. lim xS3 213 8 940 Chapter 14 Limits and Continuity By the properties of limits given in Section 14.2, lim xS3 f x 2 1 lim xS3 2x2 x 1 x 5 2x2 x 1 x 5 2 1 x2 lim xS3 lim xS3 2 lim xS3 lim xS3 3 2 1 1 213 8 3 f˛ 1 2 x 2 limit of a quotient limit of a root 1 2 limit of a polynomial Therefore, lim xS3 f ˛1 and f is continuous at x 3. ■ The facts about limits presented in Sections 14.1 and 14.2 and the definition of continuity provide justification for several assumptions about graphs that were made earlier in this book. Every polynomial function is continuous at every real number. Every rational function is continuous at every real number in its domain. Every exponential function is continuous at every real number. Every logarithmic function is continuous at every positive real number. f ˛(x) sin x number. h(x) tan x and g ˛(x) cos x are continuous at every real is continuous at every real number in its domain. Continuity on an Interval Consider continuity at an endpoint of the graph of a function f, such as a, f shown in Figure 14.4-5. b, f or b a 1 1 22 1 1 22 Continuity of Special Functions NOTE One-sided limits, which were discussed in Section 14.2.A, are a prerequisite for the material on continuity on an interval. Section 14.4 Continuity 941 (b, f(b)) (a, f(a)) Figure 14.4-5 x a is that the graph of f can be The intuitive idea of continuity at drawn at without lifting the pencil from the paper. Essentially the same analysis that was given above can be made here if we consider only values of x to the right of In An short, continuity at the endpoint and to the right of the point x a. a. a, f a, f a, f 22 22 22 means that b, f b, 1 22 f xSa f lim which leads to the 2 2 1 1 analogous discussion applies to the endpoint formal definition. 1 Continuity from the Left and Right A function f is continuous from the right at x a provided that |
xSa f(x) f(a). lim A function f is continuous from the left at x b provided that xSb f˛(x) f(b). lim Example 2 Continuity at an Endpoint Show that 2x x f ˛1 2 is continuous from the right at x 0. Solution x The function 2 from the right at f ˛1 2x, x 0 which is not defined when and because 20 0, f 0 2 1 xS0 1x 0 f lim xS0 f lim x 1 2 x 6 0, is continuous 0. 2 1 ■ The most useful functions are those that are continuous at every point in an interval. Consider the following three examples. • • • h 1 g ˛1 f ˛1 x tan x 2 is continuous at every number in the interval ln x sin x is continuous at every number in 1 is continuous at every number in 0, q 2 q, q 1 2 942 Chapter 14 Limits and Continuity Intuitively, this means that their graphs can be drawn over the entire interval without lifting the pencil from the paper. Most of the functions in this book are of this type. Continuity on an Interval A function f is said to be continuous on an open interval (a, b) provided that f is continuous at every value in the interval. A function f is said to be continuous on a closed interval [a, b] provided that f is continuous from the right at x b, continuous from the left at value in the open interval (a, b). and continuous at every x a, Analogous definitions may be given for continuity on intervals of the form a, b q, q q, b q, b a, q a, q and a Example 3 Continuity of a Function Discuss the continuity of the function shown in Figure 14.4-6. y 2 −4 −2 0 2 4 6 x −2 Figure 14.4-6 Solution x 3 The function is discontinuous at 3, 2 5, 3 on each of the intervals, and, 2, q x 2,. 1 2 3 4 1 2 but it is continuous ■ Properties of Continuous Functions Using the definition is not always the most convenient way to show that a particular function is continuous. It is often easier to establish continuity by using the following facts. Section 14.4 Continuity 943 Properties of Continuous Functions If the functions f and g are continuous at the following functions is also continuous at |
x c: x c, then each of 1. the sum function f g 2. the difference function f g 3. the product function fg f g, 4. the quotient function g (c) 0 Proof By the definition of the sum function, Because f and g are continuous at Therefore, by the first property of limits, lim xSc x c 2 1 2 f lim xSc 1 f g x 2 21 x c, and lim xSc lim xSc c 22 g 2 x 2 1 f x lim xSc lim xSc This says that f g 1 is continuous at The remaining statements are proved similarly, using limit properties 2, 3, and 4. Example 4 Continuity of Functions x f 1 sin x Assume that and Prove that the following functions are continuous at x3 5x 2 x3 5x 2 sin x x3 5x 2 g 1 b. a. x 2 2 1 x3 5x 2 2 d. 2 sin x 1 sin x x3 5x 2 x 0. 2 are continuous at x 0. c. sin x 1 2 1 Solution Because f an g are continuous at are continuous at, each of the following functions by the listed property of continuous functions. x 0 1 2 x 0 x3 5x 2 x3 5x 2 1 x3 5x 2 f gb a 1 2 1 1 2 fg 1 x 2 sin x sin x sin x 1 1 2 sin x x3 5x 2 a. b. c. d sum of continuous functions difference of continuous functions product of continuous functions quotient of continuous functions ■ 944 Chapter 14 Limits and Continuity Composite Functions Composition of functions is often used to construct new functions from given ones. Continuity of Composite Functions If the function f is continuous at continuous at continuous at x c. x f(c), x c and the function g is then the composite function g f is Example 5 Continuity of Composite Functions 2x3 3x2 x 7 is continuous at x 2. Show that h x 2 1 Solution The polynomial function 23 3 and at 5 because by limit property 5 x f ˛1 2 2 7 5. 2 2 f 2 2 1 2 1 x3 3x2 x 7 g˛1 The function is continuous at x x 2 is continuous 2x 2 lim xS5 By the box above, with which is given by g g f 21 is also continuous at x x f 1 1 1 2 g 1 22 x 2. |
2x 2lim xS5 c 2 f and x 25 g c 5, 5. 2 1 1 2 the composite function x3 3x2 x 7 2 2x3 3x2 x 7 g f, ■ The Intermediate Value Theorem This section’s introduction to continuity will close by mentioning, without proof, a very important property of continuous functions. The Intermediate Value Theorem If the function f is continuous on the closed interval [a, b] and k is any number between f(a) and f(b), then there exists at least one number c between a and b such that f(c) k. The truth of the Intermediate Value Theorem can be understood geometrically by remembering that since f is continuous on the graph 4 b, f of f can be drawn from the point without lift1 ing the pencil from the paper. As suggested in Figure 14.4-7, there is no way that this can be done unless the graph crosses the horizontal line y k, to the point a, b 3 b 1 6 k 6 f where f a, f 22 22 Section 14.4 Continuity 945 (b, f(b)) (c, f(c)) = (c, k) y f(b) k f(a) (a, f(a)) x a c b Figure 14.4-7 The first coordinate of the point where the graph crosses this line is some number c between a and b, and its second coordinate is because the point is on the graph of f. But c, f is also on the line so its sec k. ond coordinate must be k; that is, 22 c 1 The Intermediate Value Theorem further explains why the graph of a continuous function is connected and unbroken. If the function f is continuous on the interval to b a the point f and. a then its graph cannot go from the point 22 without moving through all the y values between f f c 2 1 y k, c 1 f a, b 3 22 b, f a The graphical method of solving equations that has been used throughout this book is based on the Intermediate Value Theorem, as are some root-finding features on calculators. If f is continuous on the interval a, b 4 3 f a have opposite signs, then 0 is a number between and 2 1 k 0, and there is at least one number c between a and b such that In other is a solution of the equation words, and Consequently, |
by the Intermediate Value Theorem, with 2 2 x c 0. 0. f So when a calculator shows that the graph of a continuous function f has points above and below the x-axis, there really is an x-intercept between x these points, that is, a solution of Zoom-in uses this fact by look2 ing at smaller and smaller viewing windows that contain points of the graph on both sides of the x-axis. The closer together the points are horizontally, the better the approximation of the x-intercept, or solution, that can be read from the graph. 0. f 1 946 Chapter 14 Limits and Continuity Exercises 14.4 In Exercises 1 and 2, use the graph to find all the numbers at which the function is not continuous. 4. y 1. y 2. x x 1 3 y 1 1 In Exercises 3 – 6, determine whether the function x 0, whose graph is given is continuous at and at x 2, x 3. at 3. y −1 1 3 x −1 1 3 5. y −1 1 3 6. y −1 1 3 x x x In Exercises 7–12, use the definition of continuity and the properties of limits to show that the function is continuous at the given number. x2 2x2 5x 4, x 1 2 7. 8. 9. x x f 1 g˛1 f x 1 2 2 2 10. h x 2 1 11. f x 1 2 12. k x 1 2 1 x2 5 x 2 7, x 3 2 1 x2 3x 10 2 1 x2 9 1 x2 x 6 2 1 x 3 x2 x 1 1 x2x x 6 1 28 x2 2x2 5 2 2, x 36, x 2, x 2 x2 6x 9 2 x2 1 2 1, x 2 2 In Exercises 13–18, explain why the function is not continuous at the given number. 3, x 3 13. f x 1 2 14 x2 4 x2 x 2, x 2 15. f x 1 2 x2 4x 3 x2 x 2, x 1 16. x g˛1 2 sin • 1 p x if x 0 if x 0, x 0 17. f x 1 2 x2 if x 0 if x 0 1 e, x 0 18. f x 1 2 22 x 22 x, x 0 In Exercises 19– |
24, determine whether or not the function is continuous at the given number. 19. f x 1 2 2x 4 if x 2 2x 4 if x 7 2 e, x 2 20. x g˛1 2 2x 5 if x 6 1 2x 1 if x 1 e, x 1 21. f x 1 2 x2 x if x 0 2x2 if x 7 0 e, x 0 22. g x 2 1 x3 x 1 if x 6 2 3x2 2x 1 if x 2 e, x 2 Section 14.4 Continuity 947 26. x g˛1 2 µ if x 2 x2 x 6 x2 4 if x 2 4 5 27. f x 2 1 • x2 1 if x 6 0 x if 0 6 x 2 2x 3 if x 7 2 28. h x 2 1 • if x 6 1 and x 0 1 x x2 if x 1 29. Critical Thinking For what values of b is the following function continuous at x 3? f x 2 1 bx 4 if x 3 bx2 2 if x 7 3 e 30. Critical Thinking Show that continuous at x 0. f x 2 1 2 0 x 0 is x c A function f that is not defined at is said to have x c if there is a funca removable discontinuity at x c, tion g such that g(c) is defined, g is continuous at and In Exercises 31–34, show that the function f has a removable discontinuity by finding an appropriate function g. g ˛(x) f˛(x) x c. for 31. x f˛ 1 2 x 1 x2 1 32. f x 1 2 33. f x 1 2 x2 x 0 0 2 2x 4 x x 3, x 3 23. 24, x 2 0 34. f x 1 2 sin x x Hint: See Example 1 of Section 14.1. In Exercises 25–28, determine all numbers at which the function is continuous. 35. Show that the function 25. x f˛ 1 2 µ if x 1 discontinuity at x 0 x2 x 2 x2 4x 3 3 2 if x 1 x 0 x 0x f˛ 1 that is not removable. has a 2 948 Chapter 14 Limits and Continuity 14.5 Limits Involving Infinity Objectives In the discussion that follows, it is important to remember that • Define limits involving |
infinity • Use properties of limits at infinity • Use the Limit Theorem NOTE Excursion 14.2.A is a prerequisite for some of the material that follows. q, There is no real number called “infinity,” and the symbol usually read “infinity,” does not represent any real number. which is q Nevertheless, the word “infinity” and the symbol are often used as a convenient shorthand to describe the way some functions behave under certain circumstances. Generally speaking, “infinity” indicates a situation in which some numerical quantity gets larger and larger without bound, meaning that it can be made larger than any given number. Similarly, “negative infinity,” indicates a situation in which a numerical quantity gets smaller and smaller without bound, meaning that it can be made smaller than any given negative number. q, In Section 14.1, several ways were discussed in which a function might fail to have a limit as x approaches a number c. The word “infinity” is often used to describe one such situation. Consider the function f whose graph is shown in Figure 14.5-1. y 1 3 5 x Figure 14.5-1 The graph shows that as x approaches 3 from the left or right, the corresponding values of do not get closer and closer to a particular number. Instead, they become larger and larger without bound. Although there is no limit as defined in Section 14.1, it is convenient to describe this situation symbolically by writing x f 2 1 q, x f˛ 1 as x approaches 3 is infinity.” Similarly, f which is read “the limit of does not have a limit as x approaches 1 from the left or right, because the f˛ 1 corresponding values of get smaller and smaller without bound. We x as x approaches 1 is negative infinity” and write say that “the limit of lim xS3 x f˛ 1 x 2 2 2 f˛ 1 2 lim xS1 f˛ 1 x 2 q. Section 14.5 Limits Involving Infinity 949 x 5, Near small on the right side of 5, so we write the values of f˛ 1 x 2 get very large on the left side of 5 and very xS5 f lim x 1 2 q and xS5 f lim x 1 2 q, which are read “The limit as x approaches 5 from the left is infinity” and “the limit as |
x approaches 5 from the right is negative infinity.” There are many cases like the ones illustrated above in which the language of limits and the word “infinity” can be useful for describing the behavior of a function that actually does not have a limit in the sense of Section 14.1. Example 1 Infinite Limits Describe the behavior of f x 1 2 5 x4 near x 0. Solution The graph of f is shown in Figure 14.5-2. The trace feature indicates that get small without bound as x approaches 0 from the the values of left or from the right. x f 1 2 0.5 0.5 0 500,000 Figure 14.5-2 Therefore, lim xS0 5 x4 q. Example 2 Infinite Limits ■ Describe the behavior of the function g x 1 2 8 x2 2x 8 near x 2. Solution As shown in Figure 14.5-3, the graph of g is not continuous at the left of x ues of x 2 get small without bound. To get large without bound, and the val- x 2, to the right of g 1 x 2 the values of g x 2. 1 2 950 Chapter 14 Limits and Continuity 10 3 5 10 Figure 14.5-3 Therefore, xS2 lim 8 x2 2x 8 q and xS2 lim 8 x2 2x 8 q. ■ Infinite Limits and Vertical Asymptotes The “infinite limits” considered in Figure 14.5-1 and in Examples 1 and 2 can be interpreted geometrically: Each such limit corresponds to a vertical asymptote of the graph. Vertical Asymptotes is a vertical asymptote of the graph of the x c The vertical line function f if at least one of the following is true. xSc f(x) lim xSc f(x) lim xSc f(x) lim xSc f(x) lim lim xSc lim xSc f(x) f(x) 7 Limits at Infinity 80 80 Whenever the word “limit” has been used up to now, it referred to the behavior of a function when x was near a particular number c. Now, the behavior of a function when x takes very large or very small values will be considered. That is, the end behavior of a function will be discussed. 1 The graph of f Figure 14.5-4 x 1 2 5 1 24e x 4 1 is shown in Figure 14 |
.5-4. Graphing Exploration Produce the graph shown in Figure 14.5-4 and use the trace feature as x gets larger and larger. Are the values to find values of approaching a single value? x f 1 2 As you move to the right, the graph gets very close to the horizontal In other words, as x gets larger and larger, the corresponding line y 6. NOTE Due to rounding, the trace feature on most calculators will display y 1 when x is smaller 60. than approximately However, the value of the function is always greater than 1. Why? Limits at Infinity Section 14.5 Limits Involving Infinity 951 f values of bolically as 1 x 2 get closer and closer to 6, which can be expressed sym- 2 The last statement is read “the limit of 1 xSq f lim x 6. f x 1 2 as x approaches infinity is 6.” Toward the left the graph gets very close to the horizontal line is, as x gets smaller and smaller, the corresponding values of x and closer to 1. (See note.) It is said that “the limit of negative infinity is 1,” which is written x f 1 1 2 y 1; that get closer as x approaches f 2 xSq f lim x 1. 2 1 The types of limits when x gets large or small without bound are similar to those in Section 14.1 in that the values of the function do approach a fixed number. The definition in the general case is similar: f is any function, L is a real number, and the phrase “arbitrarily close” means “as close as you want.” Let f be a function that is defined for all number a. If x 77 a for some as x takes larger and larger positive values, increasing without bound, the corresponding values of f(x) get very close, and possibly are equal, to a single real number L and the values of f(x) can be made arbitrarily close (as close as you want) to L by taking large enough values of x, then the limit of f(x) as x approaches infinity is L, which is written xS f(x) L. lim x 6 a x 7 a ” with “ Limits as x approaches negative infinity are defined analogously by replacing “ ”, “increasing” with “decreasing”, and “larger and larger positive” with � |
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