Probability Distributions

CONDENSED LESSON 13.1 Probability Distributions In this lesson, you Sketch the graph of the probability distribution for a continuous random variable Find probabilities by finding or approximating areas under a probability distribution curve Extend the definitions of mode, median, and mean to probability distributions At election time, television stations, newspapers, and magazines often conduct polls. By surveying a small sample of voters, they hope to get information about how the entire population of voters feels about a candidate or an issue. In earlier chapters you learned some statistics for example, the mean, median, and standard deviation that can be used to describe a sample. The corresponding numbers describing the entire population are called parameters. The larger the sample, the closer its statistics will be to the parameters. In the previous chapter, you worked with discrete random variables. The data had integer values, for example, 10 heads or 3 tails. Sometimes data can take on any real value within an interval. This is represented by a continuous random variable. For example, the height of a randomly selected person is a continuous random variable. A person might be 165 cm tall or 166 cm tall, but any measurement between these integer measurements is also possible (for example, 165.25 cm or 165.67897 cm). Investigation: Pencil Lengths The investigation in your book requires you to collect pencil-length data from all students in your class. The results below use these data. (Lengths are in centimeters.) {16.9, 18.7, 11.3, 13.8, 15.2, 17.0, 16.5, 16.6, 11.8, 17.2, 15.5, 15.7, 17.0, 11.4, 16.5, 16.0, 13.4, 15.7, 15.5, 14.1, 12.3, 13.8, 15.5, 15.7, 10.7, 15.6, 12.1, 14.4, 16.5, 17.9, 8.2, 17.8, 17.6, 14.1, 16.7, 14.6, 12.3, 10.0, 13.2, 14.3} Create a histogram of these data with bins representing 1 cm increments. Your histogram should look like the one at right. Divide the number of pencils in each bin by the total number of pencils. You should get these results: 8 9:.025 9 10: 0 10 11:.05 11 12:.075 12 13:.075 13 14:.1 14 15:.125 15 16:.2 16 17:.175 17 18:.15 18 19:.025 Number of pencils 10 8 6 4 2 0 8 10 12 14 16 18 20 Pencil length (cm) Make a second histogram using these new values as y-values. Your histogram should look like the one on the next page. This graph has the same shape as the one above, but the vertical scale is different. Discovering Advanced Algebra Condensed Lessons CHAPTER 13 199

Lesson 13.1 Probability Distributions The height of each bar represents the fraction of pencils with lengths in the given interval. Because the width of each bar is 1, the area of each bar also represents the fraction of pencils in the interval. Because all the pencils have been accounted for, the total area of all the bars must be 1. You can check this by adding the areas. Note that the area of each grid square is.025..25.20.15.10.05 0 Imagine that you collect and measure more and more 8 10 12 14 16 18 20 pencils and draw a histogram using the fraction of pencils Pencil length (cm) as the bin height. Sketch what a histogram of infinitely many pencil lengths would look like. Make sure you can give reasons for the shape of your histogram. Imagine you do a complete and precise survey of all the.25 pencils in the world. Assume the distribution of lengths is.20 about the same as that in the sample above. Also, assume that you use infinitely many very narrow bins. (Each bin.15 width represents an infinitely small fraction of a centimeter.) To approximate this plot, sketch over the top of your histogram with a smooth curve. Make the area between the curve and the horizontal axis about the same as the area of.10.05 0 8 10 12 14 16 18 20 the histogram. Try to make sure the extra area enclosed by Pencil length (cm) the curve above the histogram is the same as the area cut off the corners of the bins, like the curve at right. Use your curve to estimate the areas described in Step 7 in your book. These estimates are based on the curve at right. a. About.025 b. About 3(.025), or.075 c. About 32.5(.025), or.8125 d. 0 Fraction of pencils Fraction of pencils The second histogram you made in the investigation, showing the fraction of pencils in each bin, is called a relative frequency histogram. The smooth curve you drew approximates the probability distribution for a continuous random variable for the infinite set of measurements. The areas you found are the probabilities that a randomly chosen pencil length will satisfy the given condition. If x represents the continuous random variable giving the pencil lengths in centimeters, then you can write these areas as P(x 10) P(11 x 12) P(x 12.5) P(x 11) In a continuous probability distribution, the probability of any single outcome, such as the probability that a pencil length is 11 cm, is the area of a line segment, which is 0. It is theoretically possible for a pencil to be 11 cm, but the probability of choosing one outcome out of infinitely many outcomes is 0. Read Example A in your book, which illustrates how areas represent probabilities for a continuous random variable. After Example A, your book defines the mode, median, and mean of a probability distribution. These definitions are related to, but slightly different from, the definitions you learned earlier. Read them carefully, and make sure they make sense to you. Then, apply the new definitions by working through Example B. 200 CHAPTER 13 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 13.2 Normal Distributions In this lesson, you Discover that the graph of a binomial distribution is a bell-shaped curve, called a normal curve Learn the equation for a normal distribution with mean and standard deviation Use calculator functions to graph a normal curve and find areas under a portion of the curve In Chapter 12, you studied the binomial distribution for discrete random variables. In general, if an experiment has two outcomes, success and failure, with probability of success p and probability of failure q, then the probability of x successes in n trials is P(x) n C x p x q n x.note that q 1 p, so this is equivalent to P(x) n C x p x (1 p) n x.in this lesson, you will discover some properties of this probability distribution. Investigation: The Bell Complete the investigation in your book. Then, compare your results with those below. Step 1 p.5 and 1 p.5; that is, the probability of getting a head on any one trial is.5, and the probability of not getting a head (i.e., getting a tail) is.5. Theoretically, half the tosses will be heads, so 7 and 8 will give the maximum value for P(x). Step 2 Here is the completed table: Heads (x) 0 1 2 3 4 5 6 7 P(x).000.000.003.014.042.092.153.196 Heads (x) 8 9 10 11 12 13 14 15 P(x).196.153.092.042.014.003.0000.000 The sum is the total of the probabilities of all the possible outcomes of any experiment, 1. Step 3 The histogram, shown at right, is mound-shaped and symmetric. The y-values range from 0 to about.196. The maximum value occurs at x 7 and x 8. [0, 15, 1, 0, 0.25, 0.1] Step 4 The function is defined for integer values from 0 through 15. The graph consists of discrete points that form a mound-shaped, symmetric graph, with the same shape as the histogram in Step 3. The mean, median, and mode are all 7.5. [0, 18.8, 1, 0.01, 0.25, 0.1] Discovering Advanced Algebra Condensed Lessons CHAPTER 13 201

Lesson 13.2 Normal Distributions Step 5 The function describes the probability distributions when 45 coins are flipped at once. The shape is the same as Step 4 but wider and shorter. The domain is integer values from 0 to 45. The range is from 0 to about.117. The mean, median, and mode are all 22.5. Step 6 x 22.5, s 3.354 Step 7 If the number of coins increased, the domain would increase [0, 47, 1, 0.01, 0.25, 0.1] and the range would decrease. The overall shape would be the same, but the graph would be wider, smoother, and flatter. The mean, median, and mode would all move up to n,where 2 n is the number of coins. The standard deviation would be larger because of the larger spread. As n gets larger and larger, the binomial distribution looks more and more continuous until, eventually, it looks like the bell-shaped curve at right. Distributions for large populations often have this shape. The bell-shaped curve is called a normal curve, and a bell-shaped distribution is called a normal distribution. Note that you use x and s to represent the mean and standard deviation of a sample, but you use and (pronounced mew and sigma ) to represent the mean and standard deviation of an entire population. Beginning on page 735, your book discusses the equation for the graph of the normal distribution. Read that text, and work through Example A. In the example, you graph the general equation for the normal curve, write the equation for a standard normal distribution with mean and standard deviation, and then write an equation for a normal curve that fits a given binomial distribution. The general equation for a normal distribution is also given in the The Normal Distribution box on page 737. The equation for the normal distribution can be tedious to enter into a calculator. Fortunately, most calculators provide the equation as a built-in function. You have to provide only the mean and standard deviation. (See Calculator Note 13B to learn how to graph a normal distribution on your calculator.) In this chapter, the notation n(x, mean, standard deviation) is used to represent anormal distribution. So, for example, n(x, 2.6, 1.5) represents a normal distribution with mean 2.6 and standard deviation 1.5. The standard normal distribution function, that is, the function for the normal distribution with mean 0 and standard deviation 1, is simply denoted n(x). The notation N(lower, upper, mean, standard deviation) is used to represent the area under a portion of the normal curve. For example, N(2, 3, 2.6, 1.5) represents the area between x 2 and x 3 under the curve for the normal distribution with mean 2.6 and standard deviation 1.5. (See Calculator Note 13C to learn Curving downward how to find these areas on your calculator.) Read Example B in your book, following along with a pencil and paper. At points that are one standard deviation from the mean, the normal curve changes from curving downward to curving Curving upward. These points are called inflection points. You can upward estimate the standard deviation of a normal distribution by locating the inflection points of its graph. Curving upward 202 CHAPTER 13 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 13.3 z-values and Confidence Intervals In this lesson, you Discover the 68-95-99.7 rule for determining the probability that a data value is within one, two, or three standard deviations from the mean Learn how to transform x-values of a normal distribution to z-values Calculate confidence intervals Knowing how a value in a sample relates to the mean value does not tell you how typical the value is. For example, to say that the weight of a Scottish terrier is 4 lb above the mean does not tell you if this measurement is a rare event or a common event. If you knew how many standard deviations the dog s weight was from the mean, you would have a much better idea of how unusual the weight is. Investigation: Areas and Distributions Complete the investigation in your book on your own, assuming the measurement you made was 72.4 cm. Also, assume that everyone in your class got similar answers in Step 2. When you are finished, compare your results with those below. Step 1 Your sketch should be a bell-shaped curve, symmetric about x 72.4, with inflection points at about 71.6 and 73.2 (the x-values that are one standard deviation from the mean). Step 2 a. N(71.6, 73.2, 72.4, 0.8).683, or about 68% b. N(72.4 2(0.8), 72.4 2(.8), 72.4, 0.8) N(70.8, 74, 72.4, 0.8).954, or about 95% c. N(72.4 3(0.8), 72.4 3(.8), 72.4, 0.8) N(70, 74.8, 72.4, 0.8).997, or about 99.7% Step 3 For a normal distribution, there is a 68% probability that a value will be within one standard deviation of the mean, a 95% probability that a value will be within two standard deviations of the mean, and a 99.7% probability that a value will be within three standard deviations of the mean. In a normal distribution, the z-value of x is the number of standard deviations that x is from the mean. In the investigation you found that the probability that a new measurement will have a z-value between 1 and 1 is 68%, the probability that it will have a z-value between 2 and 2 is 95%, and the probability that it will have a z-value between 3 and 3 is 99.7%. You can think of the z-value of x as the image of x under a transformation that translates and either stretches or shrinks the normal distribution to the standard normal distribution n(x) with mean 0 and standard deviation 1. Transforming x-values to z-values is called standardizing the variable and can be calculated with the equation z x, where and are the mean and standard deviation of the normal distribution of x. Work through Example A in your book, which illustrates standardizing the variable. Note that when using trial and error in part c you must test only intervals that are symmetric about the mean. Discovering Advanced Algebra Condensed Lessons CHAPTER 13 203

Lesson 13.3 z-values and Confidence Intervals There is no way to know for certain how close the mean of the normally distributed population is to the mean of a sample. However, you can describe how confident you are that the population mean lies in a given interval centered at the sample mean. A p% confidence interval is an interval about the sample mean, x, in which you can be p% confident the population mean,, lies. Specifically, if z is the number of standard deviations from the mean within which p% of normally distributed data lie, then the p% confidence interval for a sample of size n is x z n x z n In many real-world situations, you will not know the population standard deviation. However, if the sample size is large enough, generally n 30, you may use the sample standard deviation, s, in place of. The 68-95-99.7 rule tells you what z-values to use if you want to be 68%, 95%, or 99.7% confident. The table on page 748 of your book gives z-values for some other commonly used confidence intervals. In Example B in your book, confidence intervals are calculated using values from the table. Read that example, and then try to solve the problem in the example below. EXAMPLE Solution The quality-control manager at a cereal company pulled a random sample of 30 boxes of Morning Crunch cereal from the production lines and weighed the contents of each box. The mean weight for the sample was 9.8 oz and the standard deviation was 0.42 oz. a. Find the 68% and 90% confidence intervals. b. The label on the Morning Crunch box states that the weight is 10 oz. Do you think the quality-control manager should report a problem? Explain. a. Using the sample standard deviation in place of, the 68% confidence interval is 9.8 1(0.42), 9.8 1(0.42) 30 30,or about (9.72, 9.88) The quality-control manager is 68% confident that the mean weight is between 9.72 and 9.88 oz. The table in your book indicates that the z-value corresponding to a 90% confidence interval is 1.645, so the 90% confidence interval is 9.8 1.645(0.42), 9.8 1.645(0.42) 30 30,or about (9.67, 9.93) The quality-control manager is 90% confident that the weight is between 9.67 and 9.93 oz. b. According to the answer to part a, the quality-control manager can be 90% certain that the population mean (that is, the mean weight for all the boxes produced) is in the interval (9.67, 9.93). Because this interval does not include 10 oz, the weight claimed on the box, the manager should report a problem. 204 CHAPTER 13 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 13.4 The Central Limit Theorem In this lesson, you Discover how sample means are related to the population mean in cases where the population is not normally distributed Apply the Central Limit Theorem to test a claim about a population mean Use the process of inference to solve a real-world problem In Lesson 13.3, you used sample statistics to estimate parameters of a normally distributed population. In this lesson, you will explore what a sample can tell you in cases where the population is not normally distributed. Investigation: Means of Samples Step 1 tells you to create a population of a particular type uniform, normal, skewed left, or skewed right. Below are results for a population that is skewed left. You should go through the steps yourself for at least one of the other possibilities. (See Calculator Note 13D to create a list of 200 values, each between 20 and 50, for the type of population you choose.) The commands on the left below generate a skewed-left population of 200 values between 20 and 50. The histogram confirms the data are indeed skewed left. (Because the values are random, you will get a different population and different results for the remaining steps if you generate a skewed-left population.) [20, 50, 2, 0, 100, 10] Step 2 The population mean,, is 43.07, and the standard deviation,, is 5.75. Step 3 The command L1(randInt(1200)) will randomly select a data value from list L1. Selecting three random values gives the sample {49.07, 46.93, 47.43}. The mean of this sample is 47.81. Adding two more randomly selected values gives {49.07, 46.93, 47.43, 46.08, 41.72}, which has mean 46.25. Adding two more values gives {49.07, 46.93, 47.43, 46.08, 41.72, 42.62, 37.82}, which has mean 44.52. The sample means are fairly close to the population mean, and they get closer to the population mean as you add more values. 2 2 Step 4 Below are graphs of y, y, and y. x x [0, 50, 5, 31.57, 54.57, 5] Discovering Advanced Algebra Condensed Lessons CHAPTER 13 205

Lesson 13.4 The Central Limit Theorem Step 5 The recursive routine on the left below adds one randomly selected value at a time to a sample and then plots the point (number sampled, mean). (In the routine, N is the number of values sampled and T is the total (sum) of the sampled values.) Executing the routine 50 times plots 50 points, as shown in the graph on the right. [0, 50, 5, 31.57, 54.57, 5] Notice that as the sample size increases, the points (which represent the sample mean) get closer and closer to the line y (which represents the population 2 2 mean), with lower and upper bounds y and y x x.in other 2 2 words, for sample size n, the sample mean is within and n n of the population mean. Step 6 If you repeat Step 5, you will get similar results. Step 7 No matter what type of population distribution you start with (uniform, normal, skewed right, and so on), you will find that as the sample size increases, the sample mean gets closer and closer to the population mean. (Make sure you check this for at least one other type of population!) You just discovered that the mean of a sample approximates the mean of the population, and that the approximation is better for larger samples. In fact, the sample means themselves are normally distributed even if the population is not. Moreover, you can use the sample means to predict the standard deviation of the population. These observations are summarized by the Central Limit Theorem. This theorem is stated on page 753 of your book. Read the statement and the paragraph that follows it. Work through Example A, which applies the Central Limit Theorem to evaluate a claim about a population mean. Note that testing the company s claim involves answering the question If the population mean is actually 324 mg, how likely would it be to select a sample with a mean of 319.96? The solution concludes that you could expect a sample with this mean or less only 5.7% of the time. The process used in Example A is called inference. Inference involves creating a hypothesis about population parameters (e.g., The mean is 324 mg ), deciding what would make the hypothesis improbable (e.g., A random sample of 25 tablets from the population has a mean that would occur less than 10% of the time ), collecting data, and either rejecting the hypothesis or letting it stand, based on probabilities. Example B takes you through the inference process step by step. Read the example carefully. Make sure you understand the meaning of null hypothesis. Then read the remainder of the lesson, which explains the meaning of simple random sample. 206 CHAPTER 13 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 13.5 Bivariate Data and Correlation In this lesson, you Determine whether two variables are correlated Use the correlation coefficient to determine how strong a correlation is Many real-world statistical problems involve predicting associations between two variables. For example, researchers may want to determine if there is an association between the number of milligrams of vitamin C a person consumes and the number of colds the person gets. The process of collecting data on two possibly related variables is called bivariate sampling. In this lesson, you will focus on determining whether there is a linear association between variables and how strong the linear association might be. A linear association between variables is called correlation. The most commonly used statistical measure of linear association is the correlation coefficient. Investigation: Looking for Connections Step 1 Look at the sample survey in Step 1 of the investigation in your book. Step 2 List at least two pairs of variables you think will have a positive correlation (as one increases, the other tends to increase). One possibility might be the number of minutes of homework (question 1) and the number of academic classes (question 5). Now, list at least two pairs of variables you think will have a negative correlation (as one increases, the other tends to decrease). One possibility is the number of academic classes (question 5) and the time spent talking, calling, e-mailing, or writing to friends (question 2). Finally, list at least two pairs of variables you think will have a weak correlation. One possibility might be the time spent communicating with friends (question 2) and the time spent watching TV or listening to music (question 3). Step 3 The table on page 208 shows the results gathered in one class. Enter the data into five calculator lists. Plot points for each pair of lists, and find the correlation coefficients. (See Calculator Note 13F.) Here are the correlation coefficients and graphs for the relationships mentioned in Step 2. However, you should make graphs and find correlation coefficients for all the possible pairings. 1 vs. 5 2 vs. 5 2 vs. 4 [ 10, 230, 30, 0, 7, 1] [ 10, 200, 30, 0, 7, 1] [ 10, 200, 30, 10, 360, 30] r.771 r.632 r.059 Discovering Advanced Algebra Condensed Lessons CHAPTER 13 207

Lesson 13.5 Bivariate Data and Correlation You should make the following observations: The graphs that increase have positive correlation coefficients, and the graphs that decrease have negative correlation coefficients. The stronger the correlation is, the closer the correlation coefficient is to 1. Weak correlations have correlation coefficients close to 0. Student # Question 1 Question 2 Question 3 Question 4 Question 5 1 0 120 60 100 3 2 20 120 90 200 4 3 80 65 80 140 6 4 55 20 220 260 5 5 10 20 155 200 4 6 15 0 145 200 4 7 90 10 80 150 6 8 215 10 0 60 6 9 100 0 140 150 6 10 60 30 120 105 5 11 65 0 120 150 6 12 10 60 300 360 4 13 120 0 0 45 6 14 30 45 285 90 4 15 40 60 150 190 4 16 0 85 150 75 3 17 0 180 30 30 4 18 80 0 0 0 6 19 90 20 0 0 6 20 45 10 180 285 5 21 10 120 0 90 4 22 40 30 100 115 5 23 0 0 360 60 5 24 30 50 45 90 5 25 60 20 30 90 6 26 45 20 30 45 5 27 20 105 20 60 4 28 0 90 0 120 4 29 50 40 0 90 5 30 40 10 0 45 4 208 CHAPTER 13 Discovering Advanced Algebra Condensed Lessons

Lesson 13.5 Bivariate Data and Correlation Step 4 Write a paragraph describing the correlations you discovered. Mention any pairs that are not correlated that you thought would be. Here are some things you might mention: The correlations between time spent on homework and number of academic classes and between time spent watching TV and bedtime are relatively strong and positive. The correlations between time spent talking to friends and number of academic classes and between time spent on homework and time spent talking to friends are relatively strong and negative. The other pairs of variables do not appear to be correlated. These data were collected from a small sample that was not very random, so they may not be good predictors of results for the entire school. The text between the investigation and Example A in your book gives information about the correlation coefficient and how its formula was derived. This text also points out that in statistics, the x- and y-variables are often called the explanatory and response variables. Read this text, and then work through Example A, which demonstrates how to calculate the correlation coefficient using the formula. It is very important not to confuse correlation with causation. The fact that two variables are strongly correlated does not mean a change in one variable causes a change in the other. Example B illustrates this point. Read that example carefully. Discovering Advanced Algebra Condensed Lessons CHAPTER 13 209

CONDENSED LESSON 13.6 The Least Squares Line In this lesson, you Learn how to fit the least squares line to a set of data Discover how the least squares line gets its name Use the root mean square error to compare a least squares line to a median-median line In Chapter 3, you learned how to fit a median-median line to data. In this lesson, you ll learn about a different line of fit, called the least squares line. The equation of the least squares line is z y rz x, where r is the correlation coefficient and z x and z y are the z-values for x and y, respectively. In practice, you want the equation to represent the relationship between x and y, not between their z-values. Using the definition of z-value, you can rewrite the equation as y s y y r x s x x,or ŷ y r s s y (x x ) x To find out more about the least squares line, read the text before Example A in your book. Then, read Example A, which illustrates how to find the least squares line for a given set of data. Investigation: Relating Variables Step 1 of the investigation in your book asks you to collect measurements from students in your class. Here are sample data for 10 students (measured in centimeters). Use these data to complete Steps 2 4, and then compare your results with those below, which look at the relationship between hand span and the length of the little finger. Student Hand Foot Length of Length Length of Length of Width of number span length little finger Height of cubit lower leg upper arm thumbnail 1 18.5 23.5 6 159 41.6 47 31 1.2 2 21.2 26 6.5 170 42 48 36 1.3 3 20.8 25.8 6.2 173 44.2 49.5 36.2 1.3 4 20.4 24.5 6.1 164 42.1 48.5 34 1.3 5 18.1 23 5.9 155 39.5 46 30.2 1.2 6 22 28.4 6.5 189 47.5 52.2 40.5 1.4 7 20.5 29 6.4 191 47.9 52.6 41 1.4 8 20.7 26.2 6.2 175 46 51 34 1.3 9 20.5 25 6.1 168 44 50 35.4 1.2 10 19.2 24.1 5.9 158 43.5 48 33.6 1.1 Step 2 The correlation coefficient for these variables is about.858, so the data are linearly related. Discovering Advanced Algebra Condensed Lessons CHAPTER 13 211

Lesson 13.6 The Least Squares Line Step 3 Using a calculator, x 20.19, y 6.18, s x 1.219, s y 0.225, and r.858. Substituting these values into the general equation gives ŷ 6.18.858 0. 225 1. 2 (x 19 20.19), or ŷ 2.981 0.158x. The line appears to be a good fit. Step 4 a. It is the same line. b. The sum of the residuals is 0. c. The least squares line is the line for which the sum of the squares of the residuals is as small as possible, and the sum of the residuals is 0. d. As hand span increases, so does the length of the little finger. The data are modeled fairly well by a straight line. [17.71, 22.39, 1, 5.798, 6.602, 1] [17.71, 22.39, 1, 5.798, 6.602, 1] You can measure the accuracy of a least squares line by calculating the root mean square error, just as you did for median-median lines in Chapter 3. Example B in your book fits both a least squares line and a median-median line to a set of data and determines which is the better fit by computing the root mean square error. Work through that example, and then try the example below. EXAMPLE Solution Find the least squares line and the median-median line for the (length of cubit, length of upper arm) data in the investigation. Then, find the root mean square error of both models. Which line is a better fit? Use your calculator to find both models and their root mean square error. Least squares line: ŷ 13.982 1.122x, root mean square error: 1.923 Median-median line: ŷ 35.904 1.610x, root mean square error: 2.446 Because the root mean square error is smaller for the least squares line, it is a better fit. You can check this visually by graphing the lines. The left graph shows the least squares model. The right graph shows the median-median model. [39, 48, 1, 30, 41, 1] [39, 48, 1, 30, 41, 1] The least squares line is often called the best-fit line because it has the smallest sum of squares of errors between data points and predictions from the line. However, because it places equal emphasis on each point, the least squares line can be affected by outliers. In contrast, the median-median model is relatively unaffected by one or two outliers. When you fit a line to data, it is always a good idea to check the line visually. Sometimes the median-median line or another line is a better fit than the least squares line. 212 CHAPTER 13 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 13.7 Nonlinear Regression In this lesson, you Linearize data to determine if a power or exponential function might fit Use your calculator to fit a variety of nonlinear functions to data Use the root mean square errors and graphs of residuals to compare how well different functions fit a set of data You have done linear regression to fit a line to data that show a strong linear trend. In this lesson, you will find models to fit data that show a clear trend that is not linear. Pages 780 and 781 of your book review the types of functions you have studied in this course, giving the general equation and graph of each type. Read the text and look over the graphs. Example A in your book shows that by looking at the graph of data you can predict the type of function that might fit (or at least rule out the types that won t fit). If you think a power or exponential model might fit your data, you can check whether or not they are good models by linearizing the data. If the fit is good, then you can use linear regression techniques to find a model. This is illustrated in Example B. Read both Example A and Example B carefully. Then, work through the example below. EXAMPLE Find a function that models these data: x 3 5 7 9 11 13 15 17 y 19.1 10.0 7.5 4.7 4.0 1.8 1.1 0.7 Solution From the graph at right, it looks like an exponential function, power, or polynomial function might fit. If the data are exponential, an equation in the form y ab x will fit. Taking the log of both sides gives log y log a x log b. This is a linear equation, where the variables are x and log y. Therefore, if the data are exponential, the graph of (x, log y) will be linear. [0, 20, 2, 0, 20, 2] The graph of (x, log y) at right appears to be fairly linear. (You can verify that a power function is not a good fit by plotting (log x, log y) and checking that the graph is not linear.) [0, 20, 2, 0.2, 1.4, 0.2] Discovering Advanced Algebra Condensed Lessons CHAPTER 13 213

Lesson 13.7 Nonlinear Regression Use linear regression to find that the least squares line for this transformed data is ŷ 1.57 0.10x. Now replace ŷ with (log y) and solve for y. log y 1.57 0.10x y 10 1.57 0.10x 10 1.57 10 0.10x 37.153 10 0.10 x Least squares line. Definition of logarithm. Multiplication property of exponents. Evaluate 10 1.57 and apply the power property of exponents. y 37.153(0.794) x Evaluate 10 0.10. The graph shows that ŷ 37.153(0.794) x fits the data well. [0, 20, 2, 0, 20, 2] The root mean square error for this equation is about 0.854, which is fairly small. Your calculator has commands to fit many types of functions, including cubic, quadratic, exponential, power, sinusoid, and logistic functions. Use calculator commands to check the function from the example above. Then, read Example C in your book, which fits a cubic function to data. Investigation: A Leaky Bottle Experiment Read the investigation in your book, and make sure you understand how the experiment works. Complete the steps using this sample data, and then compare your results with those below. Height of Time (s) water (mm) 0 150 10 134 20 120 30 109 40 97 Height of Time (s) water (mm) 50 86 60 77 70 68 80 60 Height of Time (s) water (mm) 90 54 100 48 110 43 120 39 Step 2 The data appear to be exponential, quadratic, or cubic. [0, 130, 10, 0, 150, 10] 214 CHAPTER 13 Discovering Advanced Algebra Condensed Lessons

Lesson 13.7 Nonlinear Regression Step 3 y 151.234(0.989) x ; y 0.005x 2 1.500x 149.110; y 0.000005x 3 0.006x 2 1.544x 149.456 Step 4 Below are graphs of the residuals for each model in Step 3. Exponential; root mean square error about 1.595 [0, 130, 10, 3, 3, 1] Quadratic; root mean square error about 1.095 [0, 130, 10, 3, 3, 1] Cubic; root mean square error about 1.422 [0, 130, 10, 3, 3, 1] Step 5 The quadratic curve is best. It has the least root mean square error and the least noticeable pattern in the residuals. [0, 130, 10, 0, 150, 10] Step 6 None of the equations gives a reasonable answer for when the bottle will be empty. In particular, the quadratic curve predicts that the bottle will never be empty. So none of the models are good for predicting data values outside of the domain given. Discovering Advanced Algebra Condensed Lessons CHAPTER 13 215