Scatter Plots and Lines of Fit

Download for free
Find the solutions in the app
Android iOS
Exercises marked with requires Mathleaks premium to view it's solution in the app. Download Mathleaks app on Google Play or iTunes AppStore.
Sections
Communicate Your Answer
Exercise name Free?
Communicate Your Answer 3
Communicate Your Answer 4
Monitoring Progress
Exercise name Free?
Monitoring Progress 1
Monitoring Progress 2
Monitoring Progress 3
Monitoring Progress 4
Monitoring Progress 5
Exercises
Exercise name Free?
Exercises 1 Correlation on a scatter plot is highly related to the slope of a linear function. A slope is positive if it goes up and to the right and is negative if it goes down and to the right. Similarly, a correlation is positive if it goes up and to the right and it is negative if it goes down and to the right. As you move up on a graph, the y values are increasing. As you move to the right, the x values are increasing. Therefore, when data shows a positive correlation, both the independent and dependent variables tend to increase.
Exercises 2 A line of fit can help us predict future data points in a model. It is the graph of a linear equation drawn on the same coordinate plane as the data points of a scatter plot. It is drawn as close as possible to as many data points as possible. It can also be called a line of best fit, in this case it has been algebraically proven to be the most accurate line to describe the data.
Exercises 3 To find the missing coordinate in the ordered pair, we should begin by going to the given value on the x-axis and trace a line up to the point. Here we are given that x=16.Now that we reached the point, we can trace a line to the y-axis.The missing coordinate is 6.
Exercises 4 To find the missing coordinate in the ordered pair, we should begin by going to the given value on the x-axis and trace a line up to the point. Here we are given that x=3.Now that we reached the point, we can trace a line to the y-axis.The missing coordinate is 14.
Exercises 5 To find the missing coordinate in the ordered pair, we should begin by going to the given value on the y-axis and trace a line up to the point. Here we are given that y=12.Now that we reached the point, we can trace a line to the x-axis.The missing coordinate is 7.
Exercises 6 To find the missing coordinate in the ordered pair, we should begin by going to the given value on the y-axis and trace a line up to the point. Here we are given that y=17.Now that we reached the point, we can trace a line to the x-axis.The missing coordinate is 8.
Exercises 7 aThe given scatter plot shows the hard drive capacities (in gigabytes) and the prices (in dollars) of 10 laptops.In order to find the price of the laptop with a hard drive capacity of 8 gigabytes, we will draw a horizontal segment from the point that has an x-value of 8.As we can see, the segment crosses the y-axis at the point (0,1100). Thus, the price of the laptop is $1100.bThis time, we will draw a vertical segment from the point that has a y-value of 1200 so that we can find the hard drive capacity of $1200-laptop.Since the segment crosses the x-axis at the point (12,0), the hard drive capacity of the laptop is 12 gigabytes.cExamining the graph, we can see that the plotted points go up from left to right. Thus, as the capacity of the laptops increases, the price also increases.
Exercises 8 aThe given scatter plot shows the earned run averages and the winning percentages of eight pitchers on a baseball team.In order to find the winning percentage of the pitcher with an earned run average of 4.2, we will draw a horizontal segment from the point that has an x-value of 4.2.As we can see, the segment crosses the y-axis at the point (0,0.6). Thus, the winning percentage of the pitcher is 0.6.bThis time, we will draw a vertical segment from the point that has a y-value of 0.33 so that we can find the earn run average of the pitcher with a winning percentage of 0.33.Since the segment crosses the x-axis at the point (5,0), the earn run average of the pitcher is 5.cExamining the graph, we can see that the plotted points go up from left to right. Thus, as the earn run average increases, the winning percentage also increases.
Exercises 9 When analyzing data using a scatter plot, there are three general outcomes: positive correlation, negative correlation, or no correlation.ObservationType of Correlation As x increases, y increases.Positive As x increases, y decreases.Negative There is no visible pattern.No Below we have recreated the given scatter plot. Do you see any trends?It looks like there is some kind of correlation. Let's draw a line of fit, or trend line, to help us identify the type of correlation. To do so, we'll draw a line that appears to fit the data closely.As x-values increase, y-values increase as well. This indicates a positive correlation between x and y.
Exercises 10 When analyzing data using a scatter plot, there are three general outcomes: positive correlation, negative correlation, or no correlation.ObservationType of Correlation As x increases, y increases.Positive As x increases, y decreases.Negative There is no visible pattern.No Below we have recreated the given scatter plot. Do you see any trends?Looking at the graph, it does not appear that there is a recognizable trend. Therefore, we can conclude that there is no correlation.
Exercises 11 When analyzing data using a scatter plot, there are three general outcomes: positive correlation, negative correlation, or no correlation.ObservationType of Correlation As x increases, y increases.Positive As x increases, y decreases.Negative There is no visible pattern.No Below we have recreated the given scatter plot. Do you see any trends?Looking at the graph, it does not appear that there is a recognizable trend. Therefore, we can conclude that there is no correlation.
Exercises 12 When analyzing data using a scatter plot, there are three general outcomes: positive correlation, negative correlation, or no correlation.ObservationType of Correlation As x increases, y increases.Positive As x increases, y decreases.Negative There is no visible pattern.No Below we have recreated the given scatter plot. Do you see any trends?It looks like there is some kind of correlation. Let's draw a line of fit, or trend line, to help us identify the type of correlation. To do so, we'll draw a line that appears to fit the data closely.As x-values increase, y-values decrease. This indicates a negative correlation between x and y.
Exercises 13 When analyzing data using a scatter plot, there are three general outcomes: positive correlation, negative correlation, or no correlation.ObservationType of Correlation As x increases, y increases.Positive As x increases, y decreases.Negative There is no visible pattern.No By treating the table as a set of points, we can graph the given data as a scatter plot. Do you see any trends?Looking at the graph, it does not appear that there is a recognizable trend. Therefore, we can conclude that there is no correlation.
Exercises 14 When analyzing data using a scatter plot, there are three general outcomes: positive correlation, negative correlation, or no correlation.ObservationType of Correlation As x increases, y increases.Positive As x increases, y decreases.Negative There is no visible pattern.No By treating the table as a set of points, we can graph the given data as a scatter plot. Do you see any trends?It looks like there is some kind of correlation. Let's draw a line of fit, or trend line, to help us identify the type of correlation. To do so, we'll draw a line that appears to fit the data closely.As x-values increase, y-values decrease. This indicates a negative correlation between x and y.
Exercises 15 aLet's graph birthrate versus years since 1960 on the scatter plot. We will use the data given in the table as the points.Now draw a line that seem to fit the points closely. Note that we chose the line arbitrary.Since we have the line drawn, we can find its equation. First, we determine the slope by using the slope formula, m=x2​−x1​y2​−y1​​​ where (x1​,y1​) and (x2​,y2​) represent the points that lie on the line. We can see from the graph that the line of fit passes near points (0,35) and (50,20). m=x2​−x1​y2​−y1​​Substitute (0,35) & (50,20)m=50−020−35​Subtract termsm=50-15​ba​=b/5a/5​m=10-3​Write as a decimalm=-0.3 Now, that we have slope of the function m we will substitute it together with the point (0,35) into a point-slope form and solve for y. y−y1​=m(x−x1​)​ y−y1​=m(x−x1​)x1​=0,y1​=35,m=-0.3y−35=-0.3(x−0)LHS+35=RHS+35y=-0.3x+35bThe slope of -0.3 means that every 10 years the birthrate decreases by 3 per 1000 people. The y-intercept means that in the 1960 birthrate was about 35.
Exercises 16 aLet's graph earnings versus the number of hours server works on the scatter plot. We will use the data given in the table as the points.Now draw a line that seem to fit the points closely. Note that we chose the line arbitrary. Since we have the line drawn, we can find its equation. First, we determine the slope by using the slope formula, m=x2​−x1​y2​−y1​​​ where (x1​,y1​) and (x2​,y2​) represent the points that lie on the line. We can see from the graph that the line of fit passes near points (0,0) and (2,40). m=x2​−x1​y2​−y1​​Substitute (0,0) & (2,40)m=2−040−0​Subtract termsm=240​Calculate quotientm=20 Now, that we have slope of the function m we will substitute it together with the point (0,0) into a point-slope form and solve for y. y−y1​=m(x−x1​)​ y−y1​=m(x−x1​)x1​=0,y1​=0,m=20y−0=20(x−0)Subtract termsy=20xbThe slope of 20 means that the food server earn 20$ per hour. The y-intercept means that the server doesn't earn anything if he doesn't work.
Exercises 17 For a data set to have a negative correlation, the y values must decrease as the x values decrease. There are infinitely many real-life situations that fit this requirement. One possible example has to do with temperature. Let x represent the date of a day in December and y represent the temperature on that day. Since temperature tends to decrease as the season changes from fall to winter, we can assume that a data set that fits this scenario will show a negative correlation. Suppose the temperature each day in December is taken and recorded at noon. The graph below shows the recorded data points.
Exercises 18 To see if your friend is correct or not, let's graph the points given by the table.Looking at the graph, we can easily see that the correlation is positive, it goes up and to the right. What didn't our friend consider when saying that the correlation was negative? He only looked at the y values. Yes, they are in fact decreasing, but SO ARE THE X-VALUES! If we look at the table in reverse, it is obviously a positive correlation.x2468101214 y-5-4-2-1014
Exercises 19 aWe measured five friends and here is a table with our data. Note, we rounded to the nearest inch for each measurement.FriendHeight in inches, xArm span in inches, y Joe7375 MacKenzie6666 Henrik7172 Cate6768 Dimitri6362 We can plot these points on a scatter plot so that we can visualize the relationship.Now we can draw a line of fit through the approximate center of the data points.The equation of this line is y=x.bThe equation of the line of best fit we found in Part A is y=x. We can interpret the slope and y-intercept as follows:Slope: The slope is 1. This tell us that, as humans grow taller, their arms grow at approximately the same rate. y-intercept: The y-intercept is 0. This tells us that a human with a height of 0 inches would also have an arm span of 0 inches. This makes sense because they wouldn't exist.
Exercises 20 Imagine you have a science teacher who assigns packets of questions each week. The packets always contain between 20 and 50 questions about the material covered that week. You decide to record the number of hours, rounded to the nearest half hour, that it takes you to complete each week's assignment for the first quarter of the year.WeekHours SpentNumber of Questions 10.525 2650 35.545 4441 5230 6330 7237 8437 9445 10750 When you plot those points on a graph, you get the following scatter plot.You decide that you'd like to create a line of fit for the data. You could then use this line to figure out how much time you'll need to dedicate to science homework in the remaining weeks of the year.The line of fit that you've found is y=5x+20. Note, with this system you've designed, you will need to substitute the y value and solve for x each week. While this is not ideal, you made this decision because time is almost always on the x-axis.
Exercises 21 A scatter plot can help us find an overall trend when things seem to be random. There are several inputs and outputs, the data is "scattered." The domain for scatter plot may be restricted but it is typically continuous. Here is an example of a scatter plot showing the number of car accidents in a small city each day for the first ten days of March related to the temperature in ∘C.Temperature is continuous, it can be measured as precisely as you wish. The data shows a negative correlation: as the temperature rises, the number of car accidents decreases. Presumably this is because there is less ice. A bar graph, on the other hand, is best for categorical domains, when you want to know how many there are in a certain category.The above bar graph shows the favorite colors of students in a preschool class. This would not be as easily understood on a scatter plot. A circle graph is best for describing a break down of percentages of a whole. A histogram is best for describing viewing frequencies of events happening in a continuous domain. The moral of the story is that each graph serves it's own purpose and has a situation where it would be the best choice. The best choice is a scatter plot when you have a random set of points on a continuous domain and want to look for a correlation between them.
Exercises 22 We are shown the following graph and asked to add the missing four points.We can add the four points anywhere we want so long as they are centered about and balanced on either side of the line. One example solution is shown below.
Exercises 23 A line of fit is a tool used to predict the values of future data points, either through extrapolation or interpolation. When a data set has no correlation, it means that there is no pattern within the data. It is completely random points with no way to predict future points. In this situation, a line of fit would be a useless tool. Technically, you could write any of several lines through the data, but they would have no purpose and wouldn't actually be "lines of fit." Let's look at the graph below, it has no correlation.If we wanted to, we could try to draw lines of fit to represent the data.However, none of these lines actually explain the data better than the others. You cannot find a line of fit for data with no correlation.
Exercises 24 Let's begin by making a scatter plot using the data provided in the tables.The data appears to have a U-shaped pattern, it gradually declines until x=0 and then gradually inclines. In order to write an equation for a line of fit, our data needs to have a linear positive or linear negative correlation. This data does not have a linear relationship and, therefore, cannot have a line of fit.
Exercises 25 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-3. g(x)=6xx=-3g(-3)=6(-3)a(-b)=-a⋅bg(-3)=-18 When x=-3, the function's value is -18. We will evaluate this function for x=0 and x=4 in the same way using the table below.x6xg(x) -36(-3)-18 06(0)0 46(4)24
Exercises 26 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-3. h(x)=-10xx=-3h(-3)=-10(-3)-a(-b)=a⋅bh(-3)=30 When x=-3, the function's value is 30. We will evaluate this function for x=0 and x=4 in the same way using the table below.x-10xh(x) -3-10(-3)30 0-10(0)0 4-10(4)-40
Exercises 27 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-3. f(x)=5x−8x=-3f(-3)=5(-3)−8a(-b)=-a⋅bf(-3)=-15−8Subtract termf(-3)=-23 When x=-3, the function's value is -23. We will evaluate this function for x=0 and x=4 in the same way using the table below.x5x−8f(x) -35(-3)−8-23 05(0)−8-8 45(4)−812
Exercises 28 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-3. v(x)=14−3xx=-3v(-3)=14−3(-3)-a(-b)=a⋅bv(-3)=14+9Add termsv(-3)=23 When x=-3, the function's value is 23. We will evaluate this function for x=0 and x=4 in the same way using the table below.x14−3xv(x) -314−3(-3)23 014−3(0)14 414−3(4)2