1. 2.5 – Modeling Real World Data:

2. Using Scatter Plots

3. Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.

4. Year199019921994199619982000 Price ($1000) 122.9121.5130140152.5169

5. Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years. a.Make a scatter plot of the data. Year199019921994199619982000 Price ($1000) 122.9121.5130140152.5169

7. Years Since 1990

8. Price Years Since 1990

9. Price ($1000) Years Since 1990

10. Median House Prices Price ($1000) Years Since 1990

11. Median House Prices Price ($1000) 0 Years Since 1990

12. Median House Prices Price ($1000) 0 2 4 6 8 10 Years Since 1990

13. Median House Prices Price ($1000) 0 2 4 6 8 10 Years Since 1990

14. Median House Prices Price ($1000) 120 0 2 4 6 8 10 Years Since 1990

15. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

16. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

17. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

18. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

19. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

20. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

21. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990

22. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990 b. Make a line of fit.

23. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990 b. Make a line of fit.

24. Median House Prices Price ($1000) 140 120 0 2 4 6 8 10 Years Since 1990 b. Make a line of fit.

25. c.Find a prediction equation for line of fit.

26. *Use the best two ordered pairs from b. to find the slope for the line!

27. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5)

28. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 x 2 - x 1

29. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 = 152.2 – 130 x 2 - x 1 8 – 4

30. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 = 152.2 – 130 = 22.5 x 2 - x 1 8 – 4 4

31. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 = 152.2 – 130 = 22.5 ≈ 5.63 x 2 - x 1 8 – 4 4

32. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 = 152.2 – 130 = 22.5 ≈ 5.63 x 2 - x 1 8 – 4 4 *So use x 1 = 4

33. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 = 152.2 – 130 = 22.5 ≈ 5.63 x 2 - x 1 8 – 4 4 *So use x 1 = 4, y 1 = 130

34. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 = 152.2 – 130 = 22.5 ≈ 5.63 x 2 - x 1 8 – 4 4 *So use x 1 = 4, y 1 = 130, and m ≈ 5.63

35. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 = 152.2 – 130 = 22.5 ≈ 5.63 x 2 - x 1 8 – 4 4 *So use x 1 = 4, y 1 = 130, and m ≈ 5.63 y – y 1 = m(x – x 1 )

36. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 = 152.2 – 130 = 22.5 ≈ 5.63 x 2 - x 1 8 – 4 4 *So use x 1 = 4, y 1 = 130, and m ≈ 5.63 y – y 1 = m(x – x 1 )

37. c.Find a prediction equation for line of fit. *Use the best two ordered pairs from b. to find the slope for the line! (4, 130) and (8, 152.5) m = y 2 – y 1 = 152.2 – 130 = 22.5 ≈ 5.63 x 2 - x 1 8 – 4 4 *So use x 1 = 4, y 1 = 130, and m ≈ 5.63 y – y 1 = m(x – x 1 ) y – 130 = 5.63(x – 4) y – 130 = 5.63(x) – 5.63(4) y – 130 = 5.63x – 22.52 y = 5.63x + 107.48

38. d.Predict the price in 2020.

39. 2020 means when x=30 (yrs after 1990)

40. d.Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x!

41. d.Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48

42. d.Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48 y = 5.63(30) + 107.48

43. d.Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48 y = 5.63(30) + 107.48 y = 168.9 + 107.48

44. d.Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48 y = 5.63(30) + 107.48 y = 168.9 + 107.48 y = 276.38

45. d.Predict the price in 2020. 2020 means when x=30 (yrs after 1990) *Plug 30 in for x! y = 5.63x + 107.48 y = 5.63(30) + 107.48 y = 168.9 + 107.48 y = 276.38 So, in 2020 the price will be $276,380.