A Mathematical View of Our World

1. A Mathematical View of Our World 1st ed. Parks, Musser, Trimpe, Maurer, and Maurer

2. Chapter 11 Inferential Statistics

3. Section 11.1Normal Distributions • Goals • Study normal distributions • Study standard normal distributions • Find the area under a standard normal curve

4. 11.1 Initial Problem • A class of 90 students had a mean test score of 74, with a standard deviation of 8. • If the professor curves the scores, how many students will get As and how many will get Fs? • The solution will be given at the end of the section.

5. Statistical Inference • The process of making predictions about an entire population based on information from a sample is called statistical inference.

6. Data Distributions • For large data sets, a smooth curve can often be used to approximate the histogram.

7. Data Distributions, cont’d • The larger the data set and the smaller the bin size, the better the approximation of the smooth curve.

8. Example 1 • The distribution of weights for a large sample of college men is shown.

9. Example 1, cont’d • What percent of the men have weights between: • 167 and 192 pounds? • 137 and 192 pounds? • 137 and 222 pounds?

10. Example 1, cont’d • Solution: • 167 and 192 pounds? • The area under the curve is 0.2, so 20% of the men are in this weight range.

11. Example 1, cont’d • Solution: • 137 and 192 pounds? • The area under the curve is 0.4, so 40% of the men are in this weight range.

12. Example 1, cont’d • Solution: • 137 and 222 pounds? • The area under the curve is 0.6, so 60% of the men are in this weight range.

13. Normal Distributions • Data that has a symmetric, bell-shaped distribution curve is said to have a normal distribution. • The mean and standard deviation determine the exact shape and position of the curve.

14. Example 2 • Which normal curve has the largest mean? • Which normal curve has the largest standard deviation?

15. Example 2, cont’d • Solution: The data sets are already labeled in order of smallest mean to largest mean. • Data Set III has the largest mean.

16. Example 2, cont’d • Solution: Data Set III has the largest standard deviation because it is the shortest, widest curve. • The order of the standard deviations is II, I, III.

17. Normal Distributions, cont’d • Normal distributions with various means and standard deviations are shown on the following slides.

18. Normal Distributions, cont’d

19. Normal Distributions, cont’d

20. Normal Distributions, cont’d

21. Standard Normal Distribution • The normal distribution with a mean of 0 and a standard deviation of 1 is called the standard normal distribution.

22. Standard Normal Distribution, cont’d • The areas under any normal distribution can be compared to the areas under the standard normal distribution, as shown in the figure on the next slide.

23. Standard Normal Distribution, cont’d

24. Area • One way to find the area under a region of the standard normal curve is to use a table. • Tables of values for the standard normal curve are printed in textbooks to eliminate the need to do repeated complicated calculations.

25. Example 3 • What fraction of the total area under the standard normal curve lies between a = -0.5 and b = 1.5?

26. Example 3, cont’d • Solution: Find a = -0.5 and b = 1.5 in the table.

27. Example 3, cont’d • Solution, cont’d: The value in the table is 0.6247. • A total of 62.47% of the area is shaded. • In any normal distribution, 62.47% of the data lies between 0.5 standard deviations below the mean and 1.5 standard deviations above the mean. • The probability a randomly selected data value will lie between -0.5 and 1.5 is 62.47%

28. Example 4 • What percent of the data in a standard normal distribution lies between 0.5 and 2.5?

29. Example 4, cont’d • Solution: The value in the table for a = 0.5 and b = 2.5 is 0.3023. • So 30.23% of the data in a standard normal distribution lies between 0.5 and 2.5.

30. Areas, cont’d • Because the normal curve is symmetric, the areas in the previous table are repeated.

31. Areas, cont’d Figure 11.11 and table 11.2

32. Areas, cont’d • A more common type of table:

33. Example 5 • Find the percent of data points in a standard normal distribution that lie between z = -1.8 and z = 1.3.

34. Example 5, cont’d • Solution: Find the two areas in Table 11.3 and add them together. • The area from 0 to 1.3 is 0.4032. • The area from 0 to -1.8 is 0.4641. • The total shaded area is 0.4032 + 0.4641 = 0.8673.

35. Example 6 • Find the percent of data points in a standard normal distribution that lie between z = 1.2 and z = 1.7.

36. Example 6, cont’d • Solution: Find the two areas in Table 11.3 and subtract them. • The area from 0 to 1.2 is 0.3849. • The area from 0 to 1.7 is 0.4554. • The total shaded area is 0.4554 - 0.3849 = 0.0705.

37. Question: The value from the table associated with z = 2.1 is 0.4821. To find the percentage of data values less than -2.1 in a standard normal distribution, what do you need to do? a. Add the table value to 0.5. b. Subtract the table value from 0.5. c. The table value is the answer. d. Divide the table value in half.

38. Question: What percentage of data values lie between z = -1.2 and z = -0.7 in a standard normal distribution? a. 11.51% b. 62.49% c. 12.69% d. 24.20%

39. 11.1 Initial Problem Solution • A class of 90 students had a mean test score of 74 with a standard deviation of 8 points. • The test will be curved so that all students whose scores are at least 1.5 standard deviations above or below the mean will receive As and Fs, respectively. • How many students will get As and how many will get Fs?

40. Initial Problem Solution, cont’d • Because the class is large, it is likely the scores have a normal distribution. • If the scores are curved: • The mean of 74 will correspond to a score of 0 in the standard normal distribution. • A score that is 1.5 standard deviations above the mean will correspond to a score of +1.5 in the standard normal distribution, while a score that is 1.5 standard deviations below the mean will correspond to a score of -1.5.

41. Initial Problem Solution, cont’d • The percentage of As is the same as the area to the right of z = 1.5 in the standard normal distribution. • Approximately 43.32% of the area is between 0 and 1.5. • Since 50% of the area is to the right of 0, the area above 1.5 is 50% - 43.32% = 6.68% • Thus, 6.68% of the students, or approximately 6 students, will receive As.

42. Initial Problem Solution, cont’d • The percentage of Fs is the same as the area to the left of z = -1.5 in the standard normal distribution. • Because of the symmetry of the normal distribution, this is the same as the area above z = 1.5, so the calculations are the same as in the last step. • Thus, 6.68% of the students, or approximately 6 students, will receive Fs.

43. Section 11.2Applications of Normal Distributions • Goals • Study normal distribution applications • Use the 68-95-99.7 Rule • Use the population z-score

44. 11.2 Initial Problem • Two suppliers make an engine part. • Supplier A charges $120 for 100 parts which have a standard deviation of 0.004 mm from the mean size. • Supplier B charges $90 for 100 parts which have a standard deviation of 0.012 mm from the mean size. • Which supplier is a better choice? • The solution will be given at the end of the section.

45. Normal Distributions • If a data set is represented by a normal distribution with mean μ and standard deviation σ, the percentage of the data between μ+ rσ and μ+ sσ is the same as the percentage of the data in a standard normal distribution that lies between r and s.

46. Normal Distributions, cont’d

47. Example 1 • Approximately 10% of the data in a standard normal distribution lies within 1/8 of a standard deviation from the mean. • Within 1/8 means between -0.125 and 0.125. • Suppose the measurements of a certain population are normally distributed with a mean of 112 and standard deviation of 24. What values correspond to the interval given above?

48. Example 1, cont’d • Solution: In the standard normal distribution we are considering the interval from r = -0.125 to s = 0.125. • For the nonstandard distribution, the interval will be 112 + (-0.125)(24) = 109 to 112 + (0.125)(24) = 115. • We know that 10% of the data values will lie between 112 and 115.

49. Example 2 • The HDL cholesterol levels for a group of women are approximately normally distributed with a mean of 64 mg/dL and a standard deviation of 15 mg/dL. • Determine the percentage of these women that have HDL cholesterol levels between 19 and 109 mg/dL.

50. Example 2, cont’d • Solution: The mean of 64 mg/dL corresponds to 0 in the standard normal curve. • The value of 19 is 45 less than the mean, corresponding to 3 standard deviations below the mean. • The value of 64 is 45 more than the mean, corresponding to 3 standard deviations above the mean.