1. Copyright © 2016 Brooks/Cole Cengage Learning Advanced Algebra The Normal Curve Advanced Algebra The Normal Curve Ernesto Diaz Assistant Professor of Mathematics

2. Copyright © Cengage Learning. All rights reserved. 14.3 The Normal Curve

3. 3 Cumulative Distributions

4. 4 A cumulative frequency is the sum of all preceding frequencies in which some order has been established.

5. 5 Example 1 – Find the mean, median, and mode A judge ordered a survey to determine how many of the offenders appearing in her court during the past year had three or more previous appearances. The accumulated data are shown in Table 14.9. Table 14.9 Number of Previous Appearances

6. 6 Example 1 – Find the mean, median, and mode Note the last column shows the cumulative percent. This is usually called the cumulative relative frequency. Use the cumulative relative frequency to find the percent who had three or more previous appearances, and then find the mean, median, and mode. Solution: From the cumulative relative frequency we see that 70% had 2 or fewer court appearances; we see that since the total is 100%, 30% must have had 3 or more previous appearances. cont’d

7. 7 Example 1 – Solution The mean is found (using the idea of a weighted mean). Note that the sum is 100%, or 1: = 1.93 The median is the number of court appearances for which the cumulative percent first exceeds 50%; we see that this is 2 court appearances. The mode is the number of court appearances that occurs most frequently; we see that this is 1 court appearance. cont’d

8. 8 Bell-Shaped Curves

9. 9 Suppose we survey the results of 20 children’s scores on an IQ test. The scores (rounded to the nearest 5 points) are 115, 90, 100, 95, 105, 95, 105, 105, 95, 125, 120, 110, 100, 100, 90, 110, 100, 115, 105, and 80. We can find = 103 and s  10.93.

10. 10 Bell-Shaped Curves A frequency graph of these data is shown in part a of Figure 14.34. If we consider 10,000 scores instead of only 20, we might obtain the frequency distribution shown in part b of Figure 14.34. Figure 14.34 Frequency distributions for IQ scores a. IQs of 20 childrenb. IQs of 10,000 children

11. 11 The Normal Distribution: as mathematical function (pdf) Note constants:  =3.14159 e=2.71828 This is a bell shaped curve with different centers and spreads depending on  and 

12. 12 **The beauty of the normal curve: No matter what  and  are, the area between  -  and  +  is about 68%; the area between  -2  and  +2  is about 95%; and the area between  -3  and  +3  is about 99.7%. Almost all values fall within 3 standard deviations.

13. 13 68-95-99.7 Rule 68% of the data 95% of the data 99.7% of the data

14. 14 Bell-Shaped Curves The data illustrated in Figure 14.34 approximate a commonly used curve called a normal frequency curve, or simply a normal curve. (See Figure 14.35.) Figure 14.35 A normal curve

15. 15 Bell-Shaped Curves If we obtain the frequency distribution of a large number of measurements (as with IQ), the corresponding graph tends to look normal, or bell-shaped. The normal curve has some interesting properties. In it, the mean, the median, and the mode all have the same value, and all occur exactly at the center of the distribution; we denote this value by the Greek letter mu (  ). The standard deviation for this distribution is  (sigma).

16. 16 Bell-Shaped Curves The normal distribution is a continuous (rather than a discrete) distribution, and it extends indefinitely in both directions, never touching the x-axis. It is symmetric about a vertical line drawn through the mean, . Graphs of this curve for several choices of  are shown in Figure 14.36. Figure 14.36 Variations of normal curves

17. 17 Example 2 – Find a normal distribution Predict the distribution of IQ scores of 1,000 people if we assume that IQ scores are normally distributed, with a mean of 100 and a standard deviation of 15. Solution: First, find the breaking points around the mean. For  = 100 and  = 15:  +  = 100 + 15 = 115  –  = 100 – 15 = 85  + 2  = 100 + 2(15) = 130  – 2  = 100 – 2(15) = 70  + 3  = 100 + 3(15) = 145  – 3  = 100 – 3(15) = 55

18. 18 Example 2 – Solution We use Figure 14.35 to find that 34.1% of the scores will be between 100 and 115 (i.e.,between  and  + 1  ) : 0.341  1,000 = 341 cont’d Figure 14.35 A normal curve

19. 19 Example 2 – Solution About 13.6% will be between 115 and 130 (between  + 1  and  + 2  ): 0.136  1,000 = 136 About 2.2% will be between 130 and 145 (between  + 2  and  + 3  ): 0.022  1,000 = 22 About 0.1% will be above 145 (more than  + 3  ): 0.001  1,000 = 1 cont’d

20. 20 Example 2 – Solution The distribution for intervals below the mean is identical, since the normal curve is the same to the left and to the right of the mean. The distribution is shown below. cont’d

21. 21 z-Scores

22. 22 z-Scores First, we introduce some terminology. We use z-scores (sometimes called standard scores) to determine how far, in terms of standard deviations, a given score is from the mean of the distribution. We use the z-score to translate any normal curve into a standard normal curve (the particular normal curve with a mean of 0 and a standard deviation of 1) by using the definition.

23. 23 z-Scores

24. 24 Example 5 – Find probabilities in a normal distribution The Eureka Lightbulb Company tested a new line of lightbulbs and found their lifetimes to be normally distributed, with a mean life of 98 hours and a standard deviation of 13 hours. a. What percentage of bulbs will last less than 72 hours? b. What percentage of bulbs will last less than 100 hours? c. What is the probability that a bulb selected at random will last longer than 111 hours? d. What is the probability that a bulb will last between 106 and 120 hours?

25. 25 Example 5 – Solution Draw a normal curve with mean 98 and standard deviation 13, as shown in Figure 14.38. cont’d Figure 14.38 Lightbulb lifetimes are normally distributed

26. 26 Example 5 – Solution For this example, we are given  = 98 and  = 13. a. For x = 72, z = = –2. This is 2 standard deviations below the mean. cont’d

27. 27 Example 5 – Solution The percentage we seek is shown in blue in Figure 14.39a. About 2.3% (2.2% + 0.1% = 2.3%) will last less than 72 hours. We can also use the z-score to find the percentage. Figure 14.39(a) 2 standard deviations below the mean Using z -scores cont’d

28. 28 Example 5 – Solution From Table 14.10, the area to the left of z = –2 is 0.5 – 0.4772 = 0.0228, so the probability is about 2.3%. cont’d Table 14.10 Standard Normal Distribution

29. 29 Example 5 – Solution cont’d Table 14.10 Standard Normal Distribution

30. 30 Example 5 – Solution b. For x = 100, z = This is 0.15 standard deviation above the mean. From Table 14.10, we find 0.0596 (this is shown in green in Figure 14.39b). Figure 14.39(b) 0.15 standard deviations above the mean Using z-scores cont’d  0.15.

31. 31 Example 5 – Solution Since we want the percent of values less than 100, we must add 50% for the numbers below the mean (shown in blue). The percentage we seek is 0.5000 + 0.0596 = 0.5596 or about 56.0%. cont’d

32. 32 Example 5 – Solution c. For x = 111, z = = 1. This is one standard deviation above the mean, which is the same as the z-score. The percentage we seek is shown in blue (see Figure 14.39c). Figure 14.39(c) 1 standard deviations above the mean Using z-scores cont’d

33. 33 Example 5 – Solution We know that about 15.9% (13.6% + 2.2% + 0.1% = 15.9%) of the bulbs will last longer than 111 hours, so P(bulb life > 111 hours)  0.159 We can also use Table 14.10. For z = 1.00, the table entry is 0.3413, and we are looking for the area to the right, so we compute 0.5000 – 0.3413 = 0.1587 cont’d

34. 34 Example 5 – Solution d. We first find the z-scores using  = 98 and  = 13. For x = 106, z =  0.62. From Table 14.10, the area between the z-score and the mean is 0.2324 (shown in green). For x = 120, z =  1.69. cont’d

35. 35 Example 5 – Solution From Table 14.10, the area between this z-score and the mean is 0.4545. The desired answer (shown in yellow) is approximately 0.4545 – 0.2324 = 0.2221 Since percent and probability are the same, we see the probability that the life of the bulb is between 106 and 120 hours is about 22.2%. cont’d

36. 36 z-Scores

37. 37 z-Scores Sometimes data do not fall into a normal distribution, but are skewed, which means their distribution has more tail on one side or the other. For example, Figure 14.40a shows that the 1941 scores on the SAT exam (when the test was first used) were normally distributed. Figure 14.40(a) 1941 SAT scale Distribution of SAT scores

38. 38 z-Scores However, by 1990, the scale had become skewed to the left, as shown in Figure 14.40b. Figure 14.40(b) 1990 SAT scale Distribution of SAT scores

39. 39 z-Scores In a normal distribution, the mean, median, and mode all have the same value, but if the distribution is skewed, the relative positions of the mean, median, and mode would be as shown in Figure 14.41. Figure 14.41 Comparison of three distributions a. Skewed to the right (positive skew) b. Normal distributionc. Skewed to the left (negative skew)