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Advanced Algebra The Normal Curve Ernesto Diaz Copyright © 2016 Brooks/Cole Cengage Learning 14.3 The Normal Curve Copyright © Cengage Learning. All rights reserved. Bell-Shaped Curves 3 Bell-Shaped Curves 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. 4 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. a. IQs of 20 children b. IQs of 10,000 children Frequency distributions for IQ scores Figure 14.34 5 The Normal Distribution: as mathematical function (pdf) f ( x) 1 2 Note constants: =3.14159 e=2.71828 1 x 2 ( ) 2 e This is a bell shaped curve with different centers and spreads depending on and 6 **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. 7 68-95-99.7 Rule 68% of the data 95% of the data 99.7% of the data 8 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.) A normal curve Figure 14.35 9 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). 10 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. Variations of normal curves Figure 14.36 11 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 12 Example 2 – Solution cont’d 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 A normal curve Figure 14.35 13 Example 2 – Solution cont’d 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 14 Example 2 – Solution cont’d 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. 15 z-Scores 16 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. 17 z-Scores 18 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? 19 Example 5 – Solution cont’d Draw a normal curve with mean 98 and standard deviation 13, as shown in Figure 14.38. Lightbulb lifetimes are normally distributed Figure 14.38 20 Example 5 – Solution cont’d For this example, we are given = 98 and = 13 . a. For x = 72, z = = –2. This is 2 standard deviations below the mean. 21 Example 5 – Solution cont’d The percentage we seek is shown in blue in Figure 14.39a. 2 standard deviations below the mean Using z -scores Figure 14.39(a) 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. 22 Example 5 – Solution cont’d 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%. Standard Normal Distribution Table 14.10 23 Example 5 – Solution cont’d Standard Normal Distribution Table 14.10 24 Example 5 – Solution b. For x = 100, z = cont’d 0.15. 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). 0.15 standard deviations above the mean Using z-scores Figure 14.39(b) 25 Example 5 – Solution cont’d 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%. 26 Example 5 – Solution c. For x = 111, z = cont’d = 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). 1 standard deviations above the mean Using z-scores Figure 14.39(c) 27 Example 5 – Solution cont’d 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 28 Example 5 – Solution cont’d 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. 29 Example 5 – Solution cont’d 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%. 30 z-Scores 31 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. 1941 SAT scale Distribution of SAT scores Figure 14.40(a) 32 z-Scores However, by 1990, the scale had become skewed to the left, as shown in Figure 14.40b. 1990 SAT scale Distribution of SAT scores Figure 14.40(b) 33 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. a. Skewed to the right (positive skew) b. Normal distribution c. Skewed to the left (negative skew) Comparison of three distributions Figure 14.41 34