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Chapter 13 Chapter 13: Statistics 13.1 13.2 13.3 13.4 13.5 13.6 Statistics Visual Displays of Data Measures of Central Tendency Measures of Dispersion Measures of Position The Normal Distribution Regression and Correlation © 2008 Pearson Addison-Wesley. All rights reserved 13-5-2 © 2008 Pearson Addison-Wesley. All rights reserved Chapter 1 The Normal Distribution • • • • Section 13-5 Discrete and Continuous Random Variables Definition and Properties of a Normal Curve A Table of Standard Normal Curve Areas Interpreting Normal Curve Areas The Normal Distribution 13-5-3 © 2008 Pearson Addison-Wesley. All rights reserved Random Variables A variable is a characteristic or attribute that can assume different values.* A random variable is a variable with value assigned by the result of an experiment. 13-5-4 © 2008 Pearson Addison-Wesley. All rights reserved Discrete and Continuous Random Variables A random variable that can take on only certain fixed values is called a discrete random variable. A variable whose values are not restricted in this way is a continuous random variable. *Allan G. Bluman, Elementary Statistics, 4th ed,. ©2008 13-5-6 © 2008 Pearson Addison-Wesley. All rights reserved Definition and Properties of a Normal Curve Normal Curves B A normal curve is a symmetric, bell-shaped curve. Any random variable whose distribution curve has this characteristic shape is said to have a normal distribution. A S C On a normal curve, if the quantity shown on the horizontal axis is the number of standard deviations from the mean, rather than values of the random variable itself, then we call the curve the standard normal curve. 13-5-7 © 2008 Pearson Addison-Wesley. All rights reserved 0 S is standard, with mean = 0, standard deviation = 1 A has mean < 0, standard deviation = 1 B has mean = 0, standard deviation < 1 C has mean > 0, standard deviation > 1 13-5-8 © 2008 Pearson Addison-Wesley. All rights reserved Properties of Normal Curves Empirical Rule The graph of a normal curve is bell-shaped and symmetric about a vertical line through its center. The mean, median, and mode of a normal curve are all equal and occur at the center of the distribution. Empirical Rule About 68% of all data values of a normal curve lie within 1 standard deviation of the mean (in both directions), and about 95% within 2 standard deviations, and about 99.7% within 3 standard deviations. 13-5-9 © 2008 Pearson Addison-Wesley. All rights reserved 13-5-10 © 2008 Pearson Addison-Wesley. All rights reserved Example: Applying the Empirical Rule A Table of Standard Normal Curve Areas Suppose that 280 sociology students take an exam and that the distribution of their scores can be treated as normal. Find the number of scores falling within 2 standard deviations of the mean. To answer questions that involve regions other than 1, 2, or 3 standard deviations of the mean we can refer to the table on page 829 or other tools such as a computer or calculator. Solution A total of 95% of all scores lie within 2 standard deviations of the mean. (.95)(280)= 266 scores 13-5-11 © 2008 Pearson Addison-Wesley. All rights reserved 13-5-12 © 2008 Pearson Addison-Wesley. All rights reserved A Table of Standard Normal Curve Areas Recall that the z-score of a data item is the number of standard deviations away from the mean that data item is. Example: Applying the Normal Curve Table Use the table to find the percent of all scores that lie between the mean and the following values. The table on p. 829 gives the fraction of all scores in a normal distribution that lie between the mean and z standard deviations from the mean. Because of symmetry, the table can be used for values above the mean or below the mean. a) 1.5 standard deviation above the mean b) 2.62 standard deviations below the mean 13-5-14 © 2008 Pearson Addison-Wesley. All rights reserved Example: Applying the Normal Curve Table Example: Applying the Normal Curve Table Solution (continued) Solution a) Here z = 1.50. Find 1.50 in the z column. The table entry is .433, so 43.3% of all values lie between the mean and 1.5 standard deviations above the mean. b) Even though it is below the mean, the table still works because of the symmetry. Find 2.62 in the z column. The table entry is .496, so 49.6% of all values lie between the mean and 2.62 standard deviations below the mean. z = 1.5 x z = –2.62 x 13-5-15 13-5-16 © 2008 Pearson Addison-Wesley. All rights reserved © 2008 Pearson Addison-Wesley. All rights reserved Example: Applying the Normal Curve Table Example: Applying the Normal Curve Table Find the total area indicated in the region in color below. Solution z = –1.7 x Find the total area indicated in the region in color below. z = 2.55 Solution z = 2.55 leads to an area of .495 and z = 1.7 leads to an area of .455. Add these areas to get .495 + .455 = .950 z = .61 z = 2.63 leads to an area of .496 and z = .61 leads to an area of .229. Subtract these areas to get .496 – .229 = .267. 13-5-17 © 2008 Pearson Addison-Wesley. All rights reserved z = 2.63 13-5-18 © 2008 Pearson Addison-Wesley. All rights reserved Example: Applying the Normal Curve Table Interpreting Normal Curve Areas Find the total area indicated in the region in color below. In a standard normal curve, the following three quantities are equivalent. 1. Percentage (of total items that lie in an interval) 2. Probability (of a randomly chosen item lying in an interval) z = 2.14 Solution 3. Area (under the normal curve along an interval) z = 2.14 leads to an area of .484. The total area to the right of the mean is .500. Subtract these values to get .500 – .484 = .016. 13-5-19 13-5-20 © 2008 Pearson Addison-Wesley. All rights reserved © 2008 Pearson Addison-Wesley. All rights reserved Example: Finding z-Values for Given Areas Example: Probability with Volume The volumes of soda in bottles from a small company are distributed normally with mean 12 ounces and standard deviation .15 ounce. If 1 bottle is randomly selected, what is the probability that it will have more than 12.33 ounces? Assuming a normal distribution, find the z-value meeting the condition that 39% of the area is to the right of z. 11% Solution 39% Solution 12.33 corresponds to a z-value of 2.2. The probability is equal to the area above z = 2.2, which is .500 – .486 = .014 (or 1.4%). Because 50% of the area lies to the right of the mean, there must be 11% of the area between the mean and z. From the table, A = .110 corresponds to z = .28. 13-5-21 © 2008 Pearson Addison-Wesley. All rights reserved Example: Finding z-Values for Given Areas Assuming a normal distribution, find the z-value meeting the condition that 76% of the area is to the left of z. 26% Solution 50% 24% There must be 26% of the area between the mean and z. From the table, A = .260 corresponds closely to z = .71. 13-5-23 © 2008 Pearson Addison-Wesley. All rights reserved 13-5-22 © 2008 Pearson Addison-Wesley. All rights reserved