Chapter 12 Statistics 2012 Pearson Education, Inc.

Chapter 12: Statistics 12.1 Visual Displays of Data 12.2 Measures of Central Tendency 12.3 Measures of Dispersion 12.4 Measures of Position 12.5 The Normal Distribution 2012 Pearson Education, Inc.

The Normal Distribution Section 12-5 The Normal Distribution 2012 Pearson Education, Inc.

The Normal Distribution Discrete and Continuous Random Variables Definition and Properties of a Normal Curve A Table of Standard Normal Curve Areas Interpreting Normal Curve Areas 2012 Pearson Education, Inc.

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. 2012 Pearson Education, Inc.

Definition and Properties of a Normal Curve A normal curve is a symmetric, bell-shaped curve. Any random variable whose graph has this characteristic shape is said to have a normal distribution. 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. 2012 Pearson Education, Inc.

Normal Curves B S A C 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 2012 Pearson Education, Inc.

Properties of Normal Curves 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. 2012 Pearson Education, Inc.

Empirical Rule 2012 Pearson Education, Inc.

Example: Applying the Empirical Rule 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. Solution A total of 95% of all scores lie within 2 standard deviations of the mean. (.95)(280)= 266 scores 2012 Pearson Education, Inc.

A Table of Standard Normal Curve Areas 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. The table 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. 2012 Pearson Education, Inc.

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. a) 1.5 standard deviation above the mean b) 2.62 standard deviations below the mean 2012 Pearson Education, Inc.

Example: Applying the Normal Curve Table 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. z = 1.5 2012 Pearson Education, Inc.

Example: Applying the Normal Curve Table Solution (continued) 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 = –2.62 2012 Pearson Education, Inc.

Example: Applying the Normal Curve Table Find the total area indicated in the region in color below. z = –1.7 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 2012 Pearson Education, Inc.

Example: Applying the Normal Curve Table Find the total area indicated in the region in color below. z = .61 z = 2.63 Solution z = 2.61 leads to an area of .496 and z = .61 leads to an area of .229. Subtract these areas to get .496 – .229 = .267. 2012 Pearson Education, Inc.

Example: Applying the Normal Curve Table Find the total area indicated in the region in color below. z = 2.14 Solution 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. 2012 Pearson Education, Inc.

Interpreting Normal Curve Areas In a standard normal curve, the following three quantities are equivalent. Percentage (of total items that lie in an interval) Probability (of a randomly chosen item lying in an interval) Area (under the normal curve along an interval) 2012 Pearson Education, Inc.

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? 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%). 2012 Pearson Education, Inc.

Example: Finding z-Values for Given Areas 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% 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. 2012 Pearson Education, Inc.

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 24% 50% There must be 26% of the area between the mean and z. From the table, A = .260 corresponds closely to z = .71. 2012 Pearson Education, Inc.