AP Statistics Empirical Rule What does a population that is normally distributed look like? = 80 and = 10 X 50 60 70 80 90 100 110 Normal Distribution Features • Symmetric, single-peaked, and bell-shaped. • Reported: N , • Changing µ changes the location of the curve on the horizontal axis with no change in spread. • Changing changes the spread of the curve. • is located at the inflection point of the curve. 68-95-99.7 Rule 68% 95% 99.7% 68-95-99.7% RULE 68-95-99.7 Rule—restated 68% of the observations fall within 1 standard deviation of the mean in either direction. 95% of the observations fall within 2 standard deviations of the mean in either direction. 99.7% of the observations fall within 3 standard deviations of the mean in either direction. 68-95-99.7 Rule 34% 34% 68% 47.5% 47.5% 95% 49.85% 49.85% 99.7% Average American adult male height is 69 inches (5’ 9”) tall with a standard deviation of 2.5 inches. Let H~N(69, 2.5) What is the likelihood that a randomly selected adult male would have a height less than 69 inches? Answer: P(h < 69) = .50 P represents Probability h represents one adult male height Using the 68-95-99.7 Rule Let H~N(69, 2.5) What is the likelihood that a randomly selected adult male will have a height between 64 and 74 inches? P(64 < h < 74) = .95 Let H~N(69, 2.5) What is the likelihood that a randomly selected adult male would have a height of less than 66.5 inches? P(h < 66.5) = 1 – (.50 + .34) = .16 OR .50 - .34 = .16 Let H~N(69, 2.5) What is the likelihood that a randomly selected adult male would have a height of greater than 74 inches? P(h > 74) = 1 – P(h 74) = 1 – (.50 + .47.5) = 1 - .975 = .025 Let H~N(69, 2.5) What is the probability that a randomly selected adult male would have a height between 64 and 76.5 inches? P(64 < h < 76.5) = .475 + .4985 = .9735 More about Normal Curves • Percentiles divide the area of the density curve into fractions out of 100. • Median: 50th percentile. • Q1: 25th percentile. • Q3: 75th percentile. Exploring Data: Distributions The 68-95-99.7 Rule • The distribution of scores on tests such as the SAT college entrance exams is close to normal. Scores on each of the three sections (math, critical reading, writing) of the SAT are adjusted so that the mean score is about 500 and the standard deviation is about 100. Using this information answer the following questions. The 68-95-99.7 Rule • • • • • • • A) How high must a student score to fall in the top 25%? This question is asking for the third quartile. Q3 = (.67)(standard deviation) = (.67)(100) = 67 points above the mean = 500 + 67 = 567 Scores above 567 are in the top 25%. The 68-95-99.7 Rule • • • • • B) What percent of scores fall between 200 and 800? These scores are 3 standard deviations from the mean. 200 = 500 – 3(100) 800 = 500 + 3(100) The percent of scores are 99.7%. Why are Normal Curves Important? • They provide good descriptions of some distributions of real data (SAT, ACT, characteristics of biological populations). • They are a good approximation of the results of chance outcomes (sum the two die rolls). • Many statistical procedures are based on normal distributions. However! • Many distributions are not normal (income). Don’t assume every situation is approximated by a normal distribution.