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% Empirical Rule is sometimes referred to as the 68-95-99.7% Rule. Again, to use the Empirical Rule the distribution of the data must be normal bell-shaped curve. 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. Rule of Thumb: Range divide by 6 approximately the standard deviation.

68-95-99.7 Rule 34% 34% 68% 47.5% 47.5% 95% Take time to slowly click through slide. Stress that the Empirical Rule can ONLY be used with the assumption that the distribution is normal (bell-shaped curve). Sixty-eight percent of the ordered data of a normal distribution lies within one standard deviation of the mean. Ninety-five percent of the ordered data of a normal distribution lies within two standard deviations of the mean. And, 99.7% of the ordered data of a normal distribution lies within 3 standard deviations of the mean. The normal distribution here, in this example shown, has a mean of 0 and standard deviation of 1. 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 Let H~N(69, 2.5) means H represents the variable height for an adult male has a normal distribution with population mean of 69 inches and a population standard deviation of 2.5 (or variance of 2.5). Start with an easy problem using the empirical rule AND introduce proper probability notation. Explain P implies probability (likelihood of an event), H represents the random variable of height of adult male, P(h<69) is the question in statistical notation. h represents one adult male height

Using the 68-95-99.7 Rule P(64 < h < 74) = .95 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) means H represents the variable height for an adult male has a normal distribution with population mean of 69 inches and a population standard deviation of 2.5 (or variance of 2.5).

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 There are several ways to do this problem. Some students may see it as .5 - .34 to obtain .16. If students obtained an answer of .16 but did it differently please ask them to discuss their method.

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.