Math 3 Warm Up 4/23/12 Find the probability mean and standard deviation for the following data. 2, 4, 5, 6, 5, 5, 5, 2, 2, 4, 4, 3, 3, 1, 2, 2, 3, 4, 6, 5 Hint: First create a probability distribution.
Unit 6: Data Analysis Empirical Rule
What does a population that is normally distributed look like?
Empirical Rule 68-95-99.7% RULE 68% 95% 99.7% 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
Empirical Rule 68-95-99.7% RULE 68% 95% 99.7% 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
Empirical Rule 68-95-99.7% RULE 68% 95% 99.7% 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
Empirical Rule 68-95-99.7% RULE 68% 95% 99.7% 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
Empirical Rule—restated 68% of the data values fall within 1 standard deviation of the mean in either direction 95% of the data values fall within 2 standard deviation of the mean in either direction 99.7% of the data values fall within 3 standard deviation of the mean in either direction Remember values in a data set must appear to be a normal bell-shaped histogram, dotplot, or stemplot to use the Empirical Rule! Rule of Thumb: Range divide by 6 approximately the standard deviation.
Empirical Rule 34% 34% 68% 47.5% 47.5% 95% 49.85% 49.85% 99.7% 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. What does the normal distribution for this data look like?
Empirical Rule-- 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 Empirical 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 In Calculator: 2nd Vars: normalcdf(lower, upper, mean, std. dev.) 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).
Using the Empirical 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? In Calculator: 2nd Vars normalcdf(lower, upper, mean, std. dev.) For this example: Normalcdf(64, 74, 69, 2.5) = .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).
Using Empirical Rule-- 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? 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. = .16
Using Empirical Rule--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? = .025
Using Empirical Rule--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? = .9735
Assignment: Complete Normal Distribution Practice Worksheet