The Normal Distribution and the 68-95-99.7 Rule http://www.mathsnet.net/js/balldro p.html Normal means typical If the average woman is 5 feet 4 inches tall (64 inches) would you expect to see many women who are around that height? Is it common to see women who are 6 feet tall? Is it common to see women who are 4 feet tall? Describing what is Normal Mean (EVERY ONE KNOWS THIS ONE) This is the average, the center. Here it is 64. It’s located right in the middle of the normal curve. Describing what is Normal Standard Deviation (NOT MANY KNOW THIS ONE) This tells us how spread out the distribution is. Here it is 3. What is a typical woman’s height? 68% of all women are within 1 standard deviation of the mean. Here 68% of all women are within 3 inches of 64. Going out farther… 95% of all women are within 2 standard deviations of the mean. Here 95% of all women are within 6 inches of 64. Almost all women are… 99.7% of all women are within 3 standard deviations of the mean. Here 99.7% of all women are within 9 in. of 64. This is not enough information. It leaves a lot of questions… The average grade on the quiz was 30 points. Could all of the kids gotten a 30? Did half the kids get a 20 and the other half get a 40? What would be a good score on the quiz? What would be a bad score? What if we were only given the standard deviation? The standard deviation on the quiz was 5 points. So we know the spread of the grades, but from where was the quiz centered? How did the class do in general? Was my grade good or bad? This would still leave us with plenty of questions… With both Mean and St. Dev. Consider the following statement: The mean was 30 points and the standard deviation on the quiz was 5 points. Now we know that about 68% of the class scored between… 25 and 35 points Now we can say that it was rare to get a score that was above 40 points. It was extremely rare to score above 45 points. The Normal Distribution needs both statistics to survive. We can also graph the quiz scores. A typical score… was between 25 and 35 points An extremely rare score… What percent of the students did this student beat? An extremely rare score… About 99.7 percent scored between a 15 to 45. That does not include about 0.3% An extremely rare score… You would have to split this in half to get about 0.15%. So out of 2000 students, only 3 would score that well! Practice with 68-95-99.7 Rule Suppose the average height of boys at EHS is 66 inches, with a standard deviation of 4.5 inches. Draw the normal curve representing this information. Answer: Practice with 68-95-99.7 Rule Suppose the average height of boys at EHS is 66 inches, with a standard deviation of 4.5 inches. How tall is someone 2 standard deviations above the mean? Answer: 66 + 4.5 + 4.5 = 75 inches Practice with 68-95-99.7 Rule Suppose the average height of boys at EHS is 66 inches, with a standard deviation of 4.5 inches. What percent of boys are between 61.5 and 70.5 inches tall ? Answer: 68% Practice with 68-95-99.7 Rule Suppose the average height of boys at EHS is 66 inches, with a standard deviation of 4.5 inches. How many SD’s above the mean is someone who is 79.5 inches tall? Answer: 3 Practice with 68-95-99.7 Rule Suppose the average height of boys at Sweet Home is 66 inches, with a standard deviation of 4.5 inches. What percent of boys are between 57 and 70.5 inches tall? Answer: 68%+ the bit between 57 and 61.5 68%+13.5% = 81.5%