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The Normal Distribution and the 68-95-99.7 Rule. http://www.mathsnet.net/js/balldrop.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?
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The Normal Distribution and the 68-95-99.7 Rule http://www.mathsnet.net/js/balldrop.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.
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%