Stats ppt lecture

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%