Download 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%