Transcript Chapter 3 Power Point Slides

The Normal distributions
BPS chapter 3
© 2006 W.H. Freeman and Company
Objectives (BPS 3)
The Normal distributions

Density curves

Normal distributions

The 68-95-99.7 rule

The standard Normal distribution

Using the calculator to find Normal proportions

Finding a value given a proportion
Density curves
A density curve is a mathematical model of a distribution.
It is always on or above the horizontal axis.
The total area under the curve, by definition, is equal to 1, or 100%.
The area under the curve for a range of values is the proportion of all
observations for that range.
From: Relative
area enclosed by
bars of histogram
To: Relative area
under density
curve.
Density curves come in any
imaginable shape.
Some are well-known
mathematically and others aren’t.
Normal distributions
Normal—or Gaussian—distributions are a family of symmetrical, bellshaped density curves defined by a mean m (mu) and a standard
deviation s (sigma): N (m, s).
1
f ( x) 
e
2
1  xm 
 

2 s 
2
x
e = 2.71828… The base of the natural logarithm
π = pi = 3.14159…
x
The normal density curve!
Shape? Center? Spread?
Area Under Curve?
Assume Same Scale on
Horizontal and Vertical
(not drawn) Axes.
How does the standard
deviation affect the
shape of the normal
density curve?
How does the magnitude
of the standard deviation
affect a density curve?
Drawing Density Curves
Suppose the distribution of heights, X, of young women aged 18 to 24
is approximately normal with mean mu=64.5 inches and standard
deviation sigma=2.5 inches (i.e. X~N(64.5,2.5)). Lets draw the
density curve for X!
A family of density curves
Here the means are the same (m =
15) while the standard deviations
are different (s = 2, 4, and 6).
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
Here the means are different
(m = 10, 15, and 20) while the
standard deviations are the same
(s = 3).
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
All Normal curves N (m, s) share the same
properties (page 71)

About 68% of all observations are
within 1 standard deviation
Inflection point
(s) of the mean (m).

About 95% of all observations are
within 2 s of the mean m.

Almost all (99.7%) observations
are within 3 s of the mean.
This
is called the 68-95-99.7 Rule
(aka Empirical Rule)
mean µ = 64.5
Let’s
work Problem 3.6 on Page 74
standard deviation s = 2.5
N(µ, s) = N(64.5, 2.5)
Reminder: µ (mu) is the mean of the idealized curve, while x is the mean of a sample.
σ (sigma) is the standard deviation of the idealized curve, while s is the s.d. of a sample.
Example: Women heights
N(µ, s) =
N(64.5, 2.5)
Women’s heights follow the N(64.5″,2.5″)
distribution. Using the empirical rule,
Area= ???
determine what percent of women are
Area = ???
shorter than 67 inches tall (that’s 5′7″)?
mean µ = 64.5"
standard deviation s = 2.5"
x (height) = 67"
m = 64.5″ x = 67″
z =0
z =1
We calculate z, the standardized value of x:
z
(x  m)
s
, z
(67  64.5) 2.5

 1  1 stand. dev. from mean
2.5
2.5
Because of the 68-95-99.7 rule, we can conclude that the percent of women
shorter than 67″ should be, approximately, .68 + half of (1 − .68) = .84, or
84%.
How many standard
deviations from the mean
height is the height of a
woman who is 68 inches?
Who is 58 inches?
Standardizing: calculating z-scores
A z-score measures the number of standard deviations that a data
value x is from the mean m.
z
(x  m )
s
Using the previous slide, find
the z-score of a woman
whose height is 70 inches,
whose height is 60 inches.
When x is larger than the mean, z is positive.
When x is smaller than the mean, z is negative.
Now let’s go backwards. Given a z-score find the corresponding data
value x.
Using the previous slide, find
the height of a woman
whose z-score is -1.25,
whose z-score is 2.4.
x  m  zs
The standard Normal distribution
Because all Normal distributions share the same properties, we can
standardize our data to transform any Normal curve N (m, s) into the
standard Normal curve N (0,1).
N(64.5, 2.5)
N(0,1)
=>
x
z
Standardized height (no units)
For each x we calculate a new value, z (called a z-score).
The z-score is the number of standard deviations that a
data value x is from the mean.

Using Our Calculators to Find the Percent of Women Shorter
than 67”
•TI-83 Calculator Command: Distr|normalcdf
•Syntax: normalcdf(left, right, mu, sigma) = area under normal
curve (with mean mu and standard deviation sigma) from left to
right
•mu defaults to 0, sigma defaults to 1
•Infinity is 1E99 (use the EE key), Minus Infinity is -1E99
Normalcdf(-1E99,67,64.5,2.5)
Can we compute using z-scores??
Normalcdf(-1E99,1,0,1)=Normalcdf(-1E99,1)
N(µ, s) =
N(64.5”, 2.5”)
Area ≈ 0.84
Question:
Area ≈ 0.16
What percent of women are taller than
67“??
m = 64.5” x = 67”
z=1
The National Collegiate Athletic Association (NCAA) requires Division I athletes to
score at least 820 on the combined math and verbal SAT exam to compete in their
first college year. The SAT scores of 2003 were approximately normal with mean
1026 and standard deviation 209.
What proportion of all students would be NCAA qualifiers (SAT ≥ 820)?
x  820
m  1026
s  209
(x  m )
z
s
(820  1026)
209
206
z
 0.99
209
z
Draw Picture
Normalcdf(820,1E99,1026,209)
Normalcdf(-.99,1E99)
Compute
State Answer
Area right of 820
=
=
Total area
1
−
−
Area left of 820
0.1611
≈ 84%
Answer: Approximately 84% of students who took the
SAT in 2003 scored at least 820.
Note: The actual data may contain students who scored exactly 820 on the SAT.
However, the proportion of scores exactly equal to 820 being 0 for a normal
distribution is a consequence of the idealized smoothing of density curves.
The NCAA defines a “partial qualifier” eligible to practice and receive an athletic
scholarship, but not to compete, as a combined SAT score of at least 720.
What proportion of all students who take the SAT would be partial qualifiers?
That is, what proportion have scores between 720 and 820?
x  720
m  1026
s  209
(x  m )
z
Draw Picture
Compute
State Answer
s
(720  1026)
209
306
z
 1.46
209
z
About 9% of all students who take the SAT have scores
between 720 and 820.
The cool thing about working with
normally distributed data is that
we can manipulate it and then find
answers to questions that involve
comparing seemingly noncomparable distributions.
We do this by “standardizing” the
data. All this involves is changing
N(0,1)
the scale so that the mean now
equals 0 and the standard deviation
equals 1. If you do this to different
distributions, it makes them
comparable.
(x  m )
Let’s work Problem 3.8 on Page 75
z
s
Finding a value given a proportion
When you know the proportion, but you don’t know the x-value that
represents the cut-off, you have the inverse problem.
1. State the problem and draw a picture.
2. Use the invNorm key to find the corresponding x or z.
Syntax:
invNorm(area,mu,sigma)=x
•
•
x is a value of the quantitative variable
area is the area to the left of x under normal curve with mean
mu and standard deviation sigma.
3. If finding z, unstandardize to transform z back to the original
x scale by using the formula:
x  m  zs
Example: Women’s heights
Women’s heights follow the N(64.5″,2.5″)
distribution. What is the 25th percentile for
women’s heights?
mean µ = 64.5"
standard deviation s = 2.5"
proportion = area under curve=0.25
Draw a picture!
Compute:
We will use our calculator to get both x and z.
If we have computed z we can convert back to x
State Answer:
The 25th percentile for women’s heights is 62.825”, or 5’ 2.82”.
One More Problem!

Let’s Work Problem 3.14 on Page 83