Download Standardizing and the Standard Normal Curve

Standardizing and the
Standard Normal Curve
AP Statistics
The Standardized Value (the Z
score)
• If x is an observation from a distribution
that has mean and standard deviation ,
the standardized value of x is
z 
x

Example 2.4 p. 94
The heights of young women are approximately normal
with mean

=64.5 inches
• A woman 68 inches
tall for example has a
standardized height of
68  64.5
z
 1.4
2.5
• this means that the
woman is 1.4
standard deviations
from the mean.
 =2.5 inches.
The heights of young women are
approximately normal with mean
 =64.5 inches =2.5 inches.
• Now, another woman is 5
feet (60 inches tall). The
standardized height of
this girl is
60  64.5
z
 1.8
2.5
• this means that the
woman is 1.8 standard
deviations less than the
mean height.
The Standard Normal Distribution
• The Standard Normal Distribution is the normal
distribution N(0,1) with mean 0 and standard
deviation 1.
• If a variable x has any normal distribution
N( ,  ) with mean  and standard deviation
then the standardized variable
,
z 
x  

• has the standard normal distribution.
• Table A is a table for
the areas under the
standard normal
curve. The table
entry for each value
of z is the area under
the curve to the left
of z
Finding Normal Proportions
• Step 1: State the problem in terms of the observed
value x. Draw a picture of the distribution and shade the
area of interest under the curve.
Step 2: Standardize x to restate the problem in terms of
a standard normal variable z. Draw a picture to show
the area of interest under the standard normal curve.
Step 3: Find the required area under the standard
normal curve, using table A and the fact that the total
area under the curve is 1
Step 4: Writ your conclusion in the context of the
problem
The Dummy Rules: Dummy Rule 1
• If we want the probability that x is less
than a number, then it is written
# 
P ( x  #)  P ( z 
)

• Look up your z score on table A at this
point
• Example say that the distribution of
the weights of dogs is N(25, 10) and
we want the probability that a dog
weighs less than 15 pounds. then…
15  25
P( x  15)  P( z 
)  P( z  1)
10
• Now look up -1 on table A and
you will see that the probability is
• 0.1587
The Dummy Rules: Dummy Rule 2
• If we want the probability that x is
greater than a number, then it is
written
P( x  #)  1  P( x  #)  1  P( z 
# 
)

• Look up your z score on table A at this
•
point
Example say that the distribution of
the weights of dogs is N(25, 10) and
we want the probability that a dog
weighs more than 30. then…
P( x  30)  1  P( x  30)  1  P( z 
30  25
)  1  P( z  0.5)
10
• Now look up 0.5 on table A and you
•
will see that the probability is
0.6915
The Dummy Rules: Dummy rule 3
• If we want the probability that x is in
between two numbers, then it is
written (Let our numbers be
represented by A and B)
P( A  x  B)  P( x  B)  P( x  A)  P( z 
B
A 
)  P( z 
)


• Look up both your z score on table A
•
and subtract
Example say that the distribution of
the weights of dogs is N(25, 10) and
we want the probability that a dog
weighs less than 30 but more than 2.
then…
P(2  x  30)  P( x  30)  P( x  2)  P( z 
• Now look up 0.5 and -2.3 on table A
•
•
30  25
2  25
)  P( z 
)  P( z  0.5)  P( z  2.3)
10
10
then subtract
0.6915 – 0.0107 = 0.6808
So the probability that a dog weighs
less than 30 but more than 2 is 0.6808
Example 2.9 Working with an interval
The distribution of blood cholesterol levels is roughly
normal for 14-year old boys with mean
=170 and
=30


• What percent of 14-year old boys have
blood cholesterol between 170 and 240?
• Step 1: State the problem: We want the
proportion of boys with..
170  x  240
Continued
Example 2.9 Working with an interval
The distribution of blood cholesterol levels is roughly
normal for 14-year old boys with mean
=170 and
=30


• Step 2: Standardize
and draw a picture.
P (170  x  240)
P ( x  240)  P ( x  170)
240  170
170  170
P( z 
)  P( z 
)
30
30
P ( z  2.33)  P ( z  0)
Continued
Example 2.9 Working with an interval
The distribution of blood cholesterol levels is roughly
normal for 14-year old boys with mean
=170 and
=30

• Step 3: Use the
table: Look up 2.33
and 0 on the table:
P( z  2.33)  P( z  0)
0.9901  0.5000
0.4901

• Step 4: State your
•
conclusion in context.
About 49% of boys
have cholesterol
levels between 170
and 240 mg/dl.