Download Chapter 7 The Normal Probability Distribution

Chapter 20/21
The Normal Probability
Distribution
Properties of the Normal Distribution
Standard Normal Distribution
Applications of Normal Distribution
1
Normal Distribution
If a continuous random variable is normally
distributed or follows a normal probability
distribution, then a relative frequency histogram of the
random variable has the shape of a normal curve (bellshaped and symmetric).
Many real world variables have such a distribution: for
example: SAT scores, Height, Weight, Blood pressure,
Gas price.
Can you name a few more?
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EXAMPLE
A Normal Random Variable
The following data represent the heights (in inches) of a random
sample of 50 two-year old baby boys.
(a) Create a relative frequency distribution with the lower class
limit of the first class equal to 31.5 and a class width of 1.
(b) Draw a histogram of the data.
(c ) Do you think that the variable “height of 2-year old baby boys”
is normally distributed?
36.0
33.4
37.9
35.6
36.0
34.8
36.2
37.4
39.3
33.0
36.0
35.7
34.8
38.2
34.0
36.8
35.7
38.9
36.0
31.5
36.9
33.5
35.7
37.2
34.6
37.7
35.1
35.0
38.3
39.3
38.4
36.9
37.0
35.1
33.6
35.4
34.0
33.2
35.2
39.8
36.8
34.4
36.1
34.4
37.0
34.7
35.7
35.2
36.7
37.2
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A Normal Curve is
imposed on the histogram.
This is the evidence from
sample data that the
Height is normally
distributed. The smooth
curve represents the
pattern of the distribution
of height of all two-year
old baby boys.
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The probability that the
height is between 34.5” to
35.5” is the area
underneath the curve with
the interval of 34.5 to 35.5.
As the graph suggests that
the area would be very
close to the area the
rectangle represents.
We have a better way to
determine such a
probability when the
distribution follows a
normal curve.
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Normal distribution is a bell-shaped (mounded-shaped) curve
as shown below:
NOTE: The curve is a mathematical function that has the form:
1
[( x  m )2 ]/ 2
f ( x) 
e
2
X follows a normal distribution
with mean m and standard
deviation  is denoted as:
P(X
< m ) = P(X
NOTATION:
X ~ >Nm)
(m=, ½
)
A MUST KNOW Notation
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7. The mean and median are the same for a normal distribution.
The above properties are the MUST KNOWN properties.
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Properties of a Normal Distribution X ~ N (m, )
P( X> (m + a))
P( X < (m - a) )
ma
m
m+a
X
NOTE:
P(X > m) = P(X < m ) = .5
P( X> (m + a)) = P( X< ( m - a))
P(( m - a) < X< (m + a)) = P(( m - a) < X< ( m + a))
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An example of Normal distribution
SAT scores follow a normal distribution with mean 1090 and
s.d. 180.
Notation: use X to represent SAT score. Based on the
notation, we use the notation: X ~ (1090, 180)
The following shows the distribution graph for SAT score <
one s.d. lower than mean and SAT > 1500:
P( X> (1500))
P( X < 910 )
910
P(X < 910) is probability of
SAT scores BELOW 910
1090
1500
X
P(X > 1500) is the probability of
SAT scores ABOVE 1500
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The Empirical Rule: 68-95-99.7 Rule
• 68% of the values fall within 1 standard
deviation of the mean.
• 95% of the values fall within 2 standard
deviations of the mean.
• 99.7% of the values fall within 3 standard
deviations of the mean.
Example of the 68-95-99.7 Rule
In the 2010 winter Olympics men’s slalom, Li
Lei’s time was 120.86 sec, about 1 standard
deviation slower than the mean. Given the Normal
model, how many of the 48 skiers were slower?
Example of the 68-95-99.7 Rule
• About 68% are within 1 standard deviation of
the mean.
• 100% – 68% = 32% are outside.
• “Slower” is just the left side.
• 32% / 2 = 16% are slower.
• 16% of 48 is 7.7.
• About 7 are slower than Li Lei.
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Recall: Empirical Rule describes some properties
of the Normal distribution
This Chapter will extend the Empirical rule to allow us to
find any probability for a normal distribution.
13
How to compute and Interpret the Area Under a
Normal Curve
[Similar Exam Questions]
Example:
Every year, universities recruit students using their SAT
scores. Based on the previous information, we know that
SAT scores follows a normal curve with the mean 1000 and
standard deviation 180. In the past, MSU admits students
with SAT 1090 or higher.
Q1: What is the percent of high school students who can
receive MSU admission?
Q2: If MSU decides to raise the SAT admission limit to
only admit the top 20% of high school graduates.
What should be the new SAT admission limit?
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Solution to Q1
Q1: Call X to be the SAT scores. Then, X follows a normal distribution with
mean m  1000 and standard deviation   180.
P( X>1090)
REMINDEZR: X follows a normal
distribution with mean m and standard
deviation  is denoted as:
NOTATION: X ~ N (m , )
For this example, X ~ N(1000, 180)
X, SAT score
1000 1090
Z=(x-1000)/180
(1000-1000)/180 = 0
0.5 = (1090-1000)/180
What is Z? Z is the standard Normal distribution with m = 0 and  = 1. This
is the same as we used in Chapter Three, the Z-score.
P(X>1090) = ?
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Standard Normal Distribution
NOTATION: Z ~ N(0,1)
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7. The mean = median = 0 for the Z distribution.
The above properties for Z are MUST KNOWN properties.
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How to we find the area underneath the
Standard Normal Curve?
f ( z) 
NOTE: In the
Empirical Rule,
we had :
.34, .135, .025.
Here we have
more accurate
probabilities.
P(-1 < Z <0)
1
 z2 / 2
e
2
P(0 < Z <1)
P(1 < Z < 2)
P(-2 < Z < -1) =
P(Z > 2)
P( Z < -2) =
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An example of Standardized Normal
The standardized normal is a normal distribution with
mean 0 and s.d. 1. Notation is Z~ N(0,1). The
following shows some examples of probabilities
under the Z-curve:
P( 0 < Z < 1.96)
P(-1.25 < Z< 0)
P( X < -1.25)
-1.25
P( Z > 1.96)
0
1.96
X
Q: How do we determine these probabilities?
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Solving Problems Involving Normal Distributions:
Finding the actual X values
Every year, universities recruit students using their SAT scores. Based on the
previous information, we know that SAT scores follows a normal curve with
the mean 1000 and standard deviation 180. In the past, MSU admits
students with SAT 1090 or higher.
Q1: If MSU decides to raise the SAT admission limit to only admit the top
20% of high school graduates. What should be the new SAT admission
limit?
Q2: U of Michigan accept only the top 5% of their SAT scores. What is their
minimum SAT score?
Solution:
Q1: Let X be the SAT score. The question asks to find an SAT score, call it xo,
so that
P(SAT scores > xo) = .2, or P( X > xo ) = .2
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Using the TI-83 to Find a Normal Percentage
Always draw a
• The TI-83 provides a function named normalcdf
picture!
nd
– Press 2 , DISTR (found above VARS)
– Scroll to normalcdf ( and press ENTER, or press 2.
• If z has a standard normal distribution:
– Percent(a < z < b) = normalcdf ( a , b )
– Example: to find P( -1.2 < z < .8 ),
press 2nd, DISTR, 2, then -1.2 , .8 )
– Note that the comma between -1.2 and .8 must be
entered
– Read .6731
?
-1.2
.8
?
• To find Percent( z < a ), enter normalcdf ( -5 , a )
1.96
– Example: normalcdf( -5 , 1.96 ) gives .9750
?
• To find Percent( z > a ), enter normalcdf ( a , 5-1.645
)
– Example: normalcdf( -1.645 , 5 ) gives .9500
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Using the TI-83/84 for Normal Percentages
Without Computing z-Scores
We can let the TI find its own z-scores:
– Find Percent(90 < x < 105) if x follows the normal model with mean 100
and standard deviation 15:
• Percent(90 < x < 105) = normalcdf( 90 , 105 , 100 , 15)
= .378
Notice that this is a time-saver for this type of problem, but that you may still
need to be able to compute z-scores for other types of problems!
x1
x2
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Suppose We’re Given a normal
Percentage and Need A z-score?
• IQ scores are distributed normally with a
mean of 100 and a standard deviation of
15. What score do you need to capture
the bottom 2%?
– That is, we must find a so that Percent(x < a) = 2%
when x has a normal distribution with a mean of 100
and a standard deviation of 15.
– With the TI 83/84:
a = invNorm( .02, 100 , 15) = 69.2
x
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Standardizing a Normal Random Variable
Suppose that the random variable X is normally
distributed with mean μ and standard deviation σ.
Then the random variable
Z
Xm

is normally distributed with mean μ = 0 and
standard deviation σ = 1.The random variable Z
is said to have the standard normal distribution.
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Standard Normal Curve
7-25
The table gives the area under the standard
normal curve for values to the left of a
specified Z-score, z, as shown in the figure.
7-26
IQ scores can be modeled by a normal
distribution with μ = 100 and σ = 15.
An individual whose IQ score is 120, is 1.33
standard deviations above the mean.
z
7-27
xm

120  100

 1.33
15
The area under the standard normal curve to
the left of z = 1.33 is 0.9082.
7-28
Use the Complement Rule to find the area to
the right of z = 1.33.
7-29
Areas Under the Standard Normal Curve
7-30
EXAMPLE
Finding the Area Under the Standard Normal Curve
Find the area under the standard normal curve to the left
of z = –0.38.
Area to the left of z = –0.38 is 0.3520.
7-31
Area under the normal curve to the right
of zo = 1 – Area to the left of zo
7-32
EXAMPLE
Finding the Area Under the Standard Normal Curve
Find the area under the standard normal curve to the
right of z = 1.25.
Area right of 1.25 = 1 – area left of 1.25
= 1 – 0.8944 = 0.1056
7-33
EXAMPLE
Finding the Area Under the Standard Normal Curve
Find the area under the standard normal curve between
z = –1.02 and z = 2.94.
Area between –1.02 and 2.94
= (Area left of z = 2.94) – (area left of z = –1.02)
= 0.9984 – 0.1539
= 0.8445
7-34
Problem: Find the area to the left of x.
Approach: Shade the area to the left of x.
Solution:
•
Convert the value of x to a z-score. Use
Table V to find the row and column that
correspond to z. The area to the left of x is the
value where the row and column intersect.
•
Use technology to find the area.
7-35
Problem: Find the area to the right of x.
Approach: Shade the area to the right of x.
Solution:
•
Convert the value of x to a z-score. Use
Table V to find the area to the left of z (also is
the area to the left of x). The area to the right of
z (also x) is 1 minus the area to the left of z.
•
Use technology to find the area.
7-36
Problem: Find the area between x1 and x2.
Approach: Shade the area between x1 and x2.
Solution:
•
Convert the values of x to a z-scores. Use
Table V to find the area to the left of z1 and to
the left of z2. The area between z1 and z2 is (area
to the left of z2) – (area to the left of z1).
•
Use technology to find the area.
7-37
Procedure for Finding the Value of a Normal
Random Variable
Step 1:
Draw a normal curve and shade the
area corresponding to the proportion, probability,
or percentile.
Step 2:
Use Table V to find the z-score that
corresponds to the shaded area.
Step 3:
Obtain the normal value from the
formula x = μ + zσ.
5-38
EXAMPLE
Finding the Value of a Normal Random
Variable
The combined (verbal + quantitative reasoning)
score on the GRE is normally distributed with mean
1049 and standard deviation 189.
(Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.)
What is the score of a student whose percentile
rank is at the 85th percentile?
7-39
EXAMPLE
Finding the Value of a Normal Random
Variable
The z-score that corresponds to the 85th
percentile is the z-score such that the area under
the standard normal curve to the left is 0.85. This
z-score is 1.04.
x = µ + zσ
= 1049 + 1.04(189)
= 1246
Interpretation: A person who scores 1246 on the GRE
would rank in the 85th percentile.
7-40
EXAMPLE
Finding the Value of a Normal Random
Variable
It is known that the length of a certain steel rod is
normally distributed with a mean of 100 cm and a
standard deviation of 0.45 cm. Suppose the
manufacturer wants to accept 90% of all rods
manufactured. Determine the length of rods that make
up the middle 90% of all steel rods manufactured.
7-41
EXAMPLE
Area = 0.05
Finding the Value of a Normal Random
Variable
Area = 0.05
z1 = –1.645 and z2 = 1.645
x1 = µ + z1σ
= 100 + (–1.645)(0.45)
= 99.26 cm
x2 = µ + z2σ
= 100 + (1.645)(0.45)
= 100.74 cm
Interpretation: The length of steel rods that make
up the middle 90% of all steel rods manufactured
would have lengths between 99.26 cm and 100.74
7-42
cm.
The Notation: Za
Za is a MUST KNOW Notation. This is a Z-value that probability
that Z > Za is a
Za is always Positive, since it is
a is the upper tail probability
and is less than 0.5
For the lower tail situation, we
use - Za
Area = a
-za
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Examples of finding Za or Za
Q1: Find z0.25
NOTE: This is the same as finding z0 so that P(Z > z0) = .25. This z0 is
named as z0.25
Q2: Find Z.05
Q3: Find –Z.01
NOTE: This is the same as finding z0 so that P(Z < z0) = .01. This z0 is
negative and it is the opposite of z.01. So, we call it -z.01
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