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Chapter 6
The Normal Distribution
Chapter 6
The Normal Distribution
 The Normal Distribution
 The Standard Normal Distribution
 Applications of Normal Distributions
 Sampling Distributions
 The Central Limit Theorem (CLT)
Probability Distributions of
Continuous random
variables
X
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
P(X)
0.000075339
0.000828733
0.004419907
0.015224123
0.038060308
0.073583262
0.114462852
0.147166524
0.159430401
0.147620742
0.118096593
0.082309747
0.050300401
0.027084831
0.012897539
0.005445627
0.002042110
0.000680703
0.000201690
0.000053076
0.000012384
0.000002556
0.000000465
0.000000074
0.000000010
0.000000001
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
Recall
P( X  11)  0.08231
Guessing 33 Answers on a Test
Data
Sample size
Probability of success
33
0.25
Statistics
Mean
Variance
Standard deviation
8.25
6.1875
2.487469
Porbality Distribution of Total Number of Right Answers
Probability
Distribution of Total Number of Right Answers
0.180000000
0.160000000
0.140000000
P( X  11)  .0823 1  area
0.120000000
P(X)
0.100000000
0.080000000
0.060000000
0.040000000
0.020000000
0.000000000
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
Number of Successes
X
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
P(X)
0.000075339
0.000828733
0.004419907
0.015224123
0.038060308
0.073583262
0.114462852
0.147166524
0.159430401
0.147620742
0.118096593
0.082309747
0.050300401
0.027084831
0.012897539
0.005445627
0.002042110
0.000680703
0.000201690
0.000053076
0.000012384
0.000002556
0.000000465
0.000000074
0.000000010
0.000000001
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
Recall
P(8  X  11)  area four rectangles
 0.50746
Guessing 33 Answers on a Test
Data
Sample size
Probability of success
33
0.25
Statistics
Mean
Variance
Standard deviation
8.25
6.1875
2.487469
Porbality Distribution of Total Number of Right Answers
Probability
Distribution of Total Number of Right Answers
0.180000000
P(8  X  11)  area four rectangles
 0.50746
0.160000000
0.140000000
0.120000000
P(X)
0.100000000
0.080000000
0.060000000
0.040000000
0.020000000
0.000000000
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
Number of Successes
Continuous Probability
Distribution
A continuous random variable has an infinite number of
possible values that can be represented by an interval on
the number line. For example
Average Grade in Math 1111
0
20
40
60
80
90
100
Average Grade can
be any number
between 0 and 100.
The probability distribution of a continuous random
variable is called a continuous probability distribution.
Relative Frequency Histogram
Average Final Grade for 60 students
Relative Frequency Histogram
Average Final Grade for 450 students
Relative Frequency Histogram
Average Final Grade for 1260 students
Relative Frequency Histogram
Average Final Grade for 1961 students
Relative Frequency Histogram
Average Final Grade for 1961 students
Continuous Probability
Distribution
Average Final Grade for all students
Data
World
Theory
World
Density Curve or Density Function
Mathematical
Model
f ( x)

   x f ( x)dx

   ( x   )2 f ( x)dx
2
It is mathematically easier to work with such a curve
instead of a histogram
Density Curve or Density Function
A density curve is a curve that
1) is always on or above the horizontal axis.
2) has area exactly equal to 1 underneath it.
A density curve describes the overall pattern of a
continuous distribution.
The area under the curve and above any range of
values is the relative frequency of all observations
that fall in that range (see picture in next slide).
Density Curve or Density Function
P(a  X  b) is
equal to this area
a
b
X
Shaded area equals percentage of observations
that fall between a and b
Examples of Density Functions
Normal Distributions
and the Standard
Distribution
Properties of Normal Distributions
The most important probability distribution in statistics
is the normal distribution.
Normal curve
x
A normal distribution is a continuous probability
distribution for a random variable, X. The graph of a
normal distribution is called the normal curve.
Properties of Normal Distributions
1) The mean, median, and mode are equal.
2) The normal curve is bell-shaped and symmetric about
the mean.
3) The total area under the curve is equal to one.
4) The normal curve approaches, but never touches the
x-axis as it extends farther and farther away from the
mean.
5) Between μ  σ and μ + σ the graph curves downward.
The graph curves upward to the left of μ  σ and to
the right of μ + σ.
Properties of Normal Distributions
Inflection points
Total area = 1
μ  3σ
μ  2σ
μσ
μ
μ+σ
μ + 2σ
μ + 3σ
x
If X is a continuous random variable having a normal distribution
with mean μ and standard deviation σ, you can graph a normal
curve with the equation
1
-(x - μ )2 2σ 2
y=
e
σ 2
Example of Normal Distribution
The following table displays a frequency and a relativefrequency distribution for the heights of the 3264 female
students who attend a Midwestern college.
The figure displays a relative-frequency histogram for the
same data. Included is a shaded area between 67” and 68”.
This area is P(67  Height  68) = 0.0735
We now included a normal curve that approximates the
overall shape of the distribution. The area under the curve.
between 67 and 68 is P(67  Height  68) = 0.0735
Means and Standard Deviations
A normal distribution can have any mean and any positive
standard deviation.
Inflection
points
1 2 3 4 5 6
Inflection
points
The mean gives
the location of the
line of symmetry.
x
1 2
3 4 5
6 7
8
9 10 11
Mean: μ = 3.5
Mean: μ = 6
Standard
deviation: σ  1.3
Standard
deviation: σ  1.9
The standard deviation describes the spread of the data.
x
Means and Standard Deviations
Example:
1. Which curve has the greater mean?
2. Which curve has the greater standard deviation?
B
A
1
3
5
7
9
11
13
x
The line of symmetry of curve A occurs at x = 5. The line of symmetry
of curve B occurs at x = 9. Curve B has the greater mean.
Curve B is more spread out than curve A, so curve B has the greater
standard deviation.
The Standard Normal Distribution
The standard normal distribution is a normal distribution
with a mean of 0 and a standard deviation of 1.
The horizontal scale
corresponds to z-scores.
3
2
1
0
1
2
3
z
If a variable X has a normal distribution with mean μ and
standard deviation σ, then the standardized variable
X -μ
Z =
σ
has the standard normal distribution
Properties of the Standard
Normal Distribution
1. The cumulative area is close to 0 for z-scores close to z = 3.49.
2. The cumulative area increases as the z-scores increase.
3. The cumulative area for z = 0 is 0.5000.
4. The cumulative area is close to 1 for z-scores close to z = 3.49
Area is close to 1.
Area is close to 0.
z = 3.49
3
z
2
1
0
z=0
1
Area is 0.5000.
2
3
z = 3.49
Areas Under the Standard
Normal Curve
The Standard Normal Table
P(Z  z) =
The Standard Normal Table
Example - Find P(Z  1.40)
P(Z  1.40)=0.9192
Finding the Area to the Left
of a Specified z-Score
Finding P(Z  1.23), that is, finding the area under
the standard normal curve to the left of 1.23
Always draw
the curve!
Finding the Area to the Left
of a Specified z-Score
Finding P(Z  1.23), that is, finding the area under
the standard normal curve to the left of 1.23
Finding the Area to the Right
of a Specified z-Score
Finding P(Z ≥ 0.76), that is, finding the area under
the standard normal curve to the right of 0.76
Always draw
the curve!
Finding the Area to the Right
of a Specified z-Score
Finding P(Z ≥ 0.76), that is, finding the area under
the standard normal curve to the right of 0.76
Find the Area Between Two
Specified z-Scores
Finding P(-0.68  Z  1.82), that is, finding the
area under the standard normal curve that lies
between –0.68 and 1.82
Always draw
the curve!
Find the Area Between Two
Specified z-Scores
Finding P(-0.68  Z  1.82), that is, finding the
area under the standard normal curve that lies
between –0.68 and 1.82
Summary
Using Table II to find the area under the standard
normal curve that lies (a) to the left of a specified
z-score, (b) to the right of a specified z-score, (c)
between two specified z-scores
Find the z-Score Having a Specified
Area to Its Left
Finding the z-score having area 0.04 to its left,
that is, finding z, so that P(Z  z)=0.04
Always draw
the curve!
Second decimal place for z-score
z score
Inside the table, look for the best
approximation to the given area
Find the z-Score Having a Specified
Area to Its Left
Finding the z-score having area 0.04 to its left,
that is, finding z, so that P(Z  z)=0.04
Useful Notation
Example-Find z0.025
Example-Find z0.025
Finding the z-Scores for a Specified
Area
Find the two z-scores that divide the area under
the standard normal curve into a middle 0.95 area
and two outside 0.025 areas, as depicted in the
figure.
Finding the z-Scores for a Specified
Area
Find the two z-scores that divide the area under
the standard normal curve into a middle 0.95 area
and two outside 0.025 areas, as depicted in the
figure.
Normal distribution
calculations
Three nonstandard normal
distributions
Standardizing three nonstandard
normal distributions
Normal Distribution Calculations
The process of Standardizing preserves the areas in
the sense indicated in above.
Normal Distribution Calculations
and areas are probabilities, that is
Normal Distribution Calculations
P ( a  X  b)  P ( a    X    b   )
a X  b 
 P








b 
a
 P
Z





Normal Distribution Calculations
P ( a  X  b)  P ( a    X    b   )
a X  b 
 P








b 
a
 P
Z





Normal Distribution Calculations
P ( a  X  b)  P ( a    X    b   )
a X  b 
 P








b 
a
 P
Z





Normal Distribution Calculations
P ( a  X  b)  P ( a    X    b   )
a X  b 
 P








b 
a
 P
Z





Normal Distribution Calculations
b 
a
P ( a  X  b)  P 
Z 

 
 
z-score for a
z-score for b
Percentage of people having IQs
between 115 and 140
Percentage of people having IQs
between 115 and 140
Percentage of people having IQs
between 115 and 140
P(115  X  140)  P  0.94  Z  2.50  0.1674
Conclusion: The percentage of people having IQs
between 115 and 140 is 16.74%
More Examples on
Finding Probabilities
Probability and Normal Distributions
Example:
The average on a statistics test was 78 with a standard deviation of
8. If the test scores are normally distributed, find the probability
that a student receives a test score less than 90.
μ = 78
σ=8
z  x σ- μ = 90-78
 1.5
8
P(X < 90)
μ =78
90
μ =0
?
1.5
P(X < 90) = P(Z < 1.5) = 0.9332
x
z
The probability that a student
receives a test score less than
90 is 0.9332.
Probability and Normal Distributions
Example:
The average on a statistics test was 78 with a standard deviation of
8. If the test scores are normally distributed, find the probability
that a student receives a test score greater than than 85.
z = x σ- μ = 85-78
 0.88
8
μ = 78
σ=8
P(X > 85)
μ =78 85
μ =0 0.88
?
x
z
The probability that a student
receives a test score greater
than 85 is 0.1894.
P(X > 85) = P(Z > 0.88) = 1  P(Z < 0.88) = 1  0.8106 = 0.1894
Probability and Normal Distributions
Example:
The average on a statistics test was 78 with a standard deviation of
8. If the test scores are normally distributed, find the probability
that a student receives a test score between 60 and 80.
z1 = x σ- μ = 608- 78  2.25
P(60 < X < 80)
z 2  x σ- μ = 808- 78  0.25
μ = 78
σ=8
60
μ =78 80
2.25
μ =0 0.25
?
?
x
z
The probability that a student
receives a test score between
60 and 80 is 0.5865.
P(60 < X < 80) = P(2.25 < Z < 0.25) = P(Z < 0.25)  P(Z < 2.25)
= 0.5987  0.0122 = 0.5865
Normal Distributions:
Finding Values
Finding a z-Score Given a Percentile
Example: Find the z-score that corresponds to P75.
That is P75 = 0.67
Area = 0.75
μ =0
?
0.67
z
The z-score that corresponds to P75 is the same z-score
that corresponds to an area of 0.75.
Transforming a z-score to an x-score
To transform a standard z-score to a data value, x, in a
given population, use the formula
x  μ + z
Example:
The monthly electric bills in a city are normally distributed
with a mean of $120 and a standard deviation of $16. Find
the x-value corresponding to a z-score of 1.60.
x  μ + z
= 120 +1.60(16)
= 145.6
We can conclude that an electric bill of $145.60 is 1.6 standard
deviations above the mean.
Finding a Specific Data Value
Example: The weights of bags of chips for a vending machine are
normally distributed with a mean of 1.25 ounces and a standard
deviation of 0.1 ounce. Bags that have weights in the lower 8%
are too light and will not work in the machine. What is the least a
bag of chips can weigh and still work in the machine?
P(Z < ?) = 0.08
8%
P(Z < 1.41) = 0.08
?
1.41
z
0
x
? 1.25
1.11
x  μ + z
 1.25  (1.41)0.1
 1.11
The least a bag can weigh and still work in the machine is 1.11 ounces.