
The Normal Distribution

The Standard Normal Distribution

Applications of Normal Distributions

Sampling Distributions

The Central Limit Theorem (CLT)

X
26
27
28
29
22
23
24
25
30
31
32
33
15
16
17
18
19
20
21
11
12
13
14
9
10
7
8
5
6
3
4
0
1
2
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
(

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

0.180000000
0.160000000
0.140000000
0.120000000
0.100000000
(

11)
   area
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
26
27
28
29
22
23
24
25
30
31
32
33
15
16
17
18
19
20
21
11
12
13
14
9
10
7
8
5
6
3
4
0
1
2
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

0.180000000
0.160000000
0.140000000
0.120000000
0.100000000
P (8

X

11)

 area four rectangles
0.50746
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


2








( x
 
) 2 f x dx
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
X a b
Shaded area equals percentage of observations that fall between a and b

Examples of Density Functions

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  3 σ μ  2 σ μ  σ μ μ + σ μ + 2 σ μ + 3 σ 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 y =
σ
1
2  e
-
( x μ ) 2 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
Inflection points
The mean gives the location of the line of symmetry.
x
1 2 3 4 5 6
Mean: μ = 3.5
Standard deviation: σ  1.3
1 2 3 4 5 6 7 8 9 10 11
Mean: μ = 6
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?
A
B x
1 3 5 7 9 11 13
The line of symmetry of curve of curve
A occurs at x = 5. The line of symmetry
B occurs at x = 9. Curve B has the greater mean.
Curve B is more spread out than curve standard deviation.
A , so curve B has the greater

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.
z
 3  2  1 0 1 2 3
If a variable X has a normal distribution with mean μ and standard deviation σ, then the standardized variable
Z =
X μ
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 0.
z =  3.49
 3  2  1 0 1 z = 0
Area is 0.5000.
2 3
Area is close to 1.
z z = 3.49

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 z
0.025

Example-Find z
0.025

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.

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
(

X
 b )

P a

X
 b


P 

 a


P 

 a






)
X

 
 b





 b







Normal Distribution Calculations
(

X
 b )

P a

X
 b


P 

 a


P 

 a






)
X

 
 b





 b







Normal Distribution Calculations
(

X
 b )

P a

X
 b


P 

 a


P 

 a






)
X

 
 b





 b







Normal Distribution Calculations
(

X
 b )

P a

X
 b


P 

 a


P 

 a






)
X

 
 b





 b







Normal Distribution Calculations
(

X
 b )

P a


 z-score for a b


 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
2.50
 
0.1674
Conclusion: The percentage of people having IQs between 115 and 140 is 16.74%

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
P ( X < 90) z  x μ
σ
=
90-78
8

1.5
x
μ =78 90
μ =0
P ( X < 90) = P ( Z < 1.5) = 0.9332 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.
μ = 78
σ = 8
P ( X > 85) z = x - μ
σ
=
85-78
8
 0.88
μ =78 85 x
The probability that a student receives a test score greater than 85 is 0.1894.
z
μ =0
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.
P (60 < X < 80)
μ = 78
σ = 8 z z
1
2
=
 x μ
σ
x μ
σ
-
=
=
-
8
-
8
 
2.25

0.25
60
 ?
μ =78 80
μ =0 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

Finding a z-Score Given a Percentile
Example : Find the z -score that corresponds to P
75
.
That is P
75
= 0.67
Area = 0.75
z
μ =0
The z -score that corresponds to P
75 is the same zscore 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
 ?
0
1.25
z x x  μ+ z 


1.11
The least a bag can weigh and still work in the machine is 1.11 ounces.