Download Relative Frequency Histogram Density Curve Continuous

Normal Distribution
Relative Frequency Histogram
Percent
Modeling Continuous Distribution
Test scores
10 20 30 40 50 60 70 80 90 100 110
Normal Distribution
1
2
Density Curve
Continuous Distribution
Density Function (Curve)
Percent
A smooth curve that fit
the distribution
1. Mathematical Formula
Shows Probability
Densities, f(x), for All
Values of x, &
Density
function, f (x)
Test scores
Probability Density
Function
(Value, Density)
f(x )
[ f(x) Is Not Probability ]
10 20 30 40 50 60 70 80 90 100 110
2. Property
Total Area Under Curve is 1.
Use a mathematical model to describe the variable.
x
Value
3
4
Continuous Random Variable
∫
P(c ≤ X ≤ d) =
d
c
Meaning of Area Under Curve
f (x ) dx
f(x)
Probability Is Area
Under Curve!
Example: What percentage of the
distribution is in between 72 and 86?
f(x)
c
X
d
P(X=72) = 0
5
72 86
X (Height)
6
Normal Distribution - 1
Normal Distribution
Uniform
Normal Distribution
f(X)
1. ‘Bell-Shaped’ &
Symmetrical
Skewed to right
2. Mean, Median,
Mode Are Equal
Symmetrical
X
3. Random Variable
Has Infinite Range
Mean
Median
Mode
−∞<x<∞
Skewed to left
7
8
Normal Probability
Density Function
Normal Distribution
σ
x − µ 22
1 −−⎛⎜⎝⎛⎜⎝ 212 ⎞⎟⎠⎞⎟⎠⎛⎜⎝⎛⎜⎝ xσ− µ ⎞⎟⎠⎞⎟⎠
f (x) =
e
σ 2π
1
• Normal Probability Density Curve
• Standard Normal Distribution
f (x ) =
µ
=
σ
=
π
=
x
=
9
µ
Density of Random Variable x
Mean of the Distribution
Standard Deviation of the Distribution
3.14159…; e = 2.71828…
Value of Random Variable (−∞ < x < ∞)
Notation: N (µ, σ) ⇒ Α normal distribution with
mean µ and standard deviation σ
10
Effect of Varying Parameters (µ & σ)
Example
N (72, 5) ⇒ Α normal distribution with mean 72
f(X)
and standard deviation 5.
Possible situations: Test scores, pulse rates, …
B
N (130, 24) ⇒ Α normal distribution with mean
130 and standard deviation 24.
Possible situations: Weight, Cholesterol levels, …
A
C
X
11
12
Normal Distribution - 2
Normal Distribution
Normal Distribution
Probability
Probability is
area under
curve!
P (c ≤ x ≤ d ) =
∫
d
c
Standard Normal Distribution
Standard Normal Distribution:
f ( x ) dx
A normal distribution with
mean = 0 and standard deviation = 1.
f(x)
f(x)
σ=1
Notation:
Z ~ N (µ = 0, σ = 1)
cc
xx
d
d
13
Z
0
Cap letter Z
14
Area under Standard Normal Curve
P (−1 < Z < 3)
Z
0
−1
z
0
3
Z
0
3
Z
P (Z > 3)
How to find the proportion of the are
under the standard normal curve below
z or say P ( Z < z ) = ?
Use Standard Normal Table!!!
15
16
Standard Normal Distribution
P(Z > 0.32) = Area above .32 = .374
Areas in the upper tail of the standard normal distribution
Z
.00
.01
.02
0.0 .500 .496
.492
0.1 .460 .456
.452
0.2 .421 .417
.413
0.3 .382 .378
.374
0 .32
17
18
Normal Distribution - 3
Normal Distribution
Standard Normal Distribution
Standard Normal Distribution
P(0 < Z < 0.32) = Area between 0 and .32 = .126
Areas in the upper tail of the standard normal distribution
Z
.00
.01
.02
0.0 .500 .496
.492
0.1 .460 .456
.452
0.2 .421 .417
.413
0.3 .382 .378
.374
P(Z< 0.32) = Area below .32 = .626
Areas in the upper tail of the standard normal distribution
Z
0 .32
Area = .5 - .374 = .126
.00
.01
.02
0.0 .500 .496
.492
0.1 .460 .456
.452
0.2 .421 .417
.413
0.3 .382 .378
.374
0 .32
Area = 1 - .374 = .626
19
P ( -1.00 < Z < 1.00 ) = _____
.682
.341
20
P ( -2.00 < Z < 2.00 ) = _____
.341
-1.00
0
1.00
-2.00
0
2.00
21
P ( -3.00 < Z < 3.00 ) = _____
22
P ( -1.40 < Z < 2.33 ) = ____
.909
.419
-3.00
0
.490
3.00
- 1.40
23
0
2.33
24
Normal Distribution - 4
Normal Distribution
Standardize the
Normal Distribution
Standardize the
Normal Distribution
Z=
Normal
Distribution
X −µ
σ
N ( 0 , 1)
N ( µ, σ)
Standard
Normal
Distribution
Normal
Distribution
Standardized
Normal Distribution
σ
Z
X
σ=1
µ
a
a−µ
b
0
σ
µ
µ= 0
X
Z
One table!
25
Standardizing Example
σ = 10
σ
b−µ ⎞
⎛a−µ
P ( a < X < b) = P⎜
<Z<
⎟
σ ⎠
⎝ σ
26
Standardizing Example
For a normal distribution that has
a mean = 5 and s.d. = 10, what
percentage of the distribution is
between 5 and 6.2?
Normal
Distribution
b−µ
µ= 5 6.2 X
Z=
Normal
Distribution
σ = 10
X −µ
=
6.2 − 5
10
σ
P(5 ≤ X ≤ 6.2)
= P(0 ≤ Z ≤ .12)
= .12
Standardized
Normal Distribution
σ=1
µ= 0 .12 Z
µ= 5 6.2 X
27
28
Obtaining
the Probability
Example
P(3.8 ≤ X ≤ 5)
Standardized Normal
Probability Table (Portion)
Z
.00
.01
.02
0.0 .500 .496
.492
0.1 .460 .456
.452
0.2 .421 .417
.413
0.3 .382 .378 .3745
Z=
σ=1
.452
Normal
Distribution
σ = 10
X −µ
σ
=
3 .8 − 5
= − .12
10
P(3.8 ≤ X ≤ 5)
=P(−
=P(−.12 ≤ Z ≤ 0)
Standardized
Normal Distribution
σ=1
.048
µ= 0 .12 Z
Area = .5 - .452 = .048
29
3.8 µ = 5
X
-.12 µ = 0
Area = .048
Z
30
Normal Distribution - 5
Normal Distribution
Example
P(X > 8)
Example
P(2.9
X − µ ≤ 2X
.9 −≤
5 7.1)
Z=
Normal
Distribution
σ = 10
=
= − .21
10
X − µ 7 .1 − 5
Z=
=
= .21
σ
10
σ
P(2.9 ≤ X ≤ 7.1)
=P(−.21 ≤ Z ≤ .21)
Z=
Standardized
Normal Distribution
σ=1
Normal
Distribution
σ = 10
X −µ
σ
=
8−5
= .30
10
P(X > 8)
=P(Z > .30)
Standardized
Normal Distribution
σ=1
.166
2.9 5 7.1 X
-.21 0 .21
Area = .083 + .083 = .166
.382
µ=5
Z
µ = 0 .30 Z
8 X
31
32
Area = .382
Example
P(X > 8)
More on Normal Distribution
•
Normal
Distribution
σ = 10
62%
38%
Value 8 is the 62nd percentile
µ=5
The work hours per week for residents in Ohio
has a normal distribution with µ = 42 hours
& σ = 9 hours. Find the percentage of Ohio
residents whose work hours are
A. between 42 & 60 hours.
P(42 ≤ X ≤ 60) =?
B. less than 20 hours.
P(X ≤ 20) = ?
8 X
33
34
P(42X −≤µX 42≤−60)
=?
42
Z=
Normal
Distribution
σ = 200
9
Z=
σ
X −µ
σ
=
P(X ≤ 20) = ?
=0
Z=
9
60 − 42
=2
=
9
P(42 ≤ Z ≤ 60)
= P(0 ≤ Z ≤ 2)
= .477 = 47.7%
Standardized
Normal Distribution
σ=1
Normal
Distribution
σ = 200
9
.477
X −µ
σ
=
20 − 42
= − 2.44
9
P(X ≤ 20)
= P(Z ≤ −2.44)
= 0.007 = 0.7%
Standardized
Normal Distribution
σ=1
0.007
2400 X
42 60
0 2 2.0
Z
35
20
42 2400 X
-2.44 0 2.0
Z
36
Normal Distribution - 6
Normal Distribution
Finding Z Values
for Known Probabilities
Standardized Normal
Probability Table (Portion)
What is z given
P(Z < zz)) = .80 ?
.20
z = .84
Upper Tail Area = 1 - .80
= .20
z = .84
.04
.05
.264 .261
.258
0.7 .233 .230
.227
0.8 .203 .200
.198
0.9 .176 .174
.171
37
…
Z
.80
0
Finding X Values
for Known Probabilities
…
Example: The weight of new born
infants is normally distributed with a
mean 7 lb and standard deviation of
1.2 lb. Find the 80th percentile.
Area to the left of 80th percentile in 0.200.
In the table there is a area value 0.200
corresponding to a z-score of .84.
80th percentile = 7 + .84 x 1.2 = 8.008 lb
38
Finding X Values
for Known Probabilities
Finding X Values
for Known Probabilities
Example: The Body Mass Index for a
particular population is normally
distributed with a mean 22 and
standard deviation of 4. Find the 80th
percentile.
.40
Area to the left of 80th percentile in 0.200.
In the table there is a area value 0.200
corresponding to a z-score of .84.
80th percentile = 22 + .84 x 4 = 25.36
Example: The Body Mass Index for a
particular population is normally
distributed with a mean 22 and
standard deviation of 4. Find the 40th
percentile.
Z 0 .25
39
Z = −.25
Stanine Score
(Standard Nine)
Finding X Values
for Known Probabilities
Example: The Body Mass Index for a
particular population is normally
distributed with a mean 22 and
standard deviation of 4. Find the 40th
percentile.
1
4%
Area to the left of 40th percentile in 0.40.
In the table there is a area value 0.40
corresponding to a z-score of −.25.
40th percentile = 22 − .25 x 4 = 21
40
2
3
4
5
7
8
9
?%
-1.75 -1.25 -.75 -.25 0 .25
41
6
4%
.75
1.25
1.75
42
Normal Distribution - 7