#### Transcript Slide 1

```Chapter 3
The Normal
Distributions
7/21/2015
Chapter 3
1
Density Curves
• Here is a histogram
of vocabulary scores
of n = 947 seventh
graders
• The smooth curve
drawn over the
histogram is a
mathematical model
which represents
the density
function of the
distribution
7/21/2015
Chapter 3
2
Density Curves
• The shaded bars on
this histogram
corresponds to the
scores that are less
than 6.0
• This area represents
is 30.3% of the total
area of the histogram
and is equal to the
percentage in that
range
7/21/2015
Chapter 3
3
Area Under the Curve
(AUC)
• This figure shades area
under the curve (AUC)
corresponding to
scores less than 6
• This also corresponds
to the proportion in that
range: AUC =
proportion in that
range
7/21/2015
Chapter 3
4
Density Curves
7/21/2015
Chapter 3
5
Mean and Median of Density Curve
7/21/2015
Chapter 3
6
Normal Density Curves
• Normal density curves
are a family of bellshaped curves
• The mean of the density
is denoted μ (“mu”)
• The standard deviation
is denoted σ (“sigma”)
7/21/2015
Chapter 3
7
The Normal Distribution
• Mean μ defines the center of the curve
• Standard deviation σ defines the spread
• Notation is N(µ,).
7/21/2015
Chapter 3
8
Practice Drawing Curves!
• The Normal curve is symmetrical around μ
• It has infections (blue arrows) at μ ± σ
7/21/2015
Chapter 3
9
The 68-95-99.7 Rule
•
•
•
•
68% of AUC within μ ± 1σ
95% fall within μ ± 2σ
99.7% within μ ± 3σ
Memorize!
This rule applies only
to Normal curves
7/21/2015
Chapter 3
10
Application of 68-95-99.7 rule
• Male height has a Normal distribution with μ = 70.0
inches and σ = 2.8 inches
• Notation: Let X ≡ male height; X~ N(μ = 70, σ = 2.8)
68-95-99.7 rule
• 68% in µ   = 70.0  2.8 = 67.2 to 72.8
• 95% in µ  2 = 70.0  2(2.8) = 64.4 to 75.6
• 99.7% in µ  3 = 70.0  3(2.8) = 61.6 to 78.4
7/21/2015
Chapter 3
11
Application: 68-95-99.7 Rule
What proportion of men are less than 72.8 inches tall?
μ + σ = 70 + 2.8 = 72.8 (i.e., 72.8 is one σ above μ)
68%
?68%
16%
-1
(by 68-95-99.7 Rule)
16%
(total AUC = 100%)
+1
70
72.8
(height)
84%
Therefore, 84% of men are less than 72.8” tall.
7/21/2015
Chapter 3
12
Finding Normal proportions
What proportion of men are less than 68” tall? This
is equal to the AUC to the left of 68 on X~N(70,2.8)
?
68 70
(height values)
To answer this question, first determine the z-score
for a value of 68 from X~N(70,2.8)
7/21/2015
Chapter 3
13
Z score
z
x

• The z-score tells you how many standard
deviation the value falls below (negative z score)
or above (positive z score) mean μ
• The z-score of 68 when X~N(70,2.8) is:
z
x

68  70

 0.71
2.8
Thus, 68 is 0.71 standard deviations below μ.
7/21/2015
Chapter 3
14
Example: z score and associate
value
?
68 70 (height values)
-0.71
7/21/2015
0
Chapter 3
(z values)
15
Standard Normal Table
Use Table A to determine the cumulative proportion associated with the z score
See pp. 79 – 83 in your text!
7/21/2015
Chapter 3
16
Normal Cumulative
Proportions (Table A)
z
.00
.01
.02
0.8
.2119
.2090
.2061
.2420
.2389
.2358
.2743
.2709
.2676
0.7
0.6
Thus, a z score of −0.71 has a cumulative
7/21/2015
Chapter 3of .2389
proportion
17
Normal proportions
The proportion of mean less than 68” tall (z-score =
−0.71 is .2389:
.2389
68 70
-0.71
7/21/2015
0
Chapter 3
(height values)
(z scores)
18
Area to the right
(“greater than”)
Since the total AUC = 1:
AUC to the right = 1 – AUC to left
Example: What % of men are greater than 68” tall?
.2389
1.2389 =
.7611
68 70
-0.71
7/21/2015
0
Chapter 3
(height values)
(z values)
19
Normal proportions
“The key to calculating Normal proportions is to
match the area you want with the areas that
represent cumulative proportions. If you make a
sketch of the area you want, you will almost
never go wrong. Find areas for cumulative
proportions … from [Table A] (p. 79)”
Follow the “method in the picture” (see pp. 79 –
80) to determine areas in right tails and between
two points
7/21/2015
Chapter 3
20
Finding Normal values
We just covered finding proportions for Normal
variables. At other times, we may know the
proportion and need to find the Normal value.
Method for finding a Normal value:
1. State the problem
2. Sketch the curve
3. Use Table A to look up the proportion & z-score
4. Unstandardize the z-score with this formula
x    z
7/21/2015
Chapter 3
21
State the Problem & Sketch
Curve
Problem: How tall must a man be to be taller than
10% of men in the population? (This is the same as
asking how tall he has to be to be shorter than 90%
of men.)
Recall X~N(70, 2.8)
.10
? 70
7/21/2015
(height)
Chapter 3
22
Table A
Find z score for cumulative proportion ≈.10
z
.07
.08
.09
1.3
.0853
.0838
.0823
.1020
.1003
.0985
.1210
.1190
.1170
1.2
1.1
zcum_proportion = z.1003 = −1.28
7/21/2015
Chapter 3
23
Visual Relationship Between
Cumulative proportion and z-score
.10
? 70
-1.28
7/21/2015
0
Chapter 3
(height values)
(Z value)
24
Unstandardize
• x = μ + z∙σ
= 70 + (1.28 )(2.8)
= 70 + (3.58)
= 66.42
• Conclude: A man would have to be less
than 66.42 inches tall to place him in the
lowest 10% of heights
7/21/2015
Chapter 3
25
```