1. HS 67 - Intro Health Stat The Normal Distributions
Thursday, May 16, 2019
Thursday, May 16, 2019
Chapter 11
The Normal Distributions
Note: Normal distributions are the most popular probability models in statistics. Other popular models include binomial distributions, t distributions, etc.
5/16/2019
BPS Chapter 3
Normal DistributionsBPS Chapter 3 1 1

2. Recall: Rules About Density Curves
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Recall: Rules About Density Curves
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 2 2

3. Recall the relationship between AUCs and probabilities
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Recall the relationship between AUCs and probabilities
Area Under Curve = probability for that range
30% of students
had scores ≤ 6
30% of area under the curve (AUC) is shaded
30%
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 3 3

4. The Normal pdf is: A family of related bell-shaped density
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
The Normal pdf is:
A family of related bell-shaped density
Each family member has a different μ (mean) and σ (standard deviation)
μ  location
σ  spread
There are an infinite number of Normal density curves (curve changes every time µ and/or σchanges)
Notation: X~N(µ,σ) is read “X is distributed as …”
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 5 5

5. Draw Normal curves accurately!
Symmetrical around μ
Infection points at μ ± σ
Horizontal asymptotes
Drawing a half circle is not acceptable
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6. The 68-95-99.7 rule applies only to Normal densities
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Rule
μ ± 1σ contains 68% of AUC
μ ± 2σ contains 95% of AUC
μ ± 3σ contains 99.7% of AUC
The rule applies only to Normal densities
5/16/2019
BPS Chapter 3
Normal DistributionsBPS Chapter 3 7 7

7. HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Rule - Example
Assume male height (X) has Normal distribution with μ = 70.0 inches and σ = 2.8 inches
Notation: X~ N(70, 2.8)
µ   = 70.0  2.8 = 67.2 to  68% of men between 67.2 and 72.8 inches tall
µ  2 =  2(2.8) = 64.4 to 75.6 (inches)  95% in this range
µ  3 =  3(2.8) = 61.6 to 78.4 (inches)  99.7% in this range
BPS Chapter 3
Normal DistributionsBPS Chapter 3 8 8

8. 68-95-99.7 Rule example (cont.) Height in inches 70 72.8 75.6 78.4
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Rule example (cont.) 70
Height in inches
72.8
75.6
78.4
67.2
64.4
61.6
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 9 9

9. What proportion of men are less than 72.8" tall?
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Example 2 (Male Height)
What proportion of men are less than 72.8" tall?
X~ N(70, 2.8)
By rule
68%
Total AUC =100%
 32% split evenly in tails
16%
16%
16%
16%
16% + 68% = 84%
84% of AUC to the left of 72.8
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 10 10

10. Finding Normal Probabilities: Method
Draw an reasonably accurate Normal curve with landmarks at µ and µ ±σ
Place values on the curve & shade relevant area
Standardize values (i.e., z scores)
Use Table A to determine the AUC
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11. Finding Normal Probabilities: Draw & Shade
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Finding Normal Probabilities: Draw & Shade
What proportion of men are less than 68” tall? ? 68 70
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 12 12

12. Finding Normal Probabilities: Standardize
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Finding Normal Probabilities: Standardize
In X~N(70, 2.8), the value 68 has:
This indicates that 68 is 0.71 standard deviations below the mean
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 13 13

13. Finding Normal Probabilities: Standardize
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Finding Normal Probabilities: Standardize
Proportion of men less than 68” tall = Pr(Z < -0.71):
(height)
(z score)
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 14 14

14. Standard Normal “Z” Table
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Standard Normal “Z” Table
Notation: zp where p ≡ cumulative probability
= AUC to left p
Z is the Normal random variable with µ = 0 and σ= 1
Z~N(0,1)
There is only one Standard Normal Z variable
Z table in text (pp ) and in “Formulas” handout
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 15 15

15. Finding Normal Probabilities: Table A
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Finding Normal Probabilities: Table A z
.00
.02
0.8
.2119
.2090
.2061
.2420
.2358
0.6
.2743
.2709
.2676
.01
0.7
.2389
z score of −0.71 → AUC to left is .2389
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 16 16

16. Conclude: .2389 (23.89%) of men are less than 68 inches in height
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Pr(Z < −0.7) as an Image
Conclude: (23.89%) of men are less than 68 inches in height
From Table A
.2389
(height)
(z score)
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 17 17

17. Pr(Z > z)  AUC to the right
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Pr(Z > z)  AUC to the right
Ask: What proportion of men greater than 68” tall?
“Greater than” =
“area to right of”=
1 – (AUC to left)
1 – .2389
= .7611
.2389
(height)
(z score)
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 18 18

18. AUC Between Two Points Total AUC = 1. Therefore:
Pr(a < Z < b) = Pr(Z < b) – Pr(Z < a)
.6915
.1003
-1.28
0.5
0.5
-1.28
Pr(−1.28 < Z < 0.5) = Pr(Z < 0.5) – Pr(Z < −1.28)
= – .1003
= .5912

19. Finding z percentiles and associated values
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Finding z percentiles and associated values
At times, we must find the z value or X value that is associated with a given probability. To do this:
1. State the problem
2. Sketch the curve
3. Look up the related z-score is Z table
4. If you need to know X, “unstandardize” z with this formula:
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 20 20

20. State & Sketch ? 0 (z equivalent) z.10 70 (height) 70 (height)
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
State & Sketch
How tall must a man be to be taller than 10% of men (i.e., what is the 10th percentile of heights)?
Given: X~N(70, 2.8)
.10
.10 ?
70 (height)
70 (height)
0 (z equivalent)
z.10
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 21 21

21. Find z score in Table A  z.10 = -1.28 z .07 .09 1.3 .0853 .0838
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Find z score in Table A z
.07
.09
1.3
.0853
.0838
.0823
.1020
.0985
1.1
.1210
.1190
.1170
.08
1.2
.1003
 z.10 = -1.28
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BPS Chapter 3
Normal DistributionsBPS Chapter 3 22 22

22. Visual Relationship Notation: z.1003 = −1.28 X~N(70, 2.8) .10
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Visual Relationship
Notation: z.1003 = −1.28
AUC to left
z-score
X~N(70, 2.8)
.10
(height) 70
Height (inches) ?
Height (z score)
−1.28
5/16/2019
Chapter 3
BPS Chapter 3
Normal DistributionsBPS Chapter 3 23 23

23. Unstandardize & Conclude
HS 67 - Intro Health Stat
HS 67 - Intro Health Stat
Thursday, May 16, 2019
Thursday, May 16, 2019
Unstandardize & Conclude
x = μ + z∙σ = 70 + (-1.28 )(2.8) = 70 + (3.58) = 66.42
Conclude: A man who is inches in height is taller than 10% of men
We can also conclude that he is shorter than (1 – 10%) or 90% of men
5/16/2019
Chapter 3
BPS Chapter 3
Normal DistributionsBPS Chapter 3 24 24