1. HS 67 - Intro Health Stat The Normal Distributions
Saturday, April 15, 2017
Saturday, April 15, 2017
Chapter 3
The Normal Distributions
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2. HS 67 - Intro Health Stat
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Saturday, April 15, 2017
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
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3. HS 67 - Intro Health Stat
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Saturday, April 15, 2017
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
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4. Area Under the Curve (AUC)
HS 67 - Intro Health Stat
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Saturday, April 15, 2017
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
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5. Density Curves 4/15/2017 Chapter 3 HS 67 - Intro Health Stat
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Density Curves
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6. Mean and Median of Density Curve
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7. HS 67 - Intro Health Stat
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Normal Density Curves
Normal density curves are a family of bell-shaped curves
The mean of the density is denoted μ (“mu”)
The standard deviation is denoted σ (“sigma”)
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8. The Normal Distribution
HS 67 - Intro Health Stat
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Saturday, April 15, 2017
The Normal Distribution
Mean μ defines the center of the curve
Standard deviation σ defines the spread
Notation is N(µ,).
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9. Practice Drawing Curves!
The Normal curve is symmetrical around μ
It has infections (blue arrows) at μ ± σ
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10. The 68-95-99.7 Rule 68% of AUC within μ ± 1σ 95% fall within μ ± 2σ
HS 67 - Intro Health Stat
Saturday, April 15, 2017
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The 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
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11. HS 67 - Intro Health Stat
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Saturday, April 15, 2017
Application of 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)
rule
68% in µ   =  2.8 = 67.2 to 72.8
95% in µ  2 =  2(2.8) = 64.4 to 75.6
99.7% in µ  3 =  3(2.8) = 61.6 to 78.4
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12. ? Application: 68-95-99.7 Rule 70 72.8 (height) 84% 68% 16% 16% 68%
HS 67 - Intro Health Stat
Saturday, April 15, 2017
Saturday, April 15, 2017
Application: Rule
What proportion of men are less than inches tall?
μ + σ = = 72.8 (i.e., 72.8 is one σ above μ) ?
(height)
68%
(by Rule) -1 +1
16%
16%
68%
(total AUC = 100%)
84%
Therefore, 84% of men are less than 72.8” tall.
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13. Finding Normal proportions
HS 67 - Intro Health Stat
Saturday, April 15, 2017
Saturday, April 15, 2017
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) ?
(height values)
To answer this question, first determine the z-score for a value of 68 from X~N(70,2.8)
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14. Thus, 68 is 0.71 standard deviations below μ.
HS 67 - Intro Health Stat
Saturday, April 15, 2017
Saturday, April 15, 2017
Z score
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:
Thus, 68 is 0.71 standard deviations below μ.
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15. Example: z score and associate value
HS 67 - Intro Health Stat
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Saturday, April 15, 2017
Example: z score and associate value
(height values) ?
(z values)
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16. Standard Normal Table See pp. 79 – 83 in your text!
HS 67 - Intro Health Stat
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Saturday, April 15, 2017
Standard Normal Table
Use Table A to determine the cumulative proportion associated with the z score
See pp. 79 – 83 in your text!
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17. Normal Cumulative Proportions (Table A)
HS 67 - Intro Health Stat
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Saturday, April 15, 2017
Normal Cumulative Proportions (Table A) z
.00
.02
0.8
.2119
.2090
.2061
.2420
.2358
0.6
.2743
.2709
.2676
.01
0.7
.2389
Thus, a z score of −0.71 has a cumulative proportion of .2389
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18. The proportion of mean less than 68” tall (z-score = −0.71 is .2389:
HS 67 - Intro Health Stat
Saturday, April 15, 2017
Saturday, April 15, 2017
Normal proportions
The proportion of mean less than 68” tall (z-score = −0.71 is .2389:
(z scores)
(height values)
.2389
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19. Area to the right (“greater than”)
HS 67 - Intro Health Stat
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Saturday, April 15, 2017
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?
(z values)
(height values)
1.2389 = .7611
.2389
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20. 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
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21. HS 67 - Intro Health Stat
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Saturday, April 15, 2017
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
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22. State the Problem & Sketch Curve
HS 67 - Intro Health Stat
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Saturday, April 15, 2017
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
? (height)
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23. Table A Find z score for cumulative proportion ≈.10
HS 67 - Intro Health Stat
Saturday, April 15, 2017
Saturday, April 15, 2017
Table A Find z score for cumulative proportion ≈.10 z
.07
.09
1.3
.0853
.0838
.0823
.1020
.0985
1.1
.1210
.1190
.1170
.08
1.2
.1003
zcum_proportion = z.1003 = −1.28
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24. Visual Relationship Between Cumulative proportion and z-score
HS 67 - Intro Health Stat
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Saturday, April 15, 2017
Visual Relationship Between Cumulative proportion and z-score
.10
? (height values)
(Z value)
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25. Unstandardize x = μ + z∙σ = 70 + (-1.28 )(2.8) = 70 + (3.58) = 66.42
HS 67 - Intro Health Stat
Saturday, April 15, 2017
Saturday, April 15, 2017
Unstandardize
x = μ + z∙σ = 70 + (-1.28 )(2.8) = 70 + (3.58) = 66.42
Conclude: A man would have to be less than inches tall to place him in the lowest 10% of heights
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