HS 167 4B: Probability Part B 1

4B: Probability part BNormal Distributions

HS 167 4B: Probability Part B 2

The Normal distributions

Last lecture covered the most popular type of discrete random variable: binomial variables This lecture covers the most popular continuous random variable: Normal variables History of the Normal function

Recognized by de Moivre (1667–1754)

Extended by Laplace (1749–1827)

How’s my hair?

Looks good.

HS 167 4B: Probability Part B 3

Probability density function (curve)Illustrative example: vocabulary scores of 947 seventh gradersSmooth curve drawn over histogram is a density function model of the actual distributionThis is the Normal probability density function (pdf)

HS 167 4B: Probability Part B 4

Areas under curve (cont.)

Last week we introduced the idea of the area under the curve (AUC); the same principals applies hereThe darker bars in the figure represent scores ≤ 6.0, About 30% of the scores were less than or equal to 6Therefore, selecting a score at random will have probability Pr(X ≤ 6) ≈ 0.30

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Areas under curve (cont.)Now translate this to a

Normal curveAs before, the area under

the curve (AUC) = probability

The scale of the Y-axis is adjusted so the total AUC = 1

The AUC to the left of 6.0 in the figure to the right (shaded) = 0.30

Therefore, Pr(X ≤ 6) ≈ 0.30In practice, the Normal

density curve helps us work with Normal probabilities

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Density Curves

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Normal distributionsNormal distributions = a family of distributions with

common characteristics Normal distributions have two parameters

Mean µ locates center of the curve Standard deviation quantifies spread (at points of

inflection)

Arrows indicate points of inflection

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68-95-99.7 rule for Normal RVs

68% of AUC falls within 1 standard deviation of the mean (µ )

95% fall within 2 (µ 2)

99.7% fall within 3 (µ 3)

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Illustrative example: WAISWechsler adult intelligence scores (WAIS)

vary according to a Normal distribution with μ = 100 and σ = 15

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Illustrative example: male height

Adult male height is approximately

Normal with µ = 70.0 inches and

= 2.8 inches (NHANES, 1980)

Shorthand: X ~ N(70, 2.8)

Therefore: 68% of heights = µ = 70.0 2.8 = 67.2 to 72.8 95% of heights = µ 2 = 70.0 2(2.8) = 64.4 to

75.6 99.7% of heights = µ 3 = 70.0 3(2.8) = 61.6 to

78.4

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Illustrative example: male height

What proportion of men are less than 72.8 inches tall? (Note: 72.8 is one σ above μ)

?

70 72.8 (height)

+1

84%

68% (by 68-95-99.7 Rule)

16%

-1

16%

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Male Height Example

What proportion of men are less than 68 inches tall?

?

68 70 (height)

68 does not fall on a ±σ marker.To determine the AUC, we must first standardize

the value.

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Standardized value = z score

x

z

To standardize a value, simply subtract μ and divide by σ

This is now a z-score

The z-score tells you the number of standard deviations the value falls from μ

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Example: Standardize a male height of 68”

71.08.2

7068

x

z

Recall X ~ N(70,2.8)

Therefore, the value 68 is 0.71 standard deviations below the mean of the distribution

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Men’s Height (NHANES, 1980)

-0.71 0 (standardized values)68 70 (height values)

?

What proportion of men are less than 68 inches tall? = What proportion of a Standard z curve is less than –0.71?

You can now look up the AUC in a Standard Normal “Z” table.

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Using the Standard Normal table

z .00 .01 .02

0.8 .2119 .2090 .2061

0.7 .2420 .2389 .2358

0.6 .2743 .2709 .2676

Pr(Z ≤ −0.71) = .2389

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Summary (finding Normal probabilities)

Draw curve w/ landmarksShade areaStandardize value(s)Use Z table to find appropriate AUC

-0.71 0 (standardized values)68 70 (height values)

.2389

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Right tail

What proportion of men are greater than 68” tall?Greater than look at right “tail”Area in right tail = 1 – (area in left tail)

-0.71 0 (standardized values)68 70 (height values)

.2389 1- .2389 = .7611

Therefore, 76.11% of men are greater than 68 inches tall.

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Z percentiles • zp the z score with

cumulative probability p

• What is the 50th percentile on Z? ANS: z.5 = 0

• What is the 2.5th percentile on Z? ANS: z.025 = 2

• What is the 97.5th percentile on Z? ANS: z.975 = 2

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Finding Z percentile in the table

Look up the closest entry in the table

Find corresponding z score

e.g., What is the 1st percentile on Z?

z.01 = -2.33

closest cumulative proportion is .0099

z .02 .03 .04

2.3 .0102 .0099 .0096

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Unstandardizing a value

.10 ? 70 (height values)

How tall must a man be to place in the lower 10% for men aged 18 to 24?

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Use Table A

Look up the closest proportion in the table

Find corresponding standardized score

Solve for X (“un-standardize score”)

Table A:Standard Normal Table

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Table A:Standard Normal Proportion

z .07 .09

1.3 .0853 .0838 .0823

.1020 .0985

1.1 .1210 .1190 .1170

1.2

.08

.1003

Pr(Z < -1.28) = .1003

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Men’s Height Example (NHANES, 1980)

How tall must a man be to place in the lower 10% for men aged 18 to 24?

-1.28 0 (standardized values)

.10 ? 70 (height values)

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Observed Value for a Standardized Score

“Unstandardize” z-score to find associated x :

zx

zx

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Observed Value for a Standardized Score

x = μ + zσ = 70 + (1.28 )(2.8) = 70 + (3.58) = 66.42

A man would have to be approximately 66.42 inches tall or less to place in the lower 10% of the population