4B: Probability part B Normal Distributions

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4B Probability part BNormal Distributions
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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 (16671754)
• Extended by Laplace (17491827)

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Probability density function (curve)

• Illustrative example vocabulary scores of 947
  seventh graders
• Smooth curve drawn over histogram is a density
  function model of the actual distribution
• This is the Normal probability density function
  (pdf)

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Areas under curve (cont.)

• Last week we introduced the idea of the area
  under the curve (AUC) the same principals
  applies here
• The darker bars in the figure represent scores
  6.0,
• About 30 of the scores were less than or equal
  to 6
• Therefore, 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 curve
• As 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) 0.30
• In practice, the Normal density curve helps us
  work with Normal probabilities

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Density Curves
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Normal distributions

• Normal 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 WAIS

• Wechsler adult intelligence scores (WAIS) vary
  according to a Normal distribution with µ 100
  and s 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 s above µ)

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Male Height Example

• What proportion of men are less than 68 inches
  tall?

68 does not fall on a s marker. To determine the
AUC, we must first standardize the value.
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Standardized value z score

• To standardize a value, simply subtract µ and
  divide by s
• 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
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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Mens Height (NHANES, 1980)

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

-0.71 0 (standardized values)
You can now look up the AUC in a Standard Normal
Z table.
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Using the Standard Normal table
Pr(Z -0.71) .2389
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Summary (finding Normal probabilities)

• Draw curve w/ landmarks
• Shade area
• Standardize value(s)
• Use Z table to find appropriate AUC

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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)

.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

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Unstandardizing a value
How tall must a man be to place in the lower 10
for men aged 18 to 24?
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Table AStandard Normal Table

• Use Table A
• Look up the closest proportion in the table
• Find corresponding standardized score
• Solve for X (un-standardize score)

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Table AStandard Normal Proportion
.08
?1.2
.1003
Pr(Z lt -1.28) .1003
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Mens 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)
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Observed Value for a Standardized Score

• Unstandardize z-score to find associated x

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Observed Value for a Standardized Score

• x µ zs 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